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

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

Question

(B) We have $I = \int \frac{x - \sin x}{1 - \cos x} dx$. Using the identities $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \co

Vedclass · Accepted Answer

$-x \cot \frac{x}{2} + c$

Vedclass · Answer

(B) We have $I = \int \frac{x - \sin x}{1 - \cos x} dx$. Using the identities $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we get: $I = \int \frac{x}{2 \sin^2 \frac{x}{2}} dx - \int \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} dx$ $I = \frac{1}{2} \int x \csc^2 \frac{x}{2} dx - \int \cot \frac{x}{2} dx$. Applying integration by parts to the first term: Let $u = x$ and $dv = \csc^2 \frac{x}{2} dx$. Then $du = dx$ and $v = -2 \cot \frac{x}{2}$. $\int x \csc^2 \frac{x}{2} dx = x(-2 \cot \frac{x}{2}) - \int (-2 \cot \frac{x}{2}) dx = -2x \cot \frac{x}{2} + 2 \int \cot \frac{x}{2} dx$. Substituting this back into the expression for $I$: $I = \frac{1}{2} [-2x \cot \frac{x}{2} + 2 \int \cot \frac{x}{2} dx] - \int \cot \frac{x}{2} dx$ $I = -x \cot \frac{x}{2} + \int \cot \frac{x}{2} dx - \int \cot \frac{x}{2} dx + c$ $I = -x \cot \frac{x}{2} + c$.

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

Similar Questions

If $f(x) = \int \operatorname{cosec}^5 x \, dx$,then $f\left(\frac{\pi}{4}\right) = $

$\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=$

Evaluate the integral $\int \frac{\log x - (\log x)^2 + x^2}{x^3} dx$ (where $C$ is the constant of integration).

$\int {x\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} } \;dx = $

$\int \frac{dx}{(x+1) \sqrt{x^2+1}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module