The value of the definite integral int_0^1 fr | vedclass

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

Question

(D) We have the integral $I = \int_0^1 \frac{dx}{x^2 + 2x\cos \alpha + 1}$.Completing the square in the denominator: $x^2 + 2x\cos \alpha + \cos^2 \al

Vedclass · Accepted Answer

$\frac{\alpha}{2 \sin \alpha}$

Vedclass · Answer

(D) We have the integral $I = \int_0^1 \frac{dx}{x^2 + 2x\cos \alpha + 1}$.Completing the square in the denominator: $x^2 + 2x\cos \alpha + \cos^2 \alpha + 1 - \cos^2 \alpha = (x + \cos \alpha)^2 + \sin^2 \alpha$.Thus,$I = \int_0^1 \frac{dx}{(x + \cos \alpha)^2 + \sin^2 \alpha}$.Using the standard integral $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$,we get:$I = \left[ \frac{1}{\sin \alpha} 	an^{-1}\left( \frac{x + \cos \alpha}{\sin \alpha} ight) ight]_0^1$.Evaluating at the limits:$I = \frac{1}{\sin \alpha} \left[ 	an^{-1}\left( \frac{1 + \cos \alpha}{\sin \alpha} ight) - 	an^{-1}\left( \frac{\cos \alpha}{\sin \alpha} ight) ight]$.Using trigonometric identities $\frac{1 + \cos \alpha}{\sin \alpha} = \cot(\frac{\alpha}{2})$ and $\frac{\cos \alpha}{\sin \alpha} = \cot \alpha$:$I = \frac{1}{\sin \alpha} \left[ 	an^{-1}(\cot \frac{\alpha}{2}) - 	an^{-1}(\cot \alpha) ight]$.Since $	an^{-1}(\cot 	heta) = \frac{\pi}{2} - 	heta$:$I = \frac{1}{\sin \alpha} \left[ (\frac{\pi}{2} - \frac{\alpha}{2}) - (\frac{\pi}{2} - \alpha) ight] = \frac{1}{\sin \alpha} [\frac{\alpha}{2}] = \frac{\alpha}{2 \sin \alpha}$.

The value of the definite integral $\int_0^1 \frac{dx}{x^2 + 2x\cos \alpha + 1}$ for $0 < \alpha < \pi$ is equal to

Similar Questions

The value of the $\int_{0}^{\frac{\pi}{2}} \left( \frac{1 + \sin 3y}{1 + 2\sin y} \right) dy$ is equal to

If $[x]$ is the greatest integer function,then $\int_0^5 [x] \, dx =$

If $5 f(x)+3 f\left(\frac{1}{x}\right)=2-\frac{1}{x}, x \neq 0$,then $\int_1^2 f\left(\frac{1}{x}\right) d x=$

$\int_0^\infty {\frac{{{x^2}\,dx}}{{({x^2} + {a^2})({x^2} + {b^2})}}} = $

If $\int_{0}^{\frac{\pi}{3}} \frac{\tan \theta}{\sqrt{2 k \sec \theta}} d \theta = 1 - \frac{1}{\sqrt{2}}$,$(k > 0)$,then the value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module