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1. 1.Computation · 7 min

   Evaluate each integral by parts. In every case write down your $u$ and $dv$, say which LIATE family decided the choice, and check the two indefinite integrals by differentiating.

   1. (a)$\int x\cos 3xdx$
   2. (b)${\int }_1^e{x}^2\ln xdx$
   3. (c)$\int x{\sec }^{2}xdx$
   Show the solution to question 1

   (a) Algebraic beats trigonometric. Take $u=x$, $dv=\cos 3xdx$, so $du=dx$ and $v=\frac{1}{3}\sin 3x$. Then

   $$\int x\cos 3xdx=\frac{x\sin 3x}{3}-\int \frac{\sin 3x}{3}dx=\frac{x\sin 3x}{3}+\frac{\cos 3x}{9}+C.$$

   Check: the derivative of $\frac{x\sin 3x}{3}$ is $\frac{\sin 3x}{3}+x\cos 3x$, and the derivative of $\frac{\cos 3x}{9}$ is $-\frac{\sin 3x}{3}$. The sum is $x\cos 3x$.

   (b) Logarithmic beats algebraic. Take $u=\ln x$, $dv=x^{2}dx$, so $du=\frac{dx}{x}$ and $v=\frac{x^3}{3}$. The other split, $dv=\ln xdx$, would make $v$ harder than the integrand, which is why the logarithm is always $u$. The definite form of the formula gives

   $${\int }_1^e{x}^2\ln xdx=[\frac{x^3\ln x}{3}{]}_1^e-{\int }_1^e\frac{x^3}{3}\cdot \frac{dx}{x}=\frac{e^3}{3}-0-[\frac{x^3}{9}{]}_1^e=\frac{e^3}{3}-\frac{e^3-1}{9}=\frac{2e^3+1}{9}.$$

   Sanity check: on $[1,e]$ the integrand lies between $0$ and $e^2$, so the answer must lie between $0$ and $e^2(e-1)\approx 12.7$; and $\frac{2e^3+1}{9}\approx 4.57$ does.

   (c) The only factor with a clean antiderivative is ${\sec }^{2}x$. Take $u=x$ (algebraic) and $dv={\sec }^{2}xdx$, so $du=dx$ and $v=\tan x$:

   $$\int x{\sec }^{2}xdx=x\tan x-\int \tan xdx=x\tan x-\ln |\sec x|+C=x\tan x+\ln |\cos x|+C.$$

   Check: the derivative of $x\tan x$ is $\tan x+x{\sec }^{2}x$, and the derivative of $\ln |\cos x|$ is $\frac{-\sin x}{\cos x}=-\tan x$. The sum is $x{\sec }^{2}x$.

2. 2.Critique · 7 min

   A student wants $I=\int e^x\sin xdx$ and submits the following.

   Step 1. With $u=\sin x$, $dv=e^{x}dx$, so $du=\cos xdx$, $v=e^x$: $I=e^x\sin x-\int e^x\cos xdx$.

   Step 2. For the new integral take $u=e^x$, $dv=\cos xdx$, so $du=e^{x}dx$, $v=\sin x$: $\int e^x\cos xdx=e^x\sin x-\int e^x\sin xdx=e^x\sin x-I$.

   Step 3. Substituting Step 2 into Step 1 gives $I=e^x\sin x-e^x\sin x+I$, that is, $I=I$. The equation carries no information, so integration by parts cannot evaluate this integral, and $e^x\sin x$ has no elementary antiderivative.

   Steps 1 and 2 are each, on their own, correct applications of the parts formula.

   1. (a)Explain exactly what went wrong, and why the two conclusions drawn in Step 3 do not follow.
   2. (b)Repair the calculation, evaluate $I$, and check your answer by differentiating.
   Show the solution to question 2

   (a) The roles were switched between rounds. In Step 1 the trigonometric factor was differentiated ($u=\sin x$) and the exponential was integrated. In Step 2 the roles are reversed: the exponential is differentiated and the trigonometric factor is integrated. Reversing the roles exactly undoes the first round. Step 2's equation is Step 1's equation rearranged, so substituting one into the other can only return the identity $I=I$. Nothing has been computed, but nothing has gone wrong either; the calculation is merely circular.

   The first conclusion, "parts cannot evaluate this integral", is false: one particular route went in a circle, and a different choice of roles (below) succeeds. The second conclusion is a non sequitur of a more serious kind. Since $e^x\sin x$ is continuous, it has an antiderivative by the Fundamental Theorem, and whether that antiderivative is elementary is a question about the function, not about one student's calculation. A failed method proves nothing about existence.

   (b) Keep the same roles in the second round. Differentiate the trigonometric factor again: $u=\cos x$, $dv=e^{x}dx$, so $du=-\sin xdx$, $v=e^x$, and

   $$\int e^x\cos xdx=e^x\cos x+\int e^x\sin xdx=e^x\cos x+I.$$

   Substituting this into Step 1,

   $$I=e^x\sin x-e^x\cos x-I,\text{so}2I=e^x(\sin x-\cos x),$$

   and therefore

   $$\int e^x\sin xdx=\frac{e^x(\sin x-\cos x)}{2}+C.$$

   The reason this round closes and the student's did not: differentiating the trigonometric factor twice sends $\sin x\to \cos x\to -\sin x$, so the original integral returns with a minus sign, and $I=(\dots )-I$ can be solved for $I$. The constant is added once, at the end, because the equation $I=\dots -I$ is a statement about families of antiderivatives.

   Check: the derivative of $\frac{1}{2}{e}^x(\sin x-\cos x)$ is $\frac{1}{2}{e}^x(\sin x-\cos x)+\frac{1}{2}{e}^x(\cos x+\sin x)=e^x\sin x$.

3. 3.Computation · 8 min

   Let $T_n={\int }_0^{\pi /4}{\tan }^{n}xdx$ for integers $n\ge 0$. You may quote the reduction formula $\int {\tan }^{n}xdx=\frac{{\tan }^{n-1}x}{n-1}-\int {\tan }^{n-2}xdx$ for $n\ge 2$.

