Browse Questions

# Evaluate the definite integrals$\int\limits_0^\frac{\pi}{2}\sin 2x\tan^{-1}(sin x)\;dx$

Toolbox:
• (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
• (ii) In a integral function, $\int f(x)dx, let f(x)=t,\; then\; f'(x)dx=dt,$ hence $f(x)dx=\int t dt$
• (iii)$\int \frac{dx}{a^2+x^2}=\frac{1}{a}\tan ^{-1}(x/a)+c$
• (iv)$\sin 2x=2\sin x\cos x$
• (v)$\int udv=uv-\int vdu$
Given $I=\int\limits_0^\frac{\pi}{2}\sin 2x\tan^{-1}(sin x)\;dx$

we know $\sin 2x=2\sin x\cos x$

Hence $I=\int \limits_0^{\pi/2} 2\sin x \cos x \tan^{-1}(\sin x)dx$

Now, let $\sin x=t,$ on differentiating w.r.t x we get

$\cos x dx=dt$

The limit also changes when su bstitute for t,

When $x=0,\sin x=0 \qquad t=0$

When $x=\pi/2,\sin \pi/2=1 \qquad t=1$

on substituting t and dt

Therefore $I=2\int \limits_0^1 t.\tan^{-1}(t)dt$

This is the form $\int udv=uv-\int vdu$

Let $u=\tan^{-1}t$ on differentiating w.r.t to t

$du=\frac{1}{1+t^2}dt$

Let $dv=tdt$ on integrating we get,

$v=\frac{t^2}{2}$

Now substituting for u,v,du and dt,

$I=2 \bigg\{[\tan ^{-1}t.\frac{t^2}{2}]_0^1=\int \limits_0^1 \frac{t^2}{2}.\frac{1}{1+t^2}.dt\bigg\}$

$=2 \bigg\{[\frac{t^2}{2}\tan ^{-1}t]_0^1-\frac{1}{2}\int \limits_0^1 \frac{t^2}{1+t^2}.dt\bigg\}$

Consider $I_1=\int \frac {t^2}{1+t^2} dt$

Add and subtract +1 and -1 to the numerator,

$I-1=\int \limits_0^1 \frac{t^2+t-1}{1+t^2}dt$

Seperating the terms we get,

$\int \limits_0^1 \frac{t^2+1}{1+t^2}dt-\int \limits_0^1\frac{1}{1+t^2}dt$

$=\int \limits_0^1dt-\int \limits_0^1 \frac{dt}{1+t^2}$

$\int \frac{dt}{1+t^2}$ is of the form $\int \frac{dx}{a^2+x^2}=\frac{1}{a}\tan ^{-1}(x/a)+c$

Therefore on integrating we get,

$[t]_0^1-[\tan -1 (t)]_0^1$

Hence $I=2\bigg\{[\frac{t^2}{2} \tan^{-1} t]_0^1-\frac{1}{2}\bigg\{[t]_0^1+[\tan ^{-1}(t)]_0^1 \bigg\} \bigg\}$

On applying limits we get ,

$2\bigg\{\frac{1}{2} \tan^{-1} (1)-0]-\frac{1}{2}[1-0]+\frac{1}{2}[\tan ^{-1}(1)-0]\bigg\}$

$I=2\bigg\{\frac{1}{2} (\tan^{-1} (1)-\frac{1}{2}+\frac{1}{2}\tan ^{-1}(1)\bigg\}$

But $\tan^{-1}(1)=\pi/4$

$2[\frac{1}{2}(\pi/4+\pi/4)-1/2]=2[\frac{1}{2}(\frac{\pi}{2})-\frac{1}{2}]$

$2\bigg[\frac{\pi}{4}-\frac{1}{2}\bigg]$

$=\frac{\pi}{2}-1$