Question

evaluate the integral by making the given substitution. (use c for the constant of integration.)
int sin ^ { 3 } ( \theta ) cos ( \theta ) d \theta , quad u = sin ( \theta )

Explanation:

Step1: Differentiate \(u\)

Given \(u = \sin(\theta)\), then \(du=\cos(\theta)d\theta\).

Step2: Substitute into the integral

The integral \(\int\sin^{3}(\theta)\cos(\theta)d\theta\) becomes \(\int u^{3}du\).

Step3: Integrate \(u^{3}\)

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

Step4: Substitute back \(u=\sin(\theta)\)

We get \(\frac{\sin^{4}(\theta)}{4}+C\).

Answer:

\(\frac{\sin^{4}(\theta)}{4}+C\)