logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

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} $

Can you answer this question?
 
 

1 Answer

0 votes
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 meena.p
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...