Evaluate the integral by making the given substitution. (Use $C$ for the constant of integration.) $\int \cos^8(\theta) \sin(\theta) \, d\theta, \quad u = \cos(\theta)$

Neetesh Kumar | December 8, 2024

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This is the solution to Math 132
Assignment: 5.5 Question Number 2
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Step-by-step solution:

We aim to evaluate $\int \cos^8(\theta) \sin(\theta) \, d\theta$ using the substitution $u = \cos(\theta)$.

Step 1: Substitution

Let $u = \cos(\theta)$. Differentiate to find $du$:

$\frac{du}{d\theta} = -\sin(\theta) \quad \implies \quad du = -\sin(\theta) \, d\theta$

This allows us to rewrite $\sin(\theta) \, d\theta$ as:

$\sin(\theta) \, d\theta = -du$

Also, $\cos^8(\theta) = u^8$ by substitution.

Substitute into the integral:

$\int \cos^8(\theta) \sin(\theta) \, d\theta = \int u^8 \cdot (-du)$

Simplify:

$\int \cos^8(\theta) \sin(\theta) \, d\theta = -\int u^8 \, du$

Step 2: Evaluate the simplified integral

The integral of $u^8$ is:

$\int u^8 \, du = \frac{u^9}{9}$

Substitute this result:

$-\int u^8 \, du = -\frac{u^9}{9}$

Step 3: Substitute back $u = \cos(\theta)$

Substitute back $u = \cos(\theta)$ to return to the original variable:

$-\frac{u^9}{9} = -\frac{\cos^9(\theta)}{9}$

Final Answer:

$\int \cos^8(\theta) \sin(\theta) \, d\theta = \boxed{-\frac{\cos^9(\theta)}{9} + C}$

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