   1. (a)Show that $T_n+T_{n-2}=\frac{1}{n-1}$ for every $n\ge 2$. Be explicit about the boundary term.
   2. (b)Compute the base cases $T_1$ and $T_2$ directly, without the reduction formula.
   3. (c)Compute $T_5$ and $T_6$ exactly. Then explain, without computing anything further, why $T_n$ must decrease as $n$ increases, and confirm that your two values respect this.
   Show the solution to question 3

   (a) Apply the formula between the limits. For $n\ge 2$,

   $$T_n=[\frac{{\tan }^{n-1}x}{n-1}{]}_0^{\pi /4}-{\int }_0^{\pi /4}{\tan }^{n-2}xdx=\frac{1^{n-1}-0^{n-1}}{n-1}-T_{n-2}=\frac{1}{n-1}-T_{n-2},$$

   since $\tan \frac{\pi }{4}=1$ and $\tan 0=0$. Rearranging, $T_n+T_{n-2}=\frac{1}{n-1}$. The boundary term has to be evaluated at this step, and it is not zero: it is where all the numbers come from.

   (b) Base cases. The formula needs $n\ge 2$, so $T_1$ must be done by hand:

   $$T_1={\int }_0^{\pi /4}\tan xdx=[\ln |\sec x|{]}_0^{\pi /4}=\ln \sqrt{2}-\ln 1=\frac{1}{2}\ln 2.$$

   For $T_2$ use ${\tan }^{2}x={\sec }^{2}x-1$:

   $$T_2={\int }_0^{\pi /4}({\sec }^{2}x-1)dx=[\tan x-x{]}_0^{\pi /4}=1-\frac{\pi }{4}.$$

   (As a check, the formula with $n=2$ says $T_2=1-T_0$, and $T_0={\int }_0^{\pi /4}1dx=\frac{\pi }{4}$, consistent.)

   (c) Climb the staircase. Odd rungs: $T_3=\frac{1}{2}-T_1=\frac{1}{2}-\frac{1}{2}\ln 2$, then

   $$T_5=\frac{1}{4}-T_3=\frac{1}{4}-\frac{1}{2}+\frac{1}{2}\ln 2=\frac{1}{2}\ln 2-\frac{1}{4}.$$

   Even rungs: $T_4=\frac{1}{3}-T_2=\frac{1}{3}-1+\frac{\pi }{4}=\frac{\pi }{4}-\frac{2}{3}$, then

   $$T_6=\frac{1}{5}-T_4=\frac{1}{5}-\frac{\pi }{4}+\frac{2}{3}=\frac{13}{15}-\frac{\pi }{4}.$$

   Why $T_n$ decreases: on $[0,\frac{\pi }{4}]$ we have $0\le \tan x\le 1$, so multiplying by one more factor of $\tan x$ can only shrink the integrand, ${\tan }^{n}x\le {\tan }^{n-1}x$, and integrating preserves the inequality. (Part (a) gives a bound too: since $T_{n-2}>0$, $T_n<\frac{1}{n-1}$.)

   Confirmation: $T_5\approx 0.3466-0.25=0.0966$ and $T_6\approx 0.8667-0.7854=0.0813$, so $T_6<T_5<\frac{1}{4}$, as required.

4. 4.Computation · 8 min

   Evaluate. Before computing each one, state which parity rule or identity applies, and say why the other route on offer does not.

   1. (a)$\int {\sin }^{3}x{\cos }^{4}xdx$
   2. (b)${\int }_0^{\pi /4}{\tan }^{2}x{\sec }^{4}xdx$
   3. (c)$\int \cos 3x\cos 5xdx$
   Show the solution to question 4

   (a) Odd power of sine: save one sine. The cosine power is even, so saving a cosine leaves ${\cos }^{3}x$, an odd leftover that ${\cos }^{2}x=1-{\sin }^{2}x$ cannot absorb; the sine route is the one that works. Write

   $${\sin }^{3}x{\cos }^{4}x=(1-{\cos }^{2}x){\cos }^{4}x\cdot \sin x,$$

   and substitute $u=\cos x$, $du=-\sin xdx$:

   $$\int (1-u^2)u^4(-du)=-\int (u^4-u^6)du=-\frac{u^5}{5}+\frac{u^7}{7}+C=-\frac{{\cos }^{5}x}{5}+\frac{{\cos }^{7}x}{7}+C.$$

   Check: differentiating gives ${\cos }^{4}x\sin x-{\cos }^{6}x\sin x=\sin x{\cos }^{4}x(1-{\cos }^{2}x)={\sin }^{3}x{\cos }^{4}x$.

   (b) Even power of secant: save ${\sec }^{2}x$. The tangent power is even, so there is no $\sec x\tan x$ to spare; the secant route is the one that works. Since ${\sec }^{4}x=(1+{\tan }^{2}x){\sec }^{2}x$, put $u=\tan x$, $du={\sec }^{2}xdx$, with limits $u=0$ to $u=1$:

   $${\int }_0^{\pi /4}{\tan }^{2}x(1+{\tan }^{2}x){\sec }^{2}xdx={\int }_0^1(u^2+u^4)du=\frac{1}{3}+\frac{1}{5}=\frac{8}{15}.$$

   Sanity check: the integrand is $0$ at $x=0$ and $1\cdot 4=4$ at $x=\frac{\pi }{4}$, over an interval of length $0.785$, so a value near $\frac{1}{2}$ is plausible.

   (c) Different frequencies: product-to-sum. No power rule applies because the two factors have different arguments. With $A=3x$, $B=5x$,

   $$\cos 3x\cos 5x=\frac{1}{2}[\cos (3x-5x)+\cos (3x+5x)]=\frac{1}{2}[\cos 2x+\cos 8x],$$

   using $\cos (-2x)=\cos 2x$. Integrating term by term,

   $$\int \cos 3x\cos 5xdx=\frac{\sin 2x}{4}+\frac{\sin 8x}{16}+C.$$

   Check: the derivative is $\frac{\cos 2x}{2}+\frac{\cos 8x}{2}$, which the identity says is $\cos 3x\cos 5x$.

