# By using the properties of definite integrals,evaluate the integral$\int\limits_0^\pi\frac{x\;dx}{1+\sin x}$

Toolbox:
• (i)$\int\limits_a^b f(x)dx=F(b)-F(a)$
• (ii)$\int \limits_0^af(x)dx=\int \limits_0^af(a-x)dx$
Given $I=\int\limits_0^\pi\frac{x\;dx}{1+\sin x}-----(1)$

Applying the property $\int \limits_0^af(x)dx=\int \limits_0^af(a-x)dx$

$I=\int\limits_0^\pi\frac{(\pi-x)}{1+\sin(\pi-x)}dx-----(1)$

But $\sin(\pi -x)=\sin x$

$I=\int\limits_0^\pi\frac{(\pi-x)}{1+\sin x}dx-----(2)$

$2I=\int\limits_0^\pi\frac{xdx}{1+\sin x}+\int\limits_0^\pi\frac{(\pi-x)}{1+\sin x}dx$

On simplifying we get,

$2I=\int\limits_0^\pi\frac{\pi}{1+\sin x}dx$

Multiply and divide by $(1-\sin x)$

$2I=\pi\int\limits_0^\pi\frac{(1-\sin x)}{1+\sin x)(1-\sin x)}dx$

$(1+\sin x)(1-\sin x)=1-\sin^2x=\cos ^2x$

Therefore $2I=\pi\int\limits_0^\pi\frac{1-\sin x}{\cos^2 x}$

On seperating the terms,

$2I=\pi\int\limits_0^\pi\large\frac{1}{\cos^2 x}-\int\limits_0^\pi\frac{\sin x}{\cos x}.\frac{1}{\cos x}dx$

$=\pi\int\limits_0^\pi \sec ^2x-\int \limits_0^\pi \sec x.\tan x$

On integrating we get

$2I=\pi\bigg\{[\tan x]_0^\pi -[\sec x]_0^\pi \bigg\}$

On applying limits,

$2I=\pi[(\tan \pi-\tan 0)-(\sec \pi-\sec 0)]$

But $\tan \pi=\tan 0=0$

and $\sec \pi=-1\;and\; \sec 0=1$

Now substituting these values,

$2I=\pi[0-(-1-1)]$

$2I=2\pi$

Therefore $I=\pi$