int_0^pi frac{dx}{1 - 2acos x + a^2} = | vedclass

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

Question

(C) Let $I = \int_0^\pi \frac{dx}{1 - 2a\cos x + a^2}$.Using the identity $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \cos^2 \frac{x}{

Vedclass · Accepted Answer

$\frac{\pi}{1 - a^2}$

Vedclass · Answer

(C) Let $I = \int_0^\pi \frac{dx}{1 - 2a\cos x + a^2}$.Using the identity $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}$,we get:$I = \int_0^\pi \frac{dx}{(1+a^2)(\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}) - 2a(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2})}$$I = \int_0^\pi \frac{dx}{(1-a)^2 \cos^2 \frac{x}{2} + (1+a)^2 \sin^2 \frac{x}{2}}$Divide numerator and denominator by $\cos^2 \frac{x}{2}$:$I = \int_0^\pi \frac{\sec^2 \frac{x}{2} dx}{(1-a)^2 + (1+a)^2 	an^2 \frac{x}{2}}$Let $t = 	an \frac{x}{2}$,then $dt = \frac{1}{2} \sec^2 \frac{x}{2} dx$,so $\sec^2 \frac{x}{2} dx = 2 dt$.As $x 	o 0, t 	o 0$ and as $x 	o \pi, t 	o \infty$.$I = \int_0^\infty \frac{2 dt}{(1-a)^2 + (1+a)^2 t^2} = \frac{2}{(1+a)^2} \int_0^\infty \frac{dt}{(\frac{1-a}{1+a})^2 + t^2}$Using $\int \frac{dx}{k^2 + x^2} = \frac{1}{k} 	an^{-1}(\frac{x}{k})$:$I = \frac{2}{(1+a)^2} \cdot \frac{1+a}{1-a} [	an^{-1}(\frac{1+a}{1-a} t)]_0^\infty = \frac{2}{1-a^2} [\frac{\pi}{2} - 0] = \frac{\pi}{1-a^2}$.

$\int_0^\pi \frac{dx}{1 - 2a\cos x + a^2} = $

Similar Questions

$\int_{-2}^3 |1-x^2| dx =$

The approximate value of $\int_2^{10} x^2 dx$ by using the trapezoidal rule with $4$ equal intervals is:

If $I_{1}=\int_{0}^{\pi / 2} x \sin x \, dx$ and $I_{2}=\int_{0}^{\pi / 2} x \cos x \, dx$,then which one of the following is true?

$\int_{-1}^{3/2} |x \sin \pi x| \, dx =$

By the definition of the definite integral,the value of $\lim _{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module