5. 5.Computation · 10 min

   Evaluate. For each integral name the substitution and the range of $\theta$ you are using, and say why that range lets you drop the absolute value on the root. In the definite integral convert the limits to $\theta$ and do not return to $x$.

   1. (a)$\int \frac{dx}{x^2\sqrt{x^2-9}}$ for $x>3$
   2. (b)${\int }_0^1\frac{x^2}{\sqrt{4-x^2}}dx$
   3. (c)$\int \frac{x+1}{\sqrt{x^2+4x+13}}dx$
   Show the solution to question 5

   (a) The $\sqrt{x^2-a^2}$ pattern with $a=3$. Put $x=3\sec \theta$ with $0\le \theta <\frac{\pi }{2}$, which is the range matching $x>3$; on it $\tan \theta \ge 0$, so $\sqrt{9{\tan }^2\theta }=3\tan \theta$ with no absolute value. Then $dx=3\sec \theta \tan \theta d\theta$ and

   $$\int \frac{3\sec \theta \tan \theta d\theta }{9{\sec }^2\theta \cdot 3\tan \theta }=\frac{1}{9}\int \frac{d\theta }{\sec \theta }=\frac{1}{9}\int \cos \theta d\theta =\frac{1}{9}\sin \theta +C.$$

   Reference triangle for $\sec \theta =\frac{x}{3}$: hypotenuse $x$, adjacent side $3$, opposite side $\sqrt{x^2-9}$, so $\sin \theta =\frac{\sqrt{x^2-9}}{x}$ and

   $$\int \frac{dx}{x^2\sqrt{x^2-9}}=\frac{\sqrt{x^2-9}}{9x}+C.$$

   Check: by the quotient rule the derivative of $\frac{\sqrt{x^2-9}}{x}$ is $\frac{x\cdot \frac{x}{\sqrt{x^2-9}}-\sqrt{x^2-9}}{x^2}=\frac{x^2-(x^2-9)}{x^2\sqrt{x^2-9}}=\frac{9}{x^2\sqrt{x^2-9}}$; dividing by $9$ gives the integrand.

   (b) The $\sqrt{a^2-x^2}$ pattern with $a=2$. Put $x=2\sin \theta$, $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$, where $\cos \theta \ge 0$ so $\sqrt{4-x^2}=2\cos \theta$; also $dx=2\cos \theta d\theta$. Limits: $x=0$ gives $\theta =0$; $x=1$ gives $\sin \theta =\frac{1}{2}$, so $\theta =\frac{\pi }{6}$. Then

   $${\int }_0^{\pi /6}\frac{4{\sin }^2\theta }{2\cos \theta }\cdot 2\cos \theta d\theta =4{\int }_0^{\pi /6}{\sin }^2\theta d\theta =4[\frac{\theta }{2}-\frac{\sin 2\theta }{4}{]}_0^{\pi /6}=4(\frac{\pi }{12}-\frac{\sqrt{3}}{8})=\frac{\pi }{3}-\frac{\sqrt{3}}{2},$$

   using the half-angle identity for the even power and $\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$.

   Sanity check: near $x=0$ the integrand is about $\frac{x^2}{2}$, whose integral over $[0,1]$ is $\frac{1}{6}\approx 0.167$, and the true integrand is a little larger; the answer $\approx 1.047-0.866=0.181$ fits.

   (c) Complete the square first. $x^2+4x+13=(x+2{)}^2+9$. Put $u=x+2$, so $x+1=u-1$ and $dx=du$:

   $$\int \frac{u-1}{\sqrt{u^2+9}}du=\int \frac{u}{\sqrt{u^2+9}}du-\int \frac{du}{\sqrt{u^2+9}}.$$

   The first piece is a plain substitution $w=u^2+9$, $dw=2udu$: it equals $\sqrt{u^2+9}$. The second is the $\sqrt{a^2+x^2}$ pattern with $a=3$: put $u=3\tan \theta$, $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$, where $\sec \theta >0$ so $\sqrt{u^2+9}=3\sec \theta$, and $du=3{\sec }^2\theta d\theta$:

   $$\int \frac{3{\sec }^2\theta }{3\sec \theta }d\theta =\int \sec \theta d\theta =\ln |\sec \theta +\tan \theta |+C=\ln |\frac{\sqrt{u^2+9}}{3}+\frac{u}{3}|+C=\ln (u+\sqrt{u^2+9})+C^{\prime },$$

   with the $-\ln 3$ absorbed into the constant and the absolute value dropped because $\sqrt{u^2+9}>|u|$. Altogether

   $$\int \frac{x+1}{\sqrt{x^2+4x+13}}dx=\sqrt{x^2+4x+13}-\ln (x+2+\sqrt{x^2+4x+13})+C.$$

   Check: the derivative of the root is $\frac{2x+4}{2\sqrt{x^2+4x+13}}=\frac{x+2}{\sqrt{x^2+4x+13}}$, and the derivative of the logarithm is $\frac{1}{\sqrt{x^2+4x+13}}$ (the standard $\int \frac{dx}{\sqrt{x^2+a^2}}$ form read backwards). Their difference is $\frac{x+1}{\sqrt{x^2+4x+13}}$.

6. 6.Computation · 10 min

   Integrate each rational function. Write out the form of the decomposition before finding any coefficient, say which coefficients the cover-up method reaches, and verify your decomposition at one convenient value of $x$ before integrating.

