int frac{sin 4x}{sin x} , dx = (where C is a | vedclass

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

Question

(C) We have the integral $I = \int \frac{\sin 4x}{\sin x} \, dx$. Using the identity $\sin 4x = 2 \sin 2x \cos 2x = 4 \sin x \cos x \cos 2x$,we get: $

Vedclass · Accepted Answer

$\frac{2 \sin 3x}{3} + 2 \sin x + C$

Vedclass · Answer

(C) We have the integral $I = \int \frac{\sin 4x}{\sin x} \, dx$. Using the identity $\sin 4x = 2 \sin 2x \cos 2x = 4 \sin x \cos x \cos 2x$,we get: $I = \int \frac{4 \sin x \cos x \cos 2x}{\sin x} \, dx = 4 \int \cos x \cos 2x \, dx$. Using the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we have $2 \cos 2x \cos x = \cos(2x+x) + \cos(2x-x) = \cos 3x + \cos x$. Thus,$I = 2 \int (\cos 3x + \cos x) \, dx$. Integrating term by term,we get: $I = 2 \left( \frac{\sin 3x}{3} + \sin x \right) + C = \frac{2}{3} \sin 3x + 2 \sin x + C$.

$\int \frac{\sin 4x}{\sin x} \, dx =$ (where $C$ is a constant of integration.)

Similar Questions

Find the following integral: $\int (\sin x + \cos x) \, dx$

If $x \neq \frac{-3}{\sqrt{2}}$,then $\int \frac{x^2}{2 x^2+6 \sqrt{2} x+9} d x=$

If $f(x) = \int \frac{2-3 \sin^2 x}{1+\cos 2x} dx$ and $f\left(\frac{\pi}{4}\right) = 1$, then $f(0) =$

$\int x(\tan^2 x) dx =$

If $\int \frac{(x - 1)^2}{(x^2 + 1)^2} dx = \tan^{-1} (x) + g(x) + k$,then $g(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module