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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} $

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  • (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$
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,
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,
=>$ 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 meena.p
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