The integral $\int \sin ^ { 2 } x \cos ^ { 2 } x d x$ is

A) $- \frac { 4 x - \sin ( 4 x ) } { 32 } + C$
B) $\frac { 4 x + \sin ( 4 x ) } { 32 } + C$
C) $\frac { 4 x - \sin ( 4 x ) } { 32 } + C$
D) $\frac { x - 4 \sin ( 4 x ) } { 32 } + C$
E) $\frac { x + 4 \sin ( 4 x ) } { 32 } + C$

Question

The integral $\int \sin ^ { 2 } x \cos ^ { 2 } x d x$ is

A) $- \frac { 4 x - \sin ( 4 x ) } { 32 } + C$
B) $\frac { 4 x + \sin ( 4 x ) } { 32 } + C$
C) $\frac { 4 x - \sin ( 4 x ) } { 32 } + C$
D) $\frac { x - 4 \sin ( 4 x ) } { 32 } + C$
E) $\frac { x + 4 \sin ( 4 x ) } { 32 } + C$

Quizplus · Accepted Answer

The integral can be solved using the identity $\sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x)$ and then using the substitution $u = 2x$. The result is $\frac{4x - \sin(4x)}{32} + C$.

The Integral ∫sin⁡2xcos⁡2xdx\int \sin ^ { 2 } x \cos ^ { 2 } x d x∫sin2xcos2xdx

The Integral $\int \sin ^ { 2 } x \cos ^ { 2 } x d x$