int {frac{{dx}}{{1 - cos x - sin x}}} = | vedclass

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

Question

(C) Let $I = \int \frac{dx}{1 - \cos x - \sin x}$.Using half-angle formulas $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$ and $\sin x = \frac{2

Vedclass · Accepted Answer

$\log |1 - \cot(x/2)| + c$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{1 - \cos x - \sin x}$.Using half-angle formulas $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$ and $\sin x = \frac{2	an(x/2)}{1 + 	an^2(x/2)}$,we get:$I = \int \frac{dx}{1 - \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)} - \frac{2	an(x/2)}{1 + 	an^2(x/2)}}$$I = \int \frac{(1 + 	an^2(x/2)) dx}{1 + 	an^2(x/2) - 1 + 	an^2(x/2) - 2	an(x/2)}$$I = \int \frac{\sec^2(x/2) dx}{2	an^2(x/2) - 2	an(x/2)}$Let $t = 	an(x/2)$,then $dt = \frac{1}{2}\sec^2(x/2) dx$,so $\sec^2(x/2) dx = 2 dt$.$I = \int \frac{2 dt}{2(t^2 - t)} = \int \frac{dt}{t(t - 1)}$Using partial fractions: $\frac{1}{t(t - 1)} = \frac{1}{t - 1} - \frac{1}{t}$.$I = \int \left( \frac{1}{t - 1} - \frac{1}{t} ight) dt = \log|t - 1| - \log|t| + c = \log\left| \frac{t - 1}{t} ight| + c$$I = \log\left| \frac{	an(x/2) - 1}{	an(x/2)} ight| + c = \log|1 - \cot(x/2)| + c$.

$\int {\frac{{dx}}{{1 - \cos x - \sin x}}} = $

Similar Questions

If $\int \frac{x^2}{(x-1)(x-2)(x-3)} dx = \log_e f(x) + C$,then $f(x) =$

$\int \frac{\cos x}{(1 + \sin x)(2 + \sin x)} \,dx = $

If $\int \frac{x+3}{(x-1)^2(2 x-1)} d x=\frac{A}{x-1}+B \log (2 x-1)+C \log (x-1)+K$, then $A+B+C=$

$\int \frac{dx}{x(x+1)}$ equals, where $c$ is an arbitrary constant.

If $\int {\frac{{2{x^2} + 3}}{{({x^2} - 1)({x^2} - 4)}}} dx = \log {\left( {\frac{{x - 2}}{{x + 2}}} \right)^a}{\left( {\frac{{x + 1}}{{x - 1}}} \right)^b} + c$,then the values of $a$ and $b$ respectively are

Vedclass Test Series

Exam Paper Generator

Online Exam Module