alleryStart learning →$\displaystyle\int_0^\pi \dfrac{x}{1 + \sin x}\, dx = ?$
- $\pi$
- $\pi^2/2$
- $\pi/2$
- $2\pi$
Show answer and solution
Correct answer: A. $\pi$
King's rule $x \to \pi - x$: integrand becomes $(\pi - x)/(1 + \sin(\pi - x)) = (\pi - x)/(1 + \sin x)$.
**Adding to original:** $2I = \pi\int_0^\pi dx/(1 + \sin x)$.
Weierstrass $t = \tan(x/2)$: $dx/(1+\sin x) = 2dt/(1+t)^2$.
$\int_0^\infty 2/(1+t)^2 dt = [-2/(1+t)]_0^\infty = 2$.
**So $2I = 2\pi$ ⇒ $I = \pi$.**