The value of int {{e^{2x}}(2sin 3x + 3cos 3x) | vedclass

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

Question

(A) We use the standard integration formula: $\int {{e^{ax}}[af(x) + f'(x)]\,dx = {e^{ax}}f(x) + C}$.Here,$a = 2$ and $f(x) = \sin 3x$.Then,$f'(x) = \

Vedclass · Accepted Answer

${e^{2x}}\sin 3x + C$

Vedclass · Answer

(A) We use the standard integration formula: $\int {{e^{ax}}[af(x) + f'(x)]\,dx = {e^{ax}}f(x) + C}$.Here,$a = 2$ and $f(x) = \sin 3x$.Then,$f'(x) = \frac{d}{dx}(\sin 3x) = 3\cos 3x$.The given integral is $\int {{e^{2x}}(2\sin 3x + 3\cos 3x)\,dx}$.Comparing this with the formula,we have $f(x) = \sin 3x$ and $f'(x) = 3\cos 3x$.Therefore,the integral is equal to ${e^{2x}}\sin 3x + C$.

The value of $\int {{e^{2x}}(2\sin 3x + 3\cos 3x)\,dx} $ is

Similar Questions

$\int \frac{\log _e x}{\left(1+\log _e x\right)^2} d x=$

$\int \frac{(x - 3)e^x}{(x - 1)^3} \, dx$ is equal to

$\int_0^1 \frac{x e^x}{(2+x)^3} d x$ is equal to

$\int \frac{(x-3) e^x}{(x-1)^3} d x=$ . . . . . . $+C$.

$\int_1^e {\frac{{{e^x}}}{x}(1 + x\log x)\,dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module