Ask Questions, Get Answers

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

Prove the following\[\int\limits_0^1\sin^{-1}x\;dx=\frac{\pi}{2}-1\]

Can you answer this question?

1 Answer

0 votes
  • (i)$\int \limits _a^b f(x)dx=F(b)-F(a)$
  • (ii) $ \int udv=uv-\int vdu$
  • (iii) If $f(x)=t,$ then $f'(x)dx=dt$,hence $\int f(x)dx=\int t dt$
Prove that $ \int \limits_0^1 \sin ^{-1} x dx=\frac{\pi}{2}-1$
Given $I=\int \limits_0^1 \sin ^{-1}x dx$
This is of the form $\int udv$
This can be solved by the method of integration by parts,
$\int udv =uv-\int vdu$
Here let $\sin ^{-1}x=x.$ on differentiating w.r.t x,
$\frac{1}{\sqrt {1-x^2}}=dx$
Let dv=dx, on integrating we get
Hence by substituting for u,v,du and dv we get,
$\int \limits_0^1 \sin ^{-1}x dx =(x \sin ^{-1}x)_0^1-\int \limits_0^1 \frac{x}{\sqrt {1-x^2}} dx$
Consider $\int \limits_0^1 \frac{x}{\sqrt {1-x^2}}dx$
Here let $1-x^2=t,$ Now differentiating w.r.t x,
$-2xdx=dt =>xdx=-\frac{dt}{2}$
As we substitute for t, the limit also change,
when x=0,t=1 and when x=1,t=0
Hence $ \int \limits_0^1 \frac{x}{\sqrt {1-x^2}}dx=\int \limits_0^1 \frac{-dt/2}{\sqrt t }$
$ =\frac{1}{2} \int \limits_0^1 \frac{dt}{\sqrt t}$
The negative symbol can be taken away by interchanging the limits
On integrating we get,
$\frac{1}{2} \int _0^1 \frac{dt}{\sqrt t}=\frac{1}{2} \bigg[\frac{t^{-1/2+1}}{-1/2+1}\bigg]_0^1$
$ =\frac {1}{2} \times \frac{2}{1} [t ^{1/2}]$
Therefore $I=(x \sin ^{-1} x)_0^1 -[t ^{1/2}]_0^1$
Now applying limits ,
$I=[1 \times \sin ^{-1}(1)-0]-[1-0]$
But $\sin ^{-1}(1)=\frac{\pi}{2}$
Therefore $I=\frac{\pi}{2}-1$
Hence proved



answered Mar 14, 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