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

If $\int \frac{\sqrt{2} \, dx}{\cos x \sqrt{\sin 2x}} = f(x) + c$, then $f(x) =$

Integrate the function $\frac{1}{x(\log x)^{m}}$,where $x > 0$ and $m \neq 1$.

If $f(x) = \int \frac{x}{(x^2+4)(x^2+9)} dx$ and $f(0) = \frac{1}{5} \log(\frac{2}{3})$, then $f(1) = $

If $\int \frac{\sin ^3 x}{\left(\cos ^4 x+3 \cos ^2 x+1\right) \tan ^{-1}(\sec x+\cos x)} d x=f(x)+C$,then $e^{f(x)}=$

Integrate the function $\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module