int frac{sin (x-a)}{sin (x-b)} d x = A x + B | vedclass

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

Question

(A) To evaluate the integral $I = \int \frac{\sin (x-a)}{\sin (x-b)} dx$,we rewrite the numerator as $\sin((x-b) + (b-a))$.Using the identity $\sin(u+

Vedclass · Accepted Answer

$(\cos (b-a), \sin (b-a))$

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{\sin (x-a)}{\sin (x-b)} dx$,we rewrite the numerator as $\sin((x-b) + (b-a))$.Using the identity $\sin(u+v) = \sin u \cos v + \cos u \sin v$,we get:$I = \int \frac{\sin(x-b)\cos(b-a) + \cos(x-b)\sin(b-a)}{\sin(x-b)} dx$$I = \int \cos(b-a) dx + \int \sin(b-a) \cot(x-b) dx$$I = x \cos(b-a) + \sin(b-a) \log |\sin(x-b)| + C$Comparing this with $Ax + B \log |\sin(x-b)| + C$,we find $A = \cos(b-a)$ and $B = \sin(b-a)$.Thus,$(A, B) = (\cos(b-a), \sin(b-a))$.

$\int \frac{\sin (x-a)}{\sin (x-b)} d x = A x + B \log |\sin (x-b)| + C \Rightarrow (A, B) = $

Similar Questions

Evaluate the integral $\int \frac{x^2 + \cos^2 x}{1 + x^2} \operatorname{cosec}^2 x \, dx$,where $c$ is the constant of integration.

Find the integral of the function $\tan^{3} 2x \sec 2x$.

$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x=$

$\int \frac{x^{2}+1}{x^{4}+x^{2}+1} d x=$

If $\int \frac{3 e^x-7 e^{-x}}{7 e^x+3 e^{-x}} d x=K x+L \log \left(e^{-2 x}+\frac{7}{3}\right)+C$,then $K+L=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module