int_0^pi frac{x , dx}{1 + sin x} is equal to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $I = \int_0^\pi \frac{x \, dx}{1 + \sin x}$ ....$(i)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$,we have:$I = \int_0^\p

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(C) Let $I = \int_0^\pi \frac{x \, dx}{1 + \sin x}$ ....$(i)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$,we have:$I = \int_0^\pi \frac{(\pi - x) \, dx}{1 + \sin(\pi - x)}$$I = \int_0^\pi \frac{(\pi - x) \, dx}{1 + \sin x}$ ....$(ii)$Adding $(i)$ and $(ii)$:$2I = \int_0^\pi \frac{x + \pi - x}{1 + \sin x} \, dx = \pi \int_0^\pi \frac{1}{1 + \sin x} \, dx$Multiply numerator and denominator by $(1 - \sin x)$:$2I = \pi \int_0^\pi \frac{1 - \sin x}{1 - \sin^2 x} \, dx = \pi \int_0^\pi \frac{1 - \sin x}{\cos^2 x} \, dx$$2I = \pi \int_0^\pi (\sec^2 x - \sec x 	an x) \, dx$Integrating the terms:$2I = \pi [	an x - \sec x]_0^\pi$$2I = \pi [(	an \pi - \sec \pi) - (	an 0 - \sec 0)]$$2I = \pi [(0 - (-1)) - (0 - 1)]$$2I = \pi [1 + 1] = 2\pi$$I = \pi$

$\int_0^\pi \frac{x \, dx}{1 + \sin x}$ is equal to

Similar Questions

$\int_{0}^{\pi / 2} \frac{d x}{1+\tan x}$ is equal to

$\int_0^{\frac{\pi}{2}} \frac{\sum_{n=0}^4 \left(\frac{n \pi}{4}+x\right)}{\cos x+\sin x} d x=$

Suppose $M = \int_{0}^{\pi / 2} \frac{\cos x}{x+2} dx$ and $N = \int_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} dx$. Then, the value of $(M - N)$ equals

$\int\limits_{-1}^{1} \frac{x^4}{1 + e^{x^7}} dx = $

If $f : R \rightarrow R$ is a continuous function satisfying $\int \limits_0^{\pi / 2} f(\sin 2x) \cdot \sin x \, dx + \alpha \int \limits_0^{\pi / 4} f(\cos 2x) \cdot \cos x \, dx = 0$,then $\alpha$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module