To find the integral of $\sin^5(x)$, we can use a reduction formula or trigonometric identities. Here's a step-by-step solution using the reduction method:

Step 1: Express the integral

We want to compute: $\int \sin^5(x) \, dx$

Step 2: Use a trigonometric identity

We can simplify powers of sine functions using the identity for even powers of sine. First, break $\sin^5(x)$ into $\sin^4(x) \cdot \sin(x)$, and then use the identity for $\sin^2(x)$ to reduce the power:

$\sin^5(x) = \sin^4(x) \cdot \sin(x) = (1 - \cos^2(x))^2 \cdot \sin(x)$

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

Let $u = \cos(x)$, which implies $du = -\sin(x) \, dx$. The integral becomes:

$\int (1 - u^2)^2 (-du)$

Simplify this expression:

$- \int (1 - u^2)^2 \, du$

Step 4: Expand and integrate

Now, expand $(1 - u^2)^2$:

$(1 - u^2)^2 = 1 - 2u^2 + u^4$

Thus, the integral becomes:

$- \int (1 - 2u^2 + u^4) \, du$

Now, integrate term by term:

$- \left( \int 1 \, du - 2 \int u^2 \, du + \int u^4 \, du \right)$

This simplifies to:

$- \left( u - \frac{2u^3}{3} + \frac{u^5}{5} \right) + C$

Step 5: Substitute back $u = \cos(x)$

Finally, substitute back $u = \cos(x)$:

$- \left( \cos(x) - \frac{2\cos^3(x)}{3} + \frac{\cos^5(x)}{5} \right) + C$

Final Answer:

$\int \sin^5(x) \, dx = - \cos(x) + \frac{2\cos^3(x)}{3} - \frac{\cos^5(x)}{5} + C$

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Would you like more details or have any questions about this solution?

Further Related Questions:

1. How can the reduction formula for trigonometric powers be applied to other functions like $\cos^5(x)$?
2. What other methods exist for solving integrals involving trigonometric powers?
3. How would you handle the integral of $\sin^7(x)$?
4. What is the general formula for the integral of $\sin^n(x)$?
5. Can you compute integrals involving higher even powers of sine, like $\sin^{10}(x)$?

Tip:

To handle integrals involving higher powers of sine or cosine functions, it's often