Question

evaluate the following integral.
int 12sin^{3}xcos^{2}x dx
int 12sin^{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 $\int12\sin x(1 - \cos^{2}x)\cos^{2}x dx$. Let $u = \cos x$, then $du=-\sin xdx$ and $\sin xdx=-du$.

Step2: Substitute $u$ into the integral

The integral $\int12\sin x(1 - \cos^{2}x)\cos^{2}x dx$ becomes $- 12\int(1 - u^{2})u^{2}du=-12\int(u^{2}-u^{4})du$.

Step3: Integrate term - by - term

Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)$, we have $-12(\frac{u^{3}}{3}-\frac{u^{5}}{5})+C=-4u^{3}+\frac{12}{5}u^{5}+C$.

Step4: Substitute back $u = \cos x$

The result is $-4\cos^{3}x+\frac{12}{5}\cos^{5}x+C$.

Answer:

$-4\cos^{3}x+\frac{12}{5}\cos^{5}x+C$