Question

$\int\sin^{3}2t\cos^{2}t\\ dt$

Explanation:

Step1: Use the double - angle formula

First, $\sin2t = 2\sin t\cos t$, so $\sin^{3}2t=(2\sin t\cos t)^{3}=8\sin^{3}t\cos^{3}t$. The integral becomes $\int8\sin^{3}t\cos^{3}t\cos^{2}t\mathrm{d}t = 8\int\sin^{3}t\cos^{5}t\mathrm{d}t$.

Step2: Rewrite $\sin^{3}t$

We know that $\sin^{3}t=\sin t(1 - \cos^{2}t)$. Then the integral is $8\int\sin t(1 - \cos^{2}t)\cos^{5}t\mathrm{d}t$.

Step3: Use substitution

Let $u = \cos t$, then $\mathrm{d}u=-\sin t\mathrm{d}t$. The integral becomes $- 8\int(1 - u^{2})u^{5}\mathrm{d}u=-8\int(u^{5}-u^{7})\mathrm{d}u$.

Step4: Integrate term - by - term

Integrating $u^{5}-u^{7}$ gives $\frac{u^{6}}{6}-\frac{u^{8}}{8}+C$.

Step5: Substitute back

Substituting $u = \cos t$ back, we get $-8(\frac{\cos^{6}t}{6}-\frac{\cos^{8}t}{8})+C=-\frac{4}{3}\cos^{6}t+\cos^{8}t + C$.

Answer:

$-\frac{4}{3}\cos^{6}t+\cos^{8}t + C$