# Evaluate the integral using substitution$\int\limits_0^1\sin^{-1}\bigg(\frac{2x}{1+x^2}\bigg)dx$

$\begin{array}{1 1} \frac{\pi}{2}-\log 2 \\\frac{\pi}{2}+\log 2 \\\frac{\pi}{4}-\log 4\\\frac{\pi}{4}+\log 4 \end{array}$

## 1 Answer

Toolbox:
• (i)$\int \limits_a^b f(x)dx=F(b)-f(a)$
• Method of integration by parts
• (ii)$\int udv=uv-\int vdu$
• (iii)$\int \frac{1}{x}dx=log|x|+c$
$I=\int\limits_0^1\sin^{-1}\bigg(\frac{2x}{1+x^2}\bigg)dx$

Let $x=\tan \theta\;=>\theta=\tan^{-1}(x)$

$I= \int \limits_0^1 \sin^{-1} \bigg(\large\frac{2 \tan \theta}{1+\tan 2\theta}\bigg) dx$

But we know $\large\frac{2 \tan \theta}{1+\tan 2 \theta}=\sin 2 \theta$

Therefore$I=\int \limits_0^1 \sin^{-1}(\sin 2 \theta)dx$

$=\int \limits_0^1 2\theta dx$

Now substitute for $\theta$

$=\int \limits_0^1 2 \tan^{-1}(x)dx$

This is in the form of $\int udv$

$\int udv=uv-\int vdu$

Where $u=\tan^{-1}x$ on integrating we get,

$du=\large\frac{1}{1+x^2}dx$

Let $dv=dx$

On integrating we get,

v=x,substituting for u,v,du and dv

Therefore$I=2 \bigg[(x \tan^{-1}x)=\int x.\frac{1}{1+x^2}dx \bigg]_0^1 -----(1)$

Consider $\int \large\frac{x}{1+x^2}dx$

Let $1+x^2=t$

On differentiating with respect to x,

$2xdx=dt$

=>$xdx=\large\frac{dt}{2}$ Now substituting t and dt

Therefore $\int \large\frac{xdx}{1+x^2}=\int \frac{dt/2}{t}$

$=\frac{1}{2}\int \large\frac{dt}{t}$

On integrating we get $\frac{1}{2} log |t|$

substituting back for t we get,

$\int \frac{x}{1+x^2}dx=\frac{1}{2}log|1+x^2|$

Now substituting this in equ(1)

$I=\bigg[[(x \tan ^{-1}x)]_0^1-[\frac{1}{2}log|1+x^2|]_0^1\bigg]$

$=(2x \tan^{-1}x)_0^1-(log(1+x^2))_0^1$

On applying the limits,

$(2 \times 1 \times \tan^{-1}-2 \times 0)-[log(1+1)-log(1+0)]$

$=2 \tan^{-1}(1)-log 2$

We know $\tan^{-1}(1)=\large\frac{\pi}{4}$

Therefore $2.\large\frac{\pi}{4}-log 2$

$I=\large\frac{\pi}{2}-log 2$

answered Feb 12, 2013 by

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