int_0^pi frac{dx}{1 + sin x} = | vedclass

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

Question

(C) We evaluate the integral $I = \int_0^\pi \frac{dx}{1 + \sin x}$.Multiply the numerator and denominator by $(1 - \sin x)$:$I = \int_0^\pi \frac{1 -

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) We evaluate the integral $I = \int_0^\pi \frac{dx}{1 + \sin x}$.Multiply the numerator and denominator by $(1 - \sin x)$:$I = \int_0^\pi \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} dx = \int_0^\pi \frac{1 - \sin x}{1 - \sin^2 x} dx = \int_0^\pi \frac{1 - \sin x}{\cos^2 x} dx$.Split the integral into two parts:$I = \int_0^\pi (\sec^2 x - \sec x 	an x) dx$.Now,integrate each term:$I = [	an x - \sec x]_0^\pi$.Evaluate at the limits:$I = (	an \pi - \sec \pi) - (	an 0 - \sec 0)$.Since $	an \pi = 0$,$\sec \pi = -1$,$	an 0 = 0$,and $\sec 0 = 1$:$I = (0 - (-1)) - (0 - 1) = 1 - (-1) = 1 + 1 = 2$.

$\int_0^\pi \frac{dx}{1 + \sin x} = $

Similar Questions

The value of $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\int_{\pi /2}^x {t\,dt} }}{{\sin (2x - \pi )}}$ is

$\int\limits_0^{\sqrt{3}} \frac{1}{2} \frac{d}{dx} \left( \tan^{-1} \frac{2x}{1-x^2} \right) dx$ equals

If $f(t) = \int_{-t}^t \frac{e^{-|x|}}{2} dx$, then $\lim_{t \rightarrow \infty} f(t)$ is equal to:

If $[x]$ denotes the greatest integer function of $x$ and $\int_{-\frac{3}{2}}^{\frac{3}{2}}[2x-3] dx = k$, then $\left|k+\frac{1}{2}\right| = $

The value of the definite integral,$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin 5x}}{{\sin x}}\,dx} $ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module