Question

compute the given integral.
$intsin^{5}x dx=$
$+c$

Explanation:

Step1: Rewrite $\sin^{5}x$

$\sin^{5}x=\sin^{4}x\cdot\sin x = (\sin^{2}x)^{2}\cdot\sin x=(1 - \cos^{2}x)^{2}\cdot\sin x$

Step2: Use substitution

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

Step3: Expand the integrand

$(1 - u^{2})^{2}=1 - 2u^{2}+u^{4}$. Then $-\int(1 - 2u^{2}+u^{4})du=-\int du+2\int u^{2}du-\int u^{4}du$.

Step4: Integrate term - by - term

$-\int du=-u$, $2\int u^{2}du = 2\times\frac{u^{3}}{3}=\frac{2u^{3}}{3}$, $-\int u^{4}du=-\frac{u^{5}}{5}$.

Step5: Substitute back $u = \cos x$

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

Answer:

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