Question

Evaluate the integral by making the given substitution. (Use C for the constant of integration.)

cos7(𝜃) sin(𝜃) d𝜃, u = cos(𝜃)

Answer 1

To evaluate the integral \(\int \cos^7(\theta) \sin(\theta) \, d\theta\) using the substitution \(u = \cos(\theta)\), we can follow these steps:

1. **Substitution**:
We have
\[
u = \cos(\theta) \implies du = -\sin(\theta) \, d\theta \quad \text{or} \quad d\theta = -\frac{du}{\sin(\theta)}.
\]
Since \(\sin(\theta) = \sqrt{1 - u^2}\) (using the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\)), we can express \(\sin(\theta)\) in terms of \(u\).

2. **Rewrite the integral**:
The sine term can be used to rewrite \(d\theta\):
\[
d\theta = -\frac{du}{\sqrt{1 - u^2}}.
\]

Therefore, the integral becomes:
\[
\int \cos^7(\theta) \sin(\theta) \, d\theta = \int u^7 \sin(\theta) \, d\theta = \int u^7 \sin(\theta) \left(-\frac{du}{\sin(\theta)}\right).
\]
As \(\sin(\theta)\) cancels, we have:
\[
= - \int u^7 \, du.
\]

3. **Integrate**:
Now, we can integrate:
\[
-\int u^7 \, du = -\left(\frac{u^8}{8}\right) + C = -\frac{u^8}{8} + C.
\]

4. **Back-substituting**:
Recall that \(u = \cos(\theta)\), so substituting back gives us:
\[
-\frac{(\cos(\theta))^8}{8} + C = -\frac{\cos^8(\theta)}{8} + C.
\]

Therefore, the final answer is:
\[
\int \cos^7(\theta) \sin(\theta) \, d\theta = -\frac{\cos^8(\theta)}{8} + C.
\]