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

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  • (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