The integral $\int \sin x \cos ( 2 x ) d x$ is

A) $\frac { 3 \cos x + \cos ( 3 x ) } { 6 } + C$
B) $\frac { - 3 \cos x - \cos ( 3 x ) } { 6 } + C$
C) $\frac { 3 \cos x - \cos ( 3 x ) } { 6 } + C$
D) $\frac { 3 \sin x - \sin ( 3 x ) } { 6 } + C$
E) $\frac { 3 \sin x + \sin ( 3 x ) } { 6 } + C$

Question

The integral $\int \sin x \cos ( 2 x ) d x$ is

A) $\frac { 3 \cos x + \cos ( 3 x ) } { 6 } + C$
B) $\frac { - 3 \cos x - \cos ( 3 x ) } { 6 } + C$
C) $\frac { 3 \cos x - \cos ( 3 x ) } { 6 } + C$
D) $\frac { 3 \sin x - \sin ( 3 x ) } { 6 } + C$
E) $\frac { 3 \sin x + \sin ( 3 x ) } { 6 } + C$

Quizplus · Accepted Answer

The integral can be solved using the product-to-sum identities: $\sin x \cos(2x) = \frac{1}{2}[\sin(3x) + \sin(-x)]$. Integrating gives $\frac{1}{6}[-\cos(3x) + 3\cos(x)] + C$.

The Integral ∫sin⁡xcos⁡(2x)dx\int \sin x \cos ( 2 x ) d x∫sinxcos(2x)dx

The Integral $\int \sin x \cos ( 2 x ) d x$