Question

By using the properties of definite integrals, evaluate the integral: $∫_0^π\frac{xdx}{1+sinx}$

Answer 1

Let I=$∫_0^π\frac{xdx}{1+sinx}.....(1)$

$⇒I=∫^π_0\frac{(π-x)}{1+sin(π-x}dx (∫^a_0ƒ(x)dx=∫^a_0ƒ(a-x)dx)$

$⇒I=∫^π_0\frac{(π-x)}{1+sinx}dx...(2)$

$Adding(1)and(2),we obtain$

$⇒2=∫^π_0\frac{(π)}{1+sinx}dx$

$⇒2I=π∫^π_0\frac{(1-sinx)}{(1+sinx)(1-sinx)}dx$

$⇒2I=π∫^π_0 \frac{1-sinx}{cos^2x}dx$

$⇒2I=π∫^π_0{sec^2x-tanxsecx}dx$

$⇒2I=π[tanx-secx]^π_0$

$⇒2I=π[2]$

$⇒I=π$