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\(\int e^{2 x}\left[\cos (3 x+4)+5 x^2\right] d x=\)
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Verified Answer
The correct answer is:
\(e^{2 x}\left[\frac{2}{13} \cos (3 x+4)+\frac{3}{13} \sin (3 x+4)+\frac{5 x^2}{2}-\frac{5 x}{2}+\frac{5}{4}\right]\)
\(\begin{aligned}
& \int e^{2 x}\left[\cos (3 x+4)+5 x^2\right] d x \\
& =\int e^{2 x} \cos (3 x+4) d x+5 \int e^{2 x} x^2 d x \\
& \left[\because \int e^{a x} \cos (b x+c) d x=\frac{e^{a x}}{a^2+b^2}(a \cos (b x+c)+b \sin (b x+c))+c\right]
\end{aligned}\)
\(\begin{aligned}
& =e^{2 x}\left[\frac{2}{13} \cos (3 x+4)+\frac{3}{13} \sin (3 x+4)\right]+\frac{5 x^2}{2} e^{2 x} \\
& =e^{2 x}\left(\frac{2}{13} \cos (3 x+4)+\frac{3}{13} \sin (3 x+4)\right) +\frac{5 x^2}{2} e^{2 x}-\frac{5}{2} x e^{2 x}+\frac{5}{2} \int e^{2 x} d x
\end{aligned}\)
\(\begin{aligned} & =e^{2 x}\left[\frac{2}{13} \cos (3 x+4)+\frac{3}{13} \sin (3 x+4)\right] +\frac{5}{2} x^2 e^{2 x}-\frac{5}{2} x e^{2 x}+\frac{5}{4} e^{2 x} \\ & =e^{2 x}\left[\frac{2}{13} \cos (3 x+4)+\frac{3}{13} \sin (3 x+4)+\frac{5 x^2}{2}-\frac{5}{2} x+\frac{5}{4}\right] +c\end{aligned}\)
Hence, option (a) is correct.
& \int e^{2 x}\left[\cos (3 x+4)+5 x^2\right] d x \\
& =\int e^{2 x} \cos (3 x+4) d x+5 \int e^{2 x} x^2 d x \\
& \left[\because \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))+c\right]
\end{aligned}\)
\(\begin{aligned}
& =e^{2 x}\left[\frac{2}{13} \cos (3 x+4)+\frac{3}{13} \sin (3 x+4)\right]+\frac{5 x^2}{2} e^{2 x} \\
& =e^{2 x}\left(\frac{2}{13} \cos (3 x+4)+\frac{3}{13} \sin (3 x+4)\right) +\frac{5 x^2}{2} e^{2 x}-\frac{5}{2} x e^{2 x}+\frac{5}{2} \int e^{2 x} d x
\end{aligned}\)
\(\begin{aligned} & =e^{2 x}\left[\frac{2}{13} \cos (3 x+4)+\frac{3}{13} \sin (3 x+4)\right] +\frac{5}{2} x^2 e^{2 x}-\frac{5}{2} x e^{2 x}+\frac{5}{4} e^{2 x} \\ & =e^{2 x}\left[\frac{2}{13} \cos (3 x+4)+\frac{3}{13} \sin (3 x+4)+\frac{5 x^2}{2}-\frac{5}{2} x+\frac{5}{4}\right] +c\end{aligned}\)
Hence, option (a) is correct.
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