int frac{dx}{1-cos x-sin x} is equal to | vedclass

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

Question

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

Vedclass · Accepted Answer

$\log \left|1-\cot \frac{x}{2}\right|+c$

Vedclass · Answer

(C) We have,$I = \int \frac{dx}{1-\cos x-\sin x}$.Using the 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)} = \frac{1}{2} \int \frac{\sec^2(x/2) dx}{	an(x/2)(	an(x/2)-1)}$.Let $z = 	an(x/2)$,then $dz = \frac{1}{2}\sec^2(x/2) dx$,so $\sec^2(x/2) dx = 2dz$.$I = \int \frac{2dz}{2z(z-1)} = \int \frac{dz}{z(z-1)}$.Using partial fractions: $\frac{1}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z}$.$I = \int \left(\frac{1}{z-1} - \frac{1}{z}ight) dz = \log|z-1| - \log|z| + C = \log\left|\frac{z-1}{z}ight| + C$.Substituting $z = 	an(x/2)$ back:$I = \log\left|\frac{	an(x/2)-1}{	an(x/2)}ight| + C = \log|1 - \cot(x/2)| + C$.

$\int \frac{dx}{1-\cos x-\sin x}$ is equal to

Similar Questions

Integrate the function: $\frac{x+3}{x^{2}-2x-5}$

If $\int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} d x=\frac{1}{2} f(x)+C$, then $f(1)-f(0)=$

If $\int \frac{5 \tan x}{\tan x-2} d x = \alpha x + \beta \log |\sin x - 2 \cos x| + \gamma$,then $\alpha - \beta =$

$\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$ is equal to

Integrate the function: $\frac{5 x-2}{1+2 x+3 x^{2}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module