   1. (a)$\int \frac{x^3+1}{x^2-1}dx$
   2. (b)$\int \frac{3x^2-4x+11}{(x-1)(x^2+4)}dx$
   3. (c)$\int \frac{2x^2+2x+5}{(x+2)(x-1{)}^2}dx$
   Show the solution to question 6

   (a) Improper, so divide first. The numerator has degree $3$ and the denominator degree $2$. Since $x\cdot (x^2-1)=x^3-x$, the remainder is $(x^3+1)-(x^3-x)=x+1$, and

   $$\frac{x^3+1}{x^2-1}=x+\frac{x+1}{x^2-1}=x+\frac{x+1}{(x-1)(x+1)}=x+\frac{1}{x-1}(x\ne -1).$$

   The remainder fraction cancels outright, so no unknown coefficients are needed at all. (The same thing is visible from $x^3+1=(x+1)(x^2-x+1)$.) Hence

   $$\int \frac{x^3+1}{x^2-1}dx=\frac{x^2}{2}+\ln |x-1|+C.$$

   (b) A simple linear factor and an irreducible quadratic. The fraction is proper ($2<3$) and $x^2+4$ has no real roots, so the form is

   $$\frac{3x^2-4x+11}{(x-1)(x^2+4)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+4}.$$

   Cover-up reaches $A$ only: cover $(x-1)$ and set $x=1$, $A=\frac{3-4+11}{1+4}=2$. For the rest clear denominators: $3x^2-4x+11=A(x^2+4)+(Bx+C)(x-1)=(A+B)x^2+(C-B)x+(4A-C)$. Comparing $x^2$: $2+B=3$, so $B=1$. Comparing constants: $8-C=11$, so $C=-3$. (The $x$ coefficient, $C-B=-4$, agrees.)

   Verify at $x=0$: left side $\frac{11}{(-1)(4)}=-\frac{11}{4}$; right side $\frac{2}{-1}+\frac{-3}{4}=-\frac{11}{4}$.

   Split the quadratic piece into a logarithm part and an arctangent part:

   $$\int (\frac{2}{x-1}+\frac{x}{x^2+4}-\frac{3}{x^2+4})dx=2\ln |x-1|+\frac{1}{2}\ln (x^2+4)-\frac{3}{2}\arctan \frac{x}{2}+C.$$

   (c) A repeated linear factor. Proper again. The factor $(x-1)$ is squared, so it contributes two terms:

   $$\frac{2x^2+2x+5}{(x+2)(x-1{)}^2}=\frac{A}{x+2}+\frac{B}{x-1}+\frac{C}{(x-1{)}^2}.$$

   Cover-up reaches $A$ and $C$. Cover $(x+2)$, set $x=-2$: $A=\frac{8-4+5}{9}=1$. Cover $(x-1{)}^2$, set $x=1$: $C=\frac{2+2+5}{3}=3$. For $B$, clear denominators: $2x^2+2x+5=A(x-1{)}^2+B(x+2)(x-1)+C(x+2)$; the $x^2$ coefficient on the right is $A+B$, so $1+B=2$ and $B=1$.

   Verify at $x=0$: left side $\frac{5}{2\cdot 1}=\frac{5}{2}$; right side $\frac{1}{2}+\frac{1}{-1}+\frac{3}{1}=\frac{5}{2}$.

   Integrate:

   $$\int (\frac{1}{x+2}+\frac{1}{x-1}+\frac{3}{(x-1{)}^2})dx=\ln |x+2|+\ln |x-1|-\frac{3}{x-1}+C.$$

   The last term is algebraic, not logarithmic: $\int \frac{3dx}{(x-1{)}^2}=\frac{3}{(1-2)(x-1)}=-\frac{3}{x-1}$. Leaving it out, or writing it as a logarithm, is the standard error of the repeated case.

7. 7.Computation · 9 min

   Rationalise each integral with a substitution of the form $u=\sqrt[n]{ax+b}$ or $u=x^{1/k}$. After substituting, confirm in writing that no $x$ survives anywhere, including in $dx$, before you integrate.

   1. (a)$\int x\sqrt{2x+1}dx$
   2. (b)${\int }_3^8\frac{dx}{x\sqrt{x+1}}$
   3. (c)$\int \frac{dx}{\sqrt{x}-\sqrt[4]{x}}$
   Show the solution to question 7

   (a) Rename the root. Put $u=\sqrt{2x+1}$, so $x=\frac{u^2-1}{2}$ and $dx=udu$. Every $x$ is now gone:

   $$\int \frac{u^2-1}{2}\cdot u\cdot udu=\frac{1}{2}\int (u^4-u^2)du=\frac{u^5}{10}-\frac{u^3}{6}+C.$$

   Back in $x$:

   $$\int x\sqrt{2x+1}dx=\frac{(2x+1{)}^{5/2}}{10}-\frac{(2x+1{)}^{3/2}}{6}+C.$$

   Check: by the chain rule $\frac{d}{dx}(2x+1{)}^{5/2}=5(2x+1{)}^{3/2}$ and $\frac{d}{dx}(2x+1{)}^{3/2}=3(2x+1{)}^{1/2}$, so the derivative is $\frac{1}{2}(2x+1{)}^{3/2}-\frac{1}{2}(2x+1{)}^{1/2}=\frac{1}{2}(2x+1{)}^{1/2}[(2x+1)-1]=x\sqrt{2x+1}$.

   (b) A root in the denominator, with limits. Put $u=\sqrt{x+1}$, so $x=u^2-1$ and $dx=2udu$; the limits become $u=2$ (from $x=3$) and $u=3$ (from $x=8$). Then

   $${\int }_3^8\frac{dx}{x\sqrt{x+1}}={\int }_2^3\frac{2udu}{(u^2-1)u}={\int }_2^3\frac{2du}{u^2-1}.$$

   By partial fractions (cover-up: $\frac{2}{1+1}=1$ at $u=1$ and $\frac{2}{-1-1}=-1$ at $u=-1$), $\frac{2}{u^2-1}=\frac{1}{u-1}-\frac{1}{u+1}$, so

   $${\int }_2^3(\frac{1}{u-1}-\frac{1}{u+1})du=[\ln \frac{u-1}{u+1}{]}_2^3=\ln \frac{2}{4}-\ln \frac{1}{3}=\ln \frac{3}{2}.$$

   Sanity check: the integrand falls from $\frac{1}{6}$ at $x=3$ to $\frac{1}{24}$ at $x=8$ over an interval of length $5$, so a value of a few tenths is right, and $\ln \frac{3}{2}\approx 0.405$.

