Question

$int 6sin(pi+x)sin(2pi-x)dx$

Explanation:

Step1: Simplify trigonometric terms

Use identities: $\sin(\pi+x)=-\sin x$, $\sin(2\pi-x)=-\sin x$
$\int 6\sin(\pi+x)\sin(2\pi-x)dx = \int 6(-\sin x)(-\sin x)dx = \int 6\sin^2x dx$

Step2: Use power-reduction identity

$\sin^2x=\frac{1-\cos2x}{2}$
$\int 6\sin^2x dx = \int 6\cdot\frac{1-\cos2x}{2}dx = \int 3(1-\cos2x)dx$

Step3: Split the integral

Separate the terms for integration
$\int 3(1-\cos2x)dx = 3\int 1dx - 3\int \cos2x dx$

Step4: Integrate each term

Integrate $\int 1dx=x$, $\int \cos2x dx=\frac{1}{2}\sin2x$
$3x - 3\cdot\frac{1}{2}\sin2x + C$

Answer:

$3x - \frac{3}{2}\sin2x + C$