The value of int sin 4x cos 3x , dx is | vedclass

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

Question

(A) Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:$\sin 4x \cos 3x = \frac{1}{2} [\sin(4x+3x) + \sin(4x-3x)] = \frac{1}{2

Vedclass · Accepted Answer

$-\frac{1}{14} \cos 7x - \frac{1}{2} \cos x + c$

Vedclass · Answer

(A) Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:$\sin 4x \cos 3x = \frac{1}{2} [\sin(4x+3x) + \sin(4x-3x)] = \frac{1}{2} [\sin 7x + \sin x]$Now, integrate the expression:$\int \sin 4x \cos 3x \, dx = \frac{1}{2} \int (\sin 7x + \sin x) \, dx$$= \frac{1}{2} [-\frac{\cos 7x}{7} - \cos x] + c$$= -\frac{1}{14} \cos 7x - \frac{1}{2} \cos x + c$

The value of $\int \sin 4x \cos 3x \, dx$ is

Similar Questions

Integrate the function: $\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}$

The value of $ \int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} dx $ is equal to

$\int \frac{dx}{\sin x+\sqrt{3} \cos x}$ equals, where $c$ is an arbitrary constant.

The value of $\int \cot x \, dx$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module