Question

By using properties of definite integrals, evaluate the integrals
${\int }_0^{\pi }\frac{x}{1+sinx}dx.$

Answer 1

Let $I={\int }_0^{\pi }\frac{x}{1+sinx}dx.$.......(i)
Then $I={\int }_0^{\pi }\frac{\pi -x}{1+sin(\pi -x)}dx[\because ({\int }_0^{a}f\left(x\right)dx={\int }_0^{0}f(a-x)dx]$
$\Rightarrow I={\int }_0^{\pi }\frac{\pi -x}{1+sinx}dx$.......(ii)
on adding Eqs. (i) and (ii) we get
2I $={\int }_0^{\pi }\frac{\pi }{(1+sinx)}dx=\pi {\int }_0^{\pi }\frac{1-sinx}{(1+sinx)(1-sinx)}dx$
(Multiply numerator and denominator by (1- sin x))
$\Rightarrow 2I=\pi {\int }_0^{\pi }\frac{1-sinx}{1-si{n}^{2}x}dx$
=$\pi {\int }_0^{\pi }\frac{1}{co{s}^{2}x}dx-\pi {\int }_0^n\frac{sinx}{co{s}^{2}x}$ $[\because si{n}^{2}x+co{s}^{2}x=1]$
$\Rightarrow 2I=\pi {\int }_0^{\pi }se{c}^{2}xdx-\pi {\int }_0^{\pi }secxtanxdx$
$\Rightarrow 2I=\pi [tanx-secx{]}_0^{\pi }\Rightarrow 2I=\pi [tan\pi -sec\pi -(tan0-sec0)]$
$\Rightarrow 2I=\pi [0+1-0+1]\Rightarrow 2I=2\pi \Rightarrow I=\pi$