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

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)/2which 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):[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)


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