Question

evaluate the following integral.
int 5sin^{3}xcos^{5}x dx
int 5sin^{3}xcos^{5}x dx=square

Explanation:

Step1: Rewrite the integrand

We know that $\sin^{3}x=\sin x\cdot\sin^{2}x=\sin x(1 - \cos^{2}x)$. So the integral $\int5\sin^{3}x\cos^{5}x dx=5\int\sin x(1 - \cos^{2}x)\cos^{5}x dx$.

Step2: Use substitution

Let $u = \cos x$, then $du=-\sin xdx$. The integral becomes $- 5\int(1 - u^{2})u^{5}du=-5\int(u^{5}-u^{7})du$.

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 $-5(\frac{u^{6}}{6}-\frac{u^{8}}{8})+C$.

Step4: Substitute back $u=\cos x$

The result is $-5(\frac{\cos^{6}x}{6}-\frac{\cos^{8}x}{8})+C=-\frac{5}{6}\cos^{6}x+\frac{5}{8}\cos^{8}x+C$.

Answer:

$-\frac{5}{6}\cos^{6}x+\frac{5}{8}\cos^{8}x+C$