Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from p.35 of https://jpmccarthymaths.wordpress.com/2011/02/01/math6037-general-information/

These questions all require integration by parts although they don’t explicitly say so.

Question 1

We must choose a $u$ and $dv$ — where we choose $u$ according to the LIATE rule. Here there are no logs, no inverses but there is an algebraic (sum of powers of $x$). Hence let $u=x$, $dv=\cos x\,dx$ (find $v$ by integrating: $\int \,dv=v$):

$\frac{du}{dx}=1\Rightarrow du=dx$; and $v=\sin x$.

Hence by the integration by parts formula:

$I=x\sin x-\int\sin x\,dx=x\sin x-(-\cos x)+c$

$=x\sin x+\cos x+c$

We check our solution by differentiating $I$ (using the product rule):

$\frac{d}{dx}I=x\cos x+\sin x-\sin x=x\cos x$,

as required.

Question 2

Again we must choose $u$ according to the LIATE rule. Hence we choose $u=x$ (note that a strict application of the LIATE — choose the most complicated part — suggests $u=2x$. Making this substitution (without applying the integration by parts formula) leads to the integral $(\int ue^u\,du)/2$which needs the $v=u$ substitution anyway…). Hence $dv=e^{2x}\,dx$ (using $\int e^{ax}\,dx=\frac{1}{a}e^{ax}$):

$\frac{du}{dx}=1\Rightarrow du=dx$; and $v=\frac{1}{2}e^{2x}$.

Using the integration by parts formula:

$I=x.\frac{1}{2}e^{2x}-\int \frac{1}{2}e^{2x}\,dx=\frac{xe^{2x}}{2}-\frac{1}{4}e^{2x}+c$.

We check our answer by differentiating $I$:

$\frac{dI}{dx}=\frac{1}{2}(x e^{2x}(2)+e^{2x})-\frac{1}{4}e^{2x}(2)=xe^{2x}$,

as required.

Question 3

Rather than having to keep track of the limits we will first evaluate the indefinite integral:

$I=\int x\cos 2x\,dx$

By the LIATE rule, let $u=x$ and $dv=\cos 2x\,dx$:

$\frac{du}{dx}=1\Rightarrow du=dx$; and $v=\frac{1}{2}\sin 2x$.

Hence by the integration by parts formula:

$I=x.\frac{1}{2}\sin 2x-\int \frac{1}{2}\sin 2x\,dx$

$=\frac{x\sin 2x}{2}-\frac{1}{2}(-\frac{1}{2}\cos 2x)+c=\frac{2x\sin 2x+\cos 2x}{4}+c$.

Now we need to evaluate this between the limits — and of course the constant of integration will cancel so we omit it:

$I=\left[\frac{2x\sin 2x+\cos 2x}{4}\right]_0^{\pi/2}$

$=\left(\frac{2(\pi/2)\sin \pi+\cos \pi}{4}\right)-\left(\frac{2(0)\sin 0+\cos 0}{4}\right)=\frac{-1}{4}-\frac{1}{4}=-\frac{1}{2}$.

Question 4

Again we compute the integral first without limits. Now by LIATE, let $u=\log x$, so $dv=dx$:

$\frac{du}{dx}=\frac{1}{x}\Rightarrow du=\frac{dx}{x}$ and $v=\int dx=x$

Now using the integration by parts formula:

$I=\log x.x-\int x.\frac{dx}{x}=x\log x-\int dx=x\log x-x+c$.

Now we need to evaluate this between the limits, again omitting the constant of integration:

$I=\left[x(\log x-1)\right]_1^2=(2(\log 2-1))-(1(\log 1-1))$

$=2\log 2-2-\log 1+1$.

Now $\log 1=0$ and $\log x^n=n\log x$ for $x\in\mathbb{R}$, $n\in\mathbb{N}$:

$I=\log 4-1$,

with $4,-1\in\mathbb{Q}$ as required.

Question 5

Again by LIATE let $u=\log 2x$, $dv=dx$:

$\frac{du}{dx}=\frac{1}{2x}2\Rightarrow du=\frac{dx}{x}$; and $v=x$.

Hence by the integration by parts formula:

$I=\log 2x.x-\int x\frac{dx}{x}=x\log 2x-\int dx=x(\log 2x-1)+c$.

Question 6

Again by LIATE we choose $u=\sin^{-1}x$ and $dv=dx$:

$\frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}\Rightarrow du=\frac{dx}{\sqrt{1-x^2}}$ and $v=x$.

Hence by the integration by parts formula:

$I=\sin^{-1}x.x-\int x\frac{dx}{\sqrt{1-x^2}}$.

