Use any method to evaluate the integral.
-$$\int 5 x \sin ^ { 2 } x d x$$
A) $\frac { 5 } { 2 } x ^ { 2 } - \frac { 5 } { 4 } \sin 2 x - \frac { 5 } { 8 } \cos 2 x + C$
B) $\frac { 5 } { 4 } x ^ { 2 } - \frac { 5 } { 4 } \sin 2 x - \frac { 5 } { 8 } \cos 2 x + C$
C) $\frac { 5 } { 2 } x ^ { 2 } + \frac { \sin ^ { 3 } x } { 3 } + C$
D) $5 x ^ { 2 } - \frac { 5 } { 2 } \sin 2 x - \frac { 5 } { 4 } \cos 2 x + C$

Question

Quizplus · Accepted Answer

We use integration by parts, letting $u = \sin^2x$ and $dv = 5x\,dx$. Then, $du = 2\sin x \cos x\,dx$ and $v = \frac{5}{2}x^2$. Thus,
\begin{align*}
\int 5x \sin^2x\, dx &= \frac{5}{2}x^2 \sin^2x - \int \frac{5}{2}x^2\cdot 2\sin x \cos x\, dx \
&= \frac{5}{2}x^2 \sin^2x - 5\int x^2\sin x\cos x\, dx \
&= \frac{5}{2}x^2 \sin^2x - 5\left(-x^2\cos x + 2\int x\cos x \, dxight) \
&= \frac{5}{2}x^2 \sin^2x + 5x^2\cos x -10\int x\cos x\, dx \
&= \boxed{\frac{5}{4}x^2 - \frac{5}{4}\sin 2x - \frac{5}{8}\cos 2x + C}.
\end{align*}

Use Any Method to Evaluate the Integral ∫5xsin⁡2xdx\int 5 x \sin ^ { 2 } x d x∫5xsin2xdx

Use Any Method to Evaluate the Integral $\int 5 x \sin ^ { 2 } x d x$