If int frac{sin x}{sin (x-alpha)} dx = px - q | vedclass

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

Question

(A) To solve the integral $I = \int \frac{\sin x}{\sin (x-\alpha)} dx$,we substitute $u = x - \alpha$,so $x = u + \alpha$ and $dx = du$. Substituting

Vedclass · Accepted Answer

$-\frac{1}{2} \sin 2\alpha$

Vedclass · Answer

(A) To solve the integral $I = \int \frac{\sin x}{\sin (x-\alpha)} dx$,we substitute $u = x - \alpha$,so $x = u + \alpha$ and $dx = du$. Substituting these into the integral,we get: $I = \int \frac{\sin (u + \alpha)}{\sin u} du$ Using the identity $\sin (A+B) = \sin A \cos B + \cos A \sin B$: $I = \int \frac{\sin u \cos \alpha + \cos u \sin \alpha}{\sin u} du$ $I = \int (\cos \alpha + \cot u \sin \alpha) du$ $I = \cos \alpha \int du + \sin \alpha \int \cot u du$ $I = u \cos \alpha + \sin \alpha \log |\sin u| + c$ Substituting back $u = x - \alpha$: $I = (x - \alpha) \cos \alpha + \sin \alpha \log |\sin (x - \alpha)| + c$ $I = x \cos \alpha - \alpha \cos \alpha + \sin \alpha \log |\sin (x - \alpha)| + c$ Comparing this with the given form $px - q \log |\sin (x - \alpha)| + c$: $p = \cos \alpha$ $-q = \sin \alpha \implies q = -\sin \alpha$ Therefore,$pq = (\cos \alpha)(-\sin \alpha) = -\sin \alpha \cos \alpha = -\frac{1}{2} (2 \sin \alpha \cos \alpha) = -\frac{1}{2} \sin 2\alpha$.

If $\int \frac{\sin x}{\sin (x-\alpha)} dx = px - q \log |\sin (x-\alpha)| + c$,then $pq =$ . . . . . . .

Similar Questions

Let $I=\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x$ and $J=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x$. Then,for an arbitrary constant $C$,the value of $J-I$ equals

Integrate the function $\frac{1}{1+\cot x}$.

$\int \frac{dx}{(1 + x^2)\sqrt{1 - x^2}} = $

Let a function $h(x)$ be defined as $h(x) = 0$, for all $x \ne 0$. Also $\int_{-\infty}^{\infty} h(x) \cdot f(x) \, dx = f(0)$, for every function $f(x)$. Then the value of the definite integral $\int_{-\infty}^{\infty} h'(x) \cdot \sin x \, dx$ is

$\int \frac{4 e^{x}+6 e^{-x}}{9 e^{x}-4 e^{-x}} d x=A x+B \log \left|9 e^{2 x}-4\right|+c$,then (Where $c$ is constant of integration)

Vedclass Test Series

Exam Paper Generator

Online Exam Module