int e^{3x} sin(4x-5) dx = . . . . . . + C | vedclass

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

Question

(C) We use the standard formula: $\int e^{ax} \sin(bx+c) dx = \frac{e^{ax}}{a^2+b^2} [a \sin(bx+c) - b \cos(bx+c)] + C$. Here,$a = 3$,$b = 4$,and the

Vedclass · Accepted Answer

$\frac{e^{3x}}{25}[3 \sin(4x-5) - 4 \cos(4x-5)]$

Vedclass · Answer

(C) We use the standard formula: $\int e^{ax} \sin(bx+c) dx = \frac{e^{ax}}{a^2+b^2} [a \sin(bx+c) - b \cos(bx+c)] + C$. Here,$a = 3$,$b = 4$,and the constant term is $-5$. Substituting these values into the formula: $\int e^{3x} \sin(4x-5) dx = \frac{e^{3x}}{3^2 + 4^2} [3 \sin(4x-5) - 4 \cos(4x-5)] + C$. $= \frac{e^{3x}}{9 + 16} [3 \sin(4x-5) - 4 \cos(4x-5)] + C$. $= \frac{e^{3x}}{25} [3 \sin(4x-5) - 4 \cos(4x-5)] + C$. Thus,the correct option is $C$.

$\int e^{3x} \sin(4x-5) dx = $ . . . . . . $+ C$

Similar Questions

Evaluate the integral: $\int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx$

Assertion $(A)$: If $I_n = \int \cot^n x \, dx$,then $I_6 + I_4 = \frac{-\cot^5 x}{5}$.
Reason $(R)$: $\int \cot^n x \, dx = \frac{-\cot^{n-1} x}{n-1} - \int \cot^{n-2} x \, dx$.

$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x=$

$\int \frac{(1-4 \sin^2 x) \cos x}{\cos (3x+2)} dx =$

$\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module