   (c) Two roots of the same $x$: use the least common multiple. The indices are $2$ and $4$, so put $u=x^{1/4}$: then $x=u^4$, $dx=4u^{3}du$, $\sqrt{x}=u^2$ and $\sqrt[4]{x}=u$. The integral becomes

   $$\int \frac{4u^{3}du}{u^2-u}=4\int \frac{u^2}{u-1}du.$$

   This is improper, so divide: $u^2=(u-1)(u+1)+1$, hence $\frac{u^2}{u-1}=u+1+\frac{1}{u-1}$ and

   $$4\int (u+1+\frac{1}{u-1})du=2u^2+4u+4\ln |u-1|+C.$$

   Back in $x$:

   $$\int \frac{dx}{\sqrt{x}-\sqrt[4]{x}}=2\sqrt{x}+4\sqrt[4]{x}+4\ln |\sqrt[4]{x}-1|+C.$$

   Check: the derivatives of the three terms are $x^{-1/2}$, $x^{-3/4}$ and $\frac{x^{-3/4}}{x^{1/4}-1}$; factoring out $x^{-3/4}$ and putting the bracket over the common denominator $x^{1/4}-1$ gives $x^{-3/4}\cdot \frac{(x^{1/4}+1)(x^{1/4}-1)+1}{x^{1/4}-1}=x^{-3/4}\cdot \frac{x^{1/2}}{x^{1/4}-1}=\frac{x^{-1/4}}{x^{1/4}-1}=\frac{1}{x^{1/2}-x^{1/4}}$.

8. 8.Computation · 9 min

   Four integrals built from the same two pieces, $x$ and $x^2+4$, each of which needs a different technique. For each one, say which step of the strategy settles it (table, substitution, divide-then-decompose, parts, trigonometric substitution) and why the technique that worked on the previous part does not carry over. Then evaluate.

   1. (a)$\int \frac{x}{x^2+4}dx$
   2. (b)$\int \frac{x^2}{x^2+4}dx$
   3. (c)$\int \frac{x^3}{\sqrt{x^2+4}}dx$
   4. (d)$\int x\arctan \frac{x}{2}dx$
   Show the solution to question 8

   (a) A factor is the derivative of something inside: substitute. The numerator $x$ is, up to the constant $2$, the derivative of the denominator. With $w=x^2+4$, $dw=2xdx$,

   $$\int \frac{x}{x^2+4}dx=\frac{1}{2}\int \frac{dw}{w}=\frac{1}{2}\ln (x^2+4)+C,$$

   with no absolute value because $x^2+4>0$.

   (b) A rational function that is improper: divide. The substitution of (a) fails because $x^{2}dx$ is not a multiple of $dw$. Degrees are equal, so divide first: $\frac{x^2}{x^2+4}=1-\frac{4}{x^2+4}$, and the remainder is already a table form. Hence

   $$\int \frac{x^2}{x^2+4}dx=x-4\cdot \frac{1}{2}\arctan \frac{x}{2}+C=x-2\arctan \frac{x}{2}+C.$$

   (c) A root of a quadratic, but with a spare factor: substitute. Trigonometric substitution $x=2\tan \theta$ would work, but the strategy says to look for a substitution first, and $x^3=x^2\cdot x$ supplies the $xdx$ that $w=x^2+4$ needs. With $x^2=w-4$ and $xdx=\frac{1}{2}dw$,

   $$\int \frac{x^3}{\sqrt{x^2+4}}dx=\frac{1}{2}\int \frac{w-4}{\sqrt{w}}dw=\frac{1}{2}\int (w^{1/2}-4w^{-1/2})dw=\frac{1}{2}(\frac{2}{3}{w}^{3/2}-8w^{1/2})+C,$$

   that is,

   $$\int \frac{x^3}{\sqrt{x^2+4}}dx=\frac{(x^2+4{)}^{3/2}}{3}-4\sqrt{x^2+4}+C.$$

   Check: the derivative is $\frac{3}{2}\cdot \frac{2x}{3}(x^2+4{)}^{1/2}-\frac{4x}{\sqrt{x^2+4}}=\frac{x(x^2+4)-4x}{\sqrt{x^2+4}}=\frac{x^3}{\sqrt{x^2+4}}$. (Dividing does not apply here: the integrand is not a rational function.)

   (d) A product of an algebraic and an inverse trigonometric factor: parts. No substitution presents itself, since the derivative of $\arctan \frac{x}{2}$ is $\frac{2}{x^2+4}$ and there is no such factor. By LIATE, $u=\arctan \frac{x}{2}$, $dv=xdx$, so $du=\frac{2dx}{x^2+4}$ and $v=\frac{x^2}{2}$:

   $$\int x\arctan \frac{x}{2}dx=\frac{x^2}{2}\arctan \frac{x}{2}-\int \frac{x^2}{2}\cdot \frac{2}{x^2+4}dx=\frac{x^2}{2}\arctan \frac{x}{2}-\int \frac{x^2}{x^2+4}dx.$$

   The leftover integral is exactly part (b), so

   $$\int x\arctan \frac{x}{2}dx=\frac{x^2}{2}\arctan \frac{x}{2}-x+2\arctan \frac{x}{2}+C=\frac{x^2+4}{2}\arctan \frac{x}{2}-x+C.$$

   Check: the derivative of $\frac{x^2+4}{2}\arctan \frac{x}{2}$ is $x\arctan \frac{x}{2}+\frac{x^2+4}{2}\cdot \frac{2}{x^2+4}=x\arctan \frac{x}{2}+1$, and the $-x$ contributes $-1$.

9. 9.Sketch · 8 min

   This question is about the three curves $y=x^{-1/2}$, $y=x^{-1}$ and $y=x^{-2}$ for $x>0$. A drawing with no reasoning beside it earns nothing: each feature must be justified.

