Find an anti-derivative (or integral) of the | vedclass

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

Question

(A) The anti-derivative of $\cos 3x$ is a function $F(x)$ such that $F'(x) = \cos 3x$. We know that the derivative of $\sin 3x$ with respect to $x$ is

Vedclass · Accepted Answer

$\frac{1}{3} \sin 3x$

Vedclass · Answer

(A) The anti-derivative of $\cos 3x$ is a function $F(x)$ such that $F'(x) = \cos 3x$. We know that the derivative of $\sin 3x$ with respect to $x$ is: $\frac{d}{dx}(\sin 3x) = 3 \cos 3x$ Dividing both sides by $3$,we get: $\frac{1}{3} \frac{d}{dx}(\sin 3x) = \cos 3x$ Using the linearity of the derivative,we can write: $\frac{d}{dx} \left( \frac{1}{3} \sin 3x \right) = \cos 3x$ Thus,the anti-derivative of $\cos 3x$ is $\frac{1}{3} \sin 3x$.

Find an anti-derivative (or integral) of the function $\cos 3x$ by the method of inspection.

Similar Questions

$\int \frac{1+\cos 4 x}{\cot x-\tan x} d x$ is equal to

For $-\frac{\pi}{2} < x < \frac{\pi}{2}$,evaluate the integral $\int \tan^{-1} \left( \sqrt{\frac{1 - \sin x}{1 + \sin x}} \right) dx$ (where $C$ is a constant of integration).

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

Integrate the function: $\frac{1}{\sqrt{x+a}+\sqrt{x+b}}$

$\text{If } \int \frac{3x+2}{4x^2+4x+5} dx = A \log(4x^2+4x+5) + B \tan^{-1}\left(x+\frac{1}{2}\right) + C, \text{ then } (A, B) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module