int e^{4 x^2+8 x-4}(x+1) cos left(3 x^2+6 x-4 | vedclass

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

Question

(B) Let $I = \int e^{4 x^2+8 x-4}(x+1) \cos \left(3 x^2+6 x-4\right) d x$. Substitute $t = x^2 + 2x$,then $dt = (2x + 2) dx = 2(x+1) dx$,which implies

Vedclass · Accepted Answer

$\frac{e^{4 x^2+8 x-4}}{50}\left[4 \cos \left(3 x^2+6 x-4\right)+3 \sin \left(3 x^2+6 x-4\right)\right]+c$

Vedclass · Answer

(B) Let $I = \int e^{4 x^2+8 x-4}(x+1) \cos \left(3 x^2+6 x-4\right) d x$. Substitute $t = x^2 + 2x$,then $dt = (2x + 2) dx = 2(x+1) dx$,which implies $(x+1) dx = \frac{dt}{2}$. The integral becomes $I = \frac{1}{2} \int e^{4t-4} \cos(3t-4) dt$. Using the standard integral formula $\int e^{ax+k} \cos(bt+m) dt = \frac{e^{ax+k}}{a^2+b^2} [a \cos(bt+m) + b \sin(bt+m)] + C$,where $a=4$ and $b=3$: $I = \frac{1}{2} \cdot \frac{e^{4t-4}}{4^2+3^2} [4 \cos(3t-4) + 3 \sin(3t-4)] + C$. $I = \frac{e^{4t-4}}{2 \cdot 25} [4 \cos(3t-4) + 3 \sin(3t-4)] + C$. Substituting $t = x^2 + 2x$ back,we get $4t-4 = 4(x^2+2x)-4 = 4x^2+8x-4$ and $3t-4 = 3(x^2+2x)-4 = 3x^2+6x-4$. Thus,$I = \frac{e^{4x^2+8x-4}}{50} [4 \cos(3x^2+6x-4) + 3 \sin(3x^2+6x-4)] + C$.

$\int e^{4 x^2+8 x-4}(x+1) \cos \left(3 x^2+6 x-4\right) d x=$

Similar Questions

If $\int \frac{\cos 4x + 1}{\cot x - \tan x} dx = k \cos 4x + c$,then $k$ is

Let $I(x) = \int \sqrt{\frac{x+7}{x}} \, dx$ and $I(9) = 12 + 7 \log_e 7$. If $I(1) = \alpha + 7 \log_e(1 + 2\sqrt{2})$,then $\alpha^4$ is equal to $..........$.

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

If $\int \frac{1-(\cot x)^{2019}}{\tan x+(\cot x)^{2020}} dx = \frac{1}{n} \ln |(f(x))^n + (g(x))^n| + c$, then the value of $n[(f(x))^4 + (g(x))^4]_{x=\frac{\pi}{3}}$ is:

$\int \frac{dx}{\sqrt{(x-1)(x-2)}}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module