Ask Questions, Get Answers

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

Evaluate:$\int\limits_0^1 x\sqrt{\large\frac{1-x^2}{1+x^2}}$$dx$

Can you answer this question?

1 Answer

0 votes
  • $\int \limits _a^b f(x)dx=F(b)-F(a)$
Step 1:
Given $\int\limits_0^1 x\sqrt{\large\frac{1-x^2}{1+x^2}}$$dx$
Let $x^2=t$ on differentiating w.r.t x we get,
$2xdx=dt \qquad => xdx=\large\frac{dt}{2}$
on substituting t and dt,
Therefore $I=\int\limits_0^1 \sqrt{\large\frac{1-t}{1+t}}$$dt/2$
$=\large\frac{1}{2} \int\limits_0^1 \sqrt{\frac{1-t}{1+t}}$
Multiply and divide by the conjugate 1-t
$I=\large\frac{1}{2} \int\limits_0^1 \sqrt {\frac{(1-t)(1-t)}{(1+t)(1-t)}}$$dt=\large\frac{1}{2}$$ \int\limits_0^1 \sqrt {(1-t)^2}dt$
$=\large\frac{1}{2} \int\limits_0^1 \frac{(1-t)}{\sqrt {1-t^2}}$$dt $
Step 2:
On seperating the term we get
$\large\frac{1}{2} \bigg\{\bigg[\int\limits_0^1 \frac{1}{\sqrt {1-t^2}}$$dt\bigg]-\bigg[\int\limits_0^1 \large\frac{t}{\sqrt {1-t^2}}$$dt \bigg] \bigg\}$
Consider $\large\frac{t}{\sqrt {1-t^2}}$. Here let $1-t^2=u$
on differentiating we get $-2tdt=du$
On substituting for u and du
we know we substitute u, the limit of t also changes
when t=0 and when t=1
when u=1 and when u=0
Therefore $\int\limits_0^1 \large\frac{t}{\sqrt {1-t^2}}$$dt=-\int\limits_0^1\large\frac{du/2}{\sqrt u}$
The negative symbol can be removed by changing the limits
hence $\int\limits_0^1 \large\frac{t}{\sqrt {1-t^2}} $$dt =\large\frac{1}{2}\int\limits_0^1 \frac{du}{\sqrt u}$
Therefore $I=\large\frac{1}{2}\bigg[\int\limits_0^1 \frac{t}{\sqrt {1-t^2}} $$dt -\large\frac{1}{2}\int\limits_0^1 \frac{du}{\sqrt u}\bigg]$
Step 3:
$=\large\frac{dx}{\sqrt {a^2-x^2}}=$$\sin ^{-1}(x/a)$
Therefore $I=\large\frac{1}{2} $$[\sin^{-1}(t)]_0^1-\large\frac{1}{4}\bigg[\frac{u^{-1/2+1}}{-1/2+1}\bigg]_0^1$
$=\large\frac{1}{2}$$(sin ^{-1}(t))_0^1-\large\frac{1}{4}$$(-2 \sqrt u)_0^1$
on applying the limits
$I=\large\frac{1}{2}$$[\sin^{-1}(1)-\sin^{-1}(0)]+\large\frac{2}{4}[(\sqrt 1)-0]$
But $\sin^{-1}(1)=\large\frac{\pi}{2}$
Therefore $I=\large\frac{1}{2}$$(\pi/2)+\large\frac{2}{4}$
answered Apr 29, 2013 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App