$\text { Find } \int \sin 5 x \cos 10 x d x \text {. }$
A) $\int \sin 5 x \cos 10 x d x = \frac { 1 } { 30 } ( 3 \cos 5 x - \cos 15 x ) + C$
B) $\int \sin 5 x \cos 10 x d x = \frac { 1 } { 15 } ( 3 \cos 14 x + 3 \cos 5 x ) + C$
C) $\int \sin 5 x \cos 10 x d x = \frac { 1 } { 31 } ( \cos 15 x - 3 \cos 5 x ) + C$
D) $\int \sin 5 x \cos 10 x d x = \frac { 1 } { 30 } ( 4 \cos 5 x + \cos 14 x ) + C$
E) $\int \sin 5 x \cos 10 x d x = \frac { 1 } { 31 } ( 3 \cos 5 x - \cos 15 x ) + C$

Question

$\text { Find } \int \sin 5 x \cos 10 x d x \text {. }$ 
A) $\int \sin 5 x \cos 10 x d x = \frac { 1 } { 30 } ( 3 \cos 5 x - \cos 15 x ) + C$
B) $\int \sin 5 x \cos 10 x d x = \frac { 1 } { 15 } ( 3 \cos 14 x + 3 \cos 5 x ) + C$
C) $\int \sin 5 x \cos 10 x d x = \frac { 1 } { 31 } ( \cos 15 x - 3 \cos 5 x ) + C$
D) $\int \sin 5 x \cos 10 x d x = \frac { 1 } { 30 } ( 4 \cos 5 x + \cos 14 x ) + C$
E) $\int \sin 5 x \cos 10 x d x = \frac { 1 } { 31 } ( 3 \cos 5 x - \cos 15 x ) + C$

Quizplus · Accepted Answer

To solve the integral $\int \sin 5x \cos 10x dx$, we can use the trigonometric identity for the product of sine and cosine, which is $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$. Applying this identity with $A = 5x$ and $B = 10x$, we get $\sin 5x \cos 10x = \frac{1}{2}[\sin(15x) + \sin(-5x)] = \frac{1}{2}[\sin(15x) - \sin(5x)]$. Integrating this expression yields $\frac{1}{2}[-\frac{1}{15}\cos(15x) + \frac{1}{5}\cos(5x)] + C$, which simplifies to $\frac{1}{30}(3\cos(5x) - \cos(15x)) + C$.

Find ∫sin⁡5xcos⁡10xdx. \text { Find } \int \sin 5 x \cos 10 x d x \text {. } Find ∫sin5xcos10xdx.

$\text { Find } \int \sin 5 x \cos 10 x d x \text {. }$