   1. (a)On one pair of axes, sketch all three curves for $0<x\le 4$. Mark the one point they share, plot at least two further points on each curve, and state with justification the order of the three curves (top to bottom) on $(0,1)$ and on $(1,\infty )$.
   2. (b)Using the sketch, explain why exactly one of the three regions under the curves on $(0,1]$ has finite area, and exactly one on $[1,\infty )$. Then compute both finite areas directly from the definition of an improper integral.
   3. (c)Explain why there is no exponent $p$ for which ${\int }_0^{\infty }{x}^{-p}dx$ converges. Then let $g(x)=x^{-1/2}$ for $0<x\le 1$ and $g(x)=x^{-2}$ for $x\ge 1$; show that ${\int }_0^{\infty }g(x)dx$ converges and find its value.
   Show the solution to question 9

   (a) The sketch. All three curves pass through $(1,1)$, since $1^{-p}=1$ for every $p$; that is the shared point. Further points: on $y=x^{-1/2}$, $(\frac{1}{4},2)$ and $(4,\frac{1}{2})$; on $y=x^{-1}$, $(\frac{1}{4},4)$ and $(4,\frac{1}{4})$; on $y=x^{-2}$, $(\frac{1}{4},16)$ and $(4,\frac{1}{16})$. All three are decreasing, positive, tend to $+\infty$ as $x\to 0^{+}$ (the $y$-axis is a vertical asymptote for each) and tend to $0$ as $x\to \infty$ (the $x$-axis is a horizontal asymptote for each).

   Order: write $t=\frac{1}{x}$, so the curves are $t^{1/2}$, $t$, $t^2$. On $(0,1)$ we have $t>1$, and a larger power of a number bigger than $1$ is bigger, so $x^{-2}>x^{-1}>x^{-1/2}$: the curve $y=x^{-2}$ is on top near the $y$-axis and rises fastest. On $(1,\infty )$ we have $0<t<1$, where a larger power is smaller, so the order reverses: $x^{-1/2}>x^{-1}>x^{-2}$, and $y=x^{-2}$ is now lowest and hugs the $x$-axis most closely. The three curves cross at $(1,1)$ and nowhere else, which is what a sketch must show: they swap order exactly once.

   (b) Finite areas. Near $0$ the problem is a vertical asymptote, and the region with the best chance of finite area is the one under the lowest curve there, $y=x^{-1/2}$. Near $\infty$ the problem is an infinite interval, and the candidate is again the lowest curve there, now $y=x^{-2}$. The $p$-integral theorems confirm that exactly these two converge: at zero, ${\int }_0^1{x}^{-p}dx$ converges only for $p<1$, which among $\frac{1}{2},1,2$ is just $p=\frac{1}{2}$; at infinity, ${\int }_1^{\infty }{x}^{-p}dx$ converges only for $p>1$, which is just $p=2$. The middle curve $y=1/x$ fails at both ends.

   From the definition:

   $${\int }_0^1{x}^{-1/2}dx=\underset{t\to 0^{+}}{\lim }{\int }_t^1{x}^{-1/2}dx=\underset{t\to 0^{+}}{\lim }[2\sqrt{x}{]}_t^1=\underset{t\to 0^{+}}{\lim }(2-2\sqrt{t})=2,$$

   $${\int }_1^{\infty }{x}^{-2}dx=\underset{t\to \infty }{\lim }{\int }_1^t{x}^{-2}dx=\underset{t\to \infty }{\lim }[-\frac{1}{x}{]}_1^t=\underset{t\to \infty }{\lim }(1-\frac{1}{t})=1.$$

   For contrast, ${\int }_t^1\frac{dx}{x}=-\ln t\to \infty$ as $t\to 0^{+}$ and ${\int }_1^t\frac{dx}{x}=\ln t\to \infty$ as $t\to \infty$, so $y=1/x$ encloses infinite area on both sides of $x=1$.

   (c) No single $p$ works. An integral with trouble at both ends must be split, and both pieces must converge: ${\int }_0^{\infty }{x}^{-p}dx={\int }_0^1{x}^{-p}dx+{\int }_1^{\infty }{x}^{-p}dx$. The first piece needs $p<1$ and the second needs $p>1$. No $p$ satisfies both, so the integral diverges for every $p$. In the picture: whichever curve is low enough at one end is too high at the other.

   The function $g$ uses the winning curve on each side. Each piece has a single problem at one end, and by (b) both converge:

   $${\int }_0^{\infty }g(x)dx={\int }_0^1{x}^{-1/2}dx+{\int }_1^{\infty }{x}^{-2}dx=2+1=3.$$

   The two pieces of $g$ agree at $x=1$ (both give $1$), so $g$ is continuous on $(0,\infty )$, and the value $3$ is the area of the region under the lower envelope of the sketch.

10. 10.Proof · 11 min

    Let $J={\int }_0^1\frac{x^4(1-x{)}^4}{1+x^2}dx$. This question proves that $\frac{22}{7}$ overestimates $\pi$, and by how much.

    1. (a)Expand the numerator and carry out the polynomial long division, showing that $\frac{x^4(1-x{)}^4}{1+x^2}=x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}$.
    2. (b)Deduce that $J=\frac{22}{7}-\pi$, and explain why this proves $\pi <\frac{22}{7}$.
    3. (c)Prove that $\frac{1}{1260}<\frac{22}{7}-\pi <\frac{1}{630}$. You will need ${\int }_0^1{x}^4(1-x{)}^{4}dx$, which you should compute, and a two-sided bound on $\frac{1}{1+x^2}$.
    Show the solution to question 10

    (a) Expand, then divide. By the binomial theorem $(1-x{)}^4=1-4x+6x^2-4x^3+x^4$, so

    $$x^4(1-x{)}^4=x^8-4x^7+6x^6-4x^5+x^4.$$

    Divide by $x^2+1$ term by term, always killing the leading term. $x^6$: subtract $x^8+x^6$, leaving $-4x^7+5x^6-4x^5+x^4$. $-4x^5$: subtract $-4x^7-4x^5$, leaving $5x^6+x^4$. $5x^4$: subtract $5x^6+5x^4$, leaving $-4x^4$. $-4x^2$: subtract $-4x^4-4x^2$, leaving $4x^2$. $4$: subtract $4x^2+4$, leaving the remainder $-4$. Hence

    $$x^4(1-x{)}^4=(x^2+1)(x^6-4x^5+5x^4-4x^2+4)-4,$$

    and dividing through by $1+x^2$ gives the stated identity. Check by multiplying back: $(x^2+1)(x^6-4x^5+5x^4-4x^2+4)=x^8-4x^7+5x^6-4x^4+4x^2+x^6-4x^5+5x^4-4x^2+4=x^8-4x^7+6x^6-4x^5+x^4+4$, and subtracting $4$ recovers the numerator.

