Question

evaluate the following integral.
int 6sin^{3}xcos^{2}x dx
int 6sin^{3}xcos^{2}x dx=square

Explanation:

Step1: Rewrite $\sin^{3}x$

We know that $\sin^{3}x=\sin x\cdot\sin^{2}x=\sin x(1 - \cos^{2}x)$. So the integral becomes $\int6\sin x(1 - \cos^{2}x)\cos^{2}x dx$.

Step2: Use substitution

Let $u = \cos x$, then $du=-\sin xdx$. The integral $\int6\sin x(1 - \cos^{2}x)\cos^{2}x dx=- 6\int(1 - u^{2})u^{2}du$.

Step3: Expand the integrand

Expand $(1 - u^{2})u^{2}=u^{2}-u^{4}$. So the integral is $-6\int(u^{2}-u^{4})du$.

Step4: Integrate term - by - term

Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $-6(\frac{u^{3}}{3}-\frac{u^{5}}{5})+C=- 2u^{3}+\frac{6}{5}u^{5}+C$.

Step5: Substitute back $u = \cos x$

The result is $-2\cos^{3}x+\frac{6}{5}\cos^{5}x + C$.

Answer:

$-2\cos^{3}x+\frac{6}{5}\cos^{5}x + C$