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}.

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