    (b) Integrate the pieces. The polynomial part integrates by the power rule and the remainder is the arctangent form:

    $$J=[\frac{x^7}{7}-\frac{4x^6}{6}+x^5-\frac{4x^3}{3}+4x{]}_0^1-4[\arctan x{]}_0^1=(\frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4)-4\cdot \frac{\pi }{4}.$$

    The bracket is $\frac{1}{7}+(1+4-\frac{2}{3}-\frac{4}{3})=\frac{1}{7}+3=\frac{22}{7}$, so $J=\frac{22}{7}-\pi$.

    Why this proves $\pi <\frac{22}{7}$: on the open interval $(0,1)$ the integrand is a quotient of strictly positive quantities, since $x^4>0$, $(1-x{)}^4>0$ and $1+x^2>0$. A continuous function that is positive on $(0,1)$ has a positive integral over $[0,1]$, so $J>0$, that is, $\frac{22}{7}-\pi >0$. $■$

    (c) Trap the integrand. First the auxiliary integral, by expanding the numerator found in (a):

    $${\int }_0^1{x}^4(1-x{)}^{4}dx={\int }_0^1(x^4-4x^5+6x^6-4x^7+x^8)dx=\frac{1}{5}-\frac{4}{6}+\frac{6}{7}-\frac{4}{8}+\frac{1}{9}=\frac{1}{5}-\frac{2}{3}+\frac{6}{7}-\frac{1}{2}+\frac{1}{9}.$$

    Over the common denominator $630$ the numerators are $126-420+540-315+70=1$, so the integral is $\frac{1}{630}$.

    Now the bound. For $0<x<1$ we have $0<x^2<1$, hence $1<1+x^2<2$ and

    $$\frac{1}{2}<\frac{1}{1+x^2}<1.$$

    Multiplying by the positive quantity $x^4(1-x{)}^4$,

    $$\frac{1}{2}{x}^4(1-x{)}^4<\frac{x^4(1-x{)}^4}{1+x^2}<x^4(1-x{)}^4(0<x<1).$$

    All three functions are continuous on $[0,1]$ and the inequalities are strict on the open interval, so integrating over $[0,1]$ preserves them strictly:

    $$\frac{1}{2}\cdot \frac{1}{630}<J<\frac{1}{630},\text{that is,}\frac{1}{1260}<\frac{22}{7}-\pi <\frac{1}{630}.■$$

    Sanity check: $\frac{22}{7}-\pi \approx 3.142857-3.141593=0.001264$, while $\frac{1}{1260}\approx 0.000794$ and $\frac{1}{630}\approx 0.001587$.

11. 11.Critique · 8 min

    A student is asked whether ${\int }_1^{\infty }\frac{{\sin }^{2}x}{x}dx$ converges and writes:

    “For all $x$, ${\sin }^{2}x\le 1$, so $0\le \frac{{\sin }^{2}x}{x}\le \frac{1}{x}$ on $[1,\infty )$. Since ${\int }_1^{\infty }\frac{dx}{x}$ diverges (it is the $p$-integral with $p=1$), the comparison theorem shows that ${\int }_1^{\infty }\frac{{\sin }^{2}x}{x}dx$ diverges.”

    The conclusion happens to be true. The argument is not.

    1. (a)State the comparison theorem, identify which of its two parts the student believes they are using, and explain precisely why neither part applies. Give a specific integrand for which the identical reasoning would produce a false conclusion.
    2. (b)Give a valid proof that the integral diverges. The identity ${\sin }^{2}x=\frac{1}{2}(1-\cos 2x)$ is a good place to start.
    Show the solution to question 11

    (a) The comparison runs the wrong way. The theorem: if $f$ and $g$ are continuous with $0\le g(x)\le f(x)$ for $x\ge a$, then (1) if ${\int }_a^{\infty }f$ converges, so does ${\int }_a^{\infty }g$; and (2) if ${\int }_a^{\infty }g$ diverges, so does ${\int }_a^{\infty }f$. Here the student has $g=\frac{{\sin }^{2}x}{x}$ (the smaller function) and $f=\frac{1}{x}$ (the larger), and the fact in hand is that the larger integral diverges. Part (1) needs the larger integral to converge; part (2) needs the smaller to diverge. Neither hypothesis is available, so the theorem says nothing. The student is using part (2) with the roles of $f$ and $g$ swapped, which is not a theorem: a divergent big integral leaves the small one free to do either.

    A counterexample to the reasoning: $0\le \frac{1}{x^2}\le \frac{1}{x}$ on $[1,\infty )$ and ${\int }_1^{\infty }\frac{dx}{x}$ diverges, yet ${\int }_1^{\infty }\frac{dx}{x^2}=1$ converges. The student's argument, applied word for word to $\frac{1}{x^2}$, would prove a falsehood.

    (b) A correct proof. Direct comparison is awkward because ${\sin }^{2}x$ is repeatedly zero, so no positive multiple of $\frac{1}{x}$ fits underneath. Instead use the half-angle identity: for $t>1$,

    $${\int }_1^t\frac{{\sin }^{2}x}{x}dx=\frac{1}{2}{\int }_1^t\frac{dx}{x}-\frac{1}{2}{\int }_1^t\frac{\cos 2x}{x}dx=\frac{1}{2}\ln t-\frac{1}{2}{\int }_1^t\frac{\cos 2x}{x}dx.$$

    The first term tends to $+\infty$. It remains to show the second term stays bounded, and in fact that ${\int }_1^{\infty }\frac{\cos 2x}{x}dx$ converges. Integrate by parts with $u=\frac{1}{x}$, $dv=\cos 2xdx$, so $du=-\frac{dx}{x^2}$ and $v=\frac{1}{2}\sin 2x$:

    $${\int }_1^t\frac{\cos 2x}{x}dx=[\frac{\sin 2x}{2x}{]}_1^t+{\int }_1^t\frac{\sin 2x}{2x^2}dx=\frac{\sin 2t}{2t}-\frac{\sin 2}{2}+{\int }_1^t\frac{\sin 2x}{2x^2}dx.$$

    As $t\to \infty$, $\left|\frac{\sin 2t}{2t}\right|\le \frac{1}{2t}\to 0$. The last integral converges absolutely: $0\le \frac{|\sin 2x|}{2x^2}\le \frac{1}{2x^2}$ and ${\int }_1^{\infty }\frac{dx}{2x^2}=\frac{1}{2}$ converges, so by the comparison theorem (part 1, now used in the right direction) and the absolute-convergence corollary, ${\int }_1^{\infty }\frac{\sin 2x}{2x^2}dx$ converges to some finite number. Hence ${\int }_1^{\infty }\frac{\cos 2x}{x}dx$ converges; call its value $L$.