Now looking at the integral:

$J= \int\frac{x}{\sqrt{1-x^2}}\,dx$

Can we do this with the integral tables, manipulation,substitution or by parts?? Whatever works will do — experience will tell us that we need to make a substition. By noticing the function-derivative pattern $1-x^2$$(-2)x$ we should make the $w=$‘function’ substitution $w=1-x^2$ (LIATE will also tell us to pick this):

$\frac{dw}{dx}=-2x\Rightarrow dx=-\frac{dw}{2x}$,

$J=\int \frac{x}{\sqrt{w}}\left(-\frac{dw}{2x}\right)=-\frac{1}{2}\int \frac{1}{w^{1/2}}\,dw$

$=-\frac{1}{2}\int w^{-1/2}\,dw=-\frac{1}{2}\left[\frac{w^{1/2}}{1/2}\right]=-(1-x^2)^{1/2}$;

$\Rightarrow I=x\sin^{-1}x+\sqrt{1-x^2}+c$

Question 7

Again by LIATE we choose $u=\log x$ and $dv=x\,dx$:

$\frac{du}{dx}=\frac{1}{x}\Rightarrow du=\frac{dx}{x}$ and $v=\frac{x^2}{2}$.

Hence by the Integration by Parts formula:

$I=\log x.\frac{x^2}{2}-\int \frac{x^2}{2}.\frac{dx}{x}=\frac{1}{2}x^2\log x-\frac{1}{2}\int x\,dx$

$=\frac{1}{2}x^2\log x-\frac{x^2}{4}+c$

Question 8

Again by LIATE  let $u=\theta$ and $dv=\sec^2\theta\,d\theta$:

$\frac{du}{d\theta}=1\Rightarrow du=d\theta$ and $v=\int\sec^2\theta\,d\theta$.

We should recognise (from the tables if need be) that $\sec^2 x$ is the derivative of $\tan x$. Hence $v=\tan \theta$. By the Integration by Parts formula:

$I=\theta \tan\theta-\int \tan\theta\,d\theta=\theta\tan\theta-\log|\sec\theta|+c$.

where the integral of $\tan\theta$ was taken from the tables.

Question 9

Apologies! Between these limits the area under the function is infinite — namely negative infinity as it is below the $x$-axis. Look at this plot from Wolfram Alpha (an absolutely unbelievable piece of kit by the by):

http://www.wolframalpha.com/input/?i=Plot[Log[x]%2Fx^2%2C{x%2C0%2C2}]

Hence we’ll just evaluate without limits.

Now by the LIATE rule choose $u=\log x$ and $dv=dx/x^2=x^{-2}\,dx$:

$\frac{du}{dx}=\frac{1}{x}\Rightarrow du=\frac{dx}{x}$ and $v=\frac{x^{-1}}{-1}=-\frac{1}{x}$.

Hence by the Integration by Parts formula:

$I=\log x\left(-\frac{1}{x}\right)+\int \frac{1}{x}\frac{dx}{x}=-\frac{\log x}{x}+\int x^{-2}\,dx$

$=-\frac{\log x}{x}+\frac{x^{-1}}{-1}=-\frac{\log x+1}{x}+c$

Question 10

Evaluating first without limits, by the LIATE rule let $u=\log t$ and $dv=\sqrt{t}\,dt=t^{1/2}\,dt$:

$\frac{du}{dt}=\frac{1}{t}\Rightarrow du=\frac{dt}{t}$ and $v=\frac{t^{3/2}}{3/2}=\frac{2t^{3/2}}{3}$.

Hence by the Integration by Parts formula:

$I=\log t.\frac{2t^{3/2}}{3}-\int \frac{2t^{3/2}}{3}.\frac{dt}{t}=\frac{2}{3}t^{3/2}\log t-\frac{2}{3}\int t^{1/2}\,dt$

$=\frac{2}{3}t^{3/2}\log t-\frac{2}{3}.\frac{t^{3/2}}{3/2}=\frac{2}{3}t^{3/2}\log t-\frac{4}{9}t^{3/2}+c$.

Now evaluating between the limits, dropping the constant of integration:

$I=\left[\frac{2}{3}t^{3/2}\log t-\frac{4}{9}t^{3/2}\right]_1^4=\left(\frac{2}{3}4^{3/2}\log 4-\frac{4}{9}4^{3/2}\right)-\left(\frac{2}{3}1^{3/2}\log 1-\frac{4}{9}1^{3/2}\right)$.

Now first off $\log 1=0$ and $a^{m/n}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m$. Hence (or else using the calculator)

$4^{3/2}=(\sqrt{4})^3=8$:

$\Rightarrow I=\frac{2}{3}8\log 4-\frac{4}{9}(8)+\frac{4}{9}=\frac{16}{3}\log 4-\frac{28}{9}$.