Use the Table of Integrals to evaluate the integral. $\int e^{4 x} \sin 2 x d x$
A) $-\frac{1}{5} e^{4 x} \sin 2 x+\frac{1}{10} e^{4 x} \cos 2 x+C$
B) $\frac{1}{5} e^{4 x} \sin 2 x+\frac{1}{10} e^{4 x} \cos 2 x+C$
C) $\frac{1}{5} e^{4 x} \sin 2 x-\frac{3}{10} e^{4 x} \cos 2 x+C$
D) $-\frac{1}{5} e^{4 x} \sin 2 x-\frac{1}{10} e^{4 x} \cos 2 x+C$
E) $\frac{1}{5} e^{4 x} \sin 2 x-\frac{1}{10} e^{4 x} \cos 2 x+C$

Question

Quizplus · Accepted Answer

Using integration by parts with $u=\sin 2x$ and $dv=e^{4x}dx$, we get $du=2\cos 2x dx$ and $v=\frac{1}{4}e^{4x}$. Thus,

$$\int e^{4 x} \sin 2 x d x=-\frac{1}{4}e^{4x}\cos 2x+\frac{1}{2}\int e^{4x}\cos 2x dx$$

Using integration by parts again with $u=\cos 2x$ and $dv=e^{4x}dx$, we get $du=-2\sin 2x dx$ and $v=\frac{1}{4}e^{4x}$.
Thus,

$$\int e^{4 x} \sin 2 x d x=-\frac{1}{4}e^{4x}\cos 2x+\frac{1}{2}\left(\frac{1}{4}e^{4x}\cos 2x+\frac{1}{2}\int e^{4x}\sin 2x dx \right)$$

Solving the integral $\int e^{4x}\sin 2x dx$ using integration by parts again, we get

$$\int e^{4 x} \sin 2 x d x=-\frac{1}{4}e^{4x}\cos 2x+\frac{1}{8}e^{4x}\sin 2 x-\frac{1}{16}e^{4x}+C$$

Simplifying the expression gives us

$$\int e^{4 x} \sin 2 x d x=\frac{1}{5} e^{4 x} \sin 2 x-\frac{1}{10} e^{4 x} \cos 2 x+C$$

Therefore, the answer is E.

Use the Table of Integrals to Evaluate the Integral ∫e4xsin⁡2xdx\int e^{4 x} \sin 2 x d x∫e4xsin2xdx

Use the Table of Integrals to Evaluate the Integral $\int e^{4 x} \sin 2 x d x$