To solve the integral

$\int \sin^3(x) \cos^5(x) \, dx,$

we can use a substitution technique. Here’s the step-by-step solution:

Step 1: Rewrite $\sin^3(x)$

Rewrite $\sin^3(x)$ as $\sin^2(x) \cdot \sin(x)$: $\int \sin^3(x) \cos^5(x) \, dx = \int \sin^2(x) \cos^5(x) \sin(x) \, dx.$

Step 2: Use a Pythagorean Identity

Since $\sin^2(x) = 1 - \cos^2(x)$, substitute $\sin^2(x)$ as follows: $= \int (1 - \cos^2(x)) \cos^5(x) \sin(x) \, dx.$

Step 3: Substitute $u = \cos(x)$

Let $u = \cos(x)$. Then, $du = -\sin(x) \, dx$, or equivalently, $-du = \sin(x) \, dx$.

Substitute $u$ and $-du$ into the integral: $= \int (1 - u^2) u^5 \cdot (-du).$

This simplifies to: $= \int (u^5 - u^7) \, (-du).$

Step 4: Distribute the Negative Sign

Rewrite as: $= \int (-u^5 + u^7) \, du.$

Step 5: Integrate Term by Term

Now integrate each term with respect to $u$: $= -\frac{u^6}{6} + \frac{u^8}{8} + C.$

Step 6: Substitute Back $u = \cos(x)$

Replace $u$ with $\cos(x)$: $= -\frac{\cos^6(x)}{6} + \frac{\cos^8(x)}{8} + C.$

Final Answer

$\int \sin^3(x) \cos^5(x) \, dx = -\frac{\cos^6(x)}{6} + \frac{\cos^8(x)}{8} + C.$

Would you like further details, or do you have any questions?

Related Questions

1. How would you integrate $\int \sin^5(x) \cos^3(x) \, dx$ using a similar technique?
2. What happens if both sine and cosine powers are even, e.g., $\int \sin^4(x) \cos^4(x) \, dx$?
3. Can we solve $\int \sin^3(x) \cos^3(x) \, dx$ using trigonometric identities?
4. How can we solve integrals involving higher powers, like $\int \sin^7(x) \cos^9(x) \, dx$?
5. What is the general form for integrating $\int \sin^m(x) \cos^n(x) \, dx$ for arbitrary $m$ and $n$?

Tip

For integrals involving products of $\sin$ and $\cos$ raised to odd powers, using trigonometric identities and substitution is often the most efficient method.