int e^{2 x}left[cos (3 x+4)+5 x^2right] d x= | vedclass

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Question

(A) We need to evaluate the integral $I = \int e^{2 x}\left[\cos (3 x+4)+5 x^2\right] d x$. Split the integral into two parts: $I = \int e^{2 x} \cos

Vedclass · Accepted Answer

$e^{2 x}\left[\frac{2}{13} \cos (3 x+4)+\frac{3}{13} \sin (3 x+4)+\frac{5 x^2}{2}-\frac{5 x}{2}+\frac{5}{4}\right]+c$

Vedclass · Answer

(A) We need to evaluate the integral $I = \int e^{2 x}\left[\cos (3 x+4)+5 x^2\right] d x$. Split the integral into two parts: $I = \int e^{2 x} \cos (3 x+4) d x + 5 \int e^{2 x} x^2 d x$. Using the formula $\int e^{a x} \cos (b x+c) d x = \frac{e^{a x}}{a^2+b^2}(a \cos (b x+c)+b \sin (b x+c))$,we get: $\int e^{2 x} \cos (3 x+4) d x = \frac{e^{2 x}}{2^2+3^2}(2 \cos (3 x+4)+3 \sin (3 x+4)) = \frac{e^{2 x}}{13}(2 \cos (3 x+4)+3 \sin (3 x+4))$. Now,evaluate $5 \int e^{2 x} x^2 d x$ using integration by parts $\int u v dx = u \int v dx - \int (u' \int v dx) dx$: $5 \int x^2 e^{2 x} d x = 5 \left[ x^2 \frac{e^{2 x}}{2} - \int 2x \frac{e^{2 x}}{2} d x \right] = \frac{5 x^2 e^{2 x}}{2} - 5 \int x e^{2 x} d x$. Applying integration by parts again: $5 \int x e^{2 x} d x = 5 \left[ x \frac{e^{2 x}}{2} - \int 1 \cdot \frac{e^{2 x}}{2} d x \right] = \frac{5 x e^{2 x}}{2} - \frac{5}{2} \cdot \frac{e^{2 x}}{2} = \frac{5 x e^{2 x}}{2} - \frac{5 e^{2 x}}{4}$. Combining all parts: $I = e^{2 x} \left[ \frac{2}{13} \cos (3 x+4) + \frac{3}{13} \sin (3 x+4) \right] + \frac{5 x^2 e^{2 x}}{2} - \left( \frac{5 x e^{2 x}}{2} - \frac{5 e^{2 x}}{4} \right) + c$. $I = e^{2 x} \left[ \frac{2}{13} \cos (3 x+4) + \frac{3}{13} \sin (3 x+4) + \frac{5 x^2}{2} - \frac{5 x}{2} + \frac{5}{4} \right] + c$. Thus,option $(A)$ is correct.

$\int e^{2 x}\left[\cos (3 x+4)+5 x^2\right] d x=$

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