    Returning to the original integral,

    $${\int }_1^t\frac{{\sin }^{2}x}{x}dx=\frac{1}{2}\ln t-\frac{1}{2}{\int }_1^t\frac{\cos 2x}{x}dx⟶\infty -\frac{L}{2}=\infty ,$$

    since the second term has a finite limit while the first is unbounded. The integral diverges. $■$

    What made the honest argument longer than the student's: the integrand is on average half of $\frac{1}{x}$, which is enough for divergence, but "on average" is exactly what a pointwise comparison cannot see; the identity converts the average into an explicit $\frac{1}{2x}$ plus an oscillating correction whose integral converges.

12. 12.Proof · 12 min

    For integers $n\ge 0$ let ${\Gamma }_n={\int }_0^{\infty }{x}^n{e}^{-x}dx$. Work from the definition of the improper integral throughout: every claim that a limit exists must be justified.

    1. (a)Show directly from the definition that ${\Gamma }_0$ converges and equals $1$.
    2. (b)Let $n\ge 1$ and suppose ${\Gamma }_{n-1}$ converges. Prove that ${\Gamma }_n$ converges and that ${\Gamma }_n=n{\Gamma }_{n-1}$. Prove any limit you need about $t^n{e}^{-t}$; do not merely assert it.
    3. (c)Conclude that ${\Gamma }_n=n!$ for every $n\ge 0$, and state precisely what form of argument you have used.
    Show the solution to question 12

    (a) The base case from the definition. For $t>0$, ${\int }_0^t{e}^{-x}dx=[-e^{-x}{]}_0^t=1-e^{-t}$. As $t\to \infty$, $e^{-t}\to 0$, so the limit exists and

    $${\Gamma }_0=\underset{t\to \infty }{\lim }(1-e^{-t})=1.$$

    (b) The inductive step, by parts on a finite interval. Fix $n\ge 1$ and $t>0$. On $[0,t]$ everything is continuous, so the definite form of integration by parts applies with $u=x^n$, $dv=e^{-x}dx$, $du=n{x}^{n-1}dx$, $v=-e^{-x}$:

    $${\int }_0^t{x}^n{e}^{-x}dx=[-x^n{e}^{-x}{]}_0^t+n{\int }_0^t{x}^{n-1}{e}^{-x}dx=-t^n{e}^{-t}+n{\int }_0^t{x}^{n-1}{e}^{-x}dx.$$

    (The lower boundary term is $0^n{e}^0=0$ because $n\ge 1$.) Now let $t\to \infty$ on the right-hand side, one term at a time.

    The boundary term. Write $t^n{e}^{-t}=\frac{t^n}{e^t}$. As $t\to \infty$ this is of the form $\frac{\infty }{\infty }$; numerator and denominator are differentiable, and the denominator's derivative $e^t$ is never zero, so L'Hôpital's Rule may be applied. Each application lowers the power in the numerator by one and leaves $e^t$ below:

    $$\underset{t\to \infty }{\lim }\frac{t^n}{e^t}=\underset{t\to \infty }{\lim }\frac{n{t}^{n-1}}{e^t}=\cdots =\underset{t\to \infty }{\lim }\frac{n!}{e^t}=0,$$

    and the chain is read from the right: the last limit exists, so the rule licenses the equality before it, and so on back to the first. Hence $t^n{e}^{-t}\to 0$.

    The integral term. By hypothesis ${\Gamma }_{n-1}$ converges, which by definition means ${\lim }_{t\to \infty }{\int }_0^t{x}^{n-1}{e}^{-x}dx$ exists and equals ${\Gamma }_{n-1}$.

    Both pieces have limits, so by the sum law the right-hand side has the limit $0+n{\Gamma }_{n-1}$. Therefore ${\lim }_{t\to \infty }{\int }_0^t{x}^n{e}^{-x}dx$ exists, which is exactly the statement that ${\Gamma }_n$ converges, and its value is

    $${\Gamma }_n=n{\Gamma }_{n-1}.■$$

    (c) Induction. Let $P(n)$ be the statement "${\Gamma }_n$ converges and ${\Gamma }_n=n!$". Part (a) is $P(0)$, since $0!=1$. Part (b) shows that $P(n-1)$ implies $P(n)$ for every $n\ge 1$: if ${\Gamma }_{n-1}$ converges with value $(n-1)!$, then ${\Gamma }_n$ converges with value $n\cdot (n-1)!=n!$. By the principle of mathematical induction, $P(n)$ holds for every integer $n\ge 0$:

    $${\int }_0^{\infty }{x}^n{e}^{-x}dx=n!(n=0,1,2,\dots ).$$

    The argument is an induction on $n$ in which the inductive step is an integration by parts followed by a limit; the convergence of each ${\Gamma }_n$ is part of what is being proved, not assumed in advance.

    Sanity check: ${\Gamma }_1=1$ says that ${\int }_0^{\infty }x{e}^{-x}dx=1$, and indeed ${\int }_0^{t}x{e}^{-x}dx=1-(t+1)e^{-t}\to 1$. Convergence alone could also have been settled by comparison: $x^n{e}^{-x/2}\to 0$ as $x\to \infty$ by the same L'Hôpital argument, so it is bounded by some constant $M_n$ on $[0,\infty )$, giving $0\le x^n{e}^{-x}\le M_n{e}^{-x/2}$ with ${\int }_0^{\infty }{e}^{-x/2}dx=2$ convergent. The proof above gets convergence and the value in one pass.