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

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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,$I = \frac{e^{4x^2+8x-4}}{50} [4 \cos(3x^2+6x-4) + 3 \sin(3x^2+6x-4)] + C$.

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

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