**Toolbox:**

- (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

v=x

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