If int frac{sin x}{sin (x-alpha)} dx = Ax + B | vedclass

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

Question

(A) Let $I = \int \frac{\sin x}{\sin (x-\alpha)} dx$.We can rewrite the numerator as $\sin x = \sin ((x-\alpha) + \alpha)$.Using the identity $\sin (A

Vedclass · Accepted Answer

$\cos \alpha, \sin \alpha$

Vedclass · Answer

(A) Let $I = \int \frac{\sin x}{\sin (x-\alpha)} dx$.We can rewrite the numerator as $\sin x = \sin ((x-\alpha) + \alpha)$.Using the identity $\sin (A+B) = \sin A \cos B + \cos A \sin B$,we get:$\sin ((x-\alpha) + \alpha) = \sin (x-\alpha) \cos \alpha + \cos (x-\alpha) \sin \alpha$.Substituting this into the integral:$I = \int \frac{\sin (x-\alpha) \cos \alpha + \cos (x-\alpha) \sin \alpha}{\sin (x-\alpha)} dx$.$I = \int \cos \alpha dx + \int \sin \alpha \frac{\cos (x-\alpha)}{\sin (x-\alpha)} dx$.$I = \cos \alpha \int dx + \sin \alpha \int \cot (x-\alpha) dx$.Integrating,we get $I = x \cos \alpha + \sin \alpha \log |\sin (x-\alpha)| + c$.Comparing this with the given form $Ax + B \log |\sin (x-\alpha)| + c$,we find $A = \cos \alpha$ and $B = \sin \alpha$.

If $\int \frac{\sin x}{\sin (x-\alpha)} dx = Ax + B \log |\sin (x-\alpha)| + c$,then the values of $A$ and $B$ are respectively (where $c$ is a constant of integration).

Similar Questions

$\int \sin^2 x \cos x \, dx$ is equal to

$\int \frac{\sec^2 x}{(\sec x + \tan x)^{5/2}} dx =$

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

The integral $\int {\frac{{{x^7} + {x^2} + 1}}{{{{\left( {3{x^8} + 8{x^3} + 24x} \right)}^{1/3}}}}dx} $ is equal to

$\int {\frac{{\sqrt {{x^2} + 1} [\log ({x^2} + 1) - 2\log x]}}{{{x^4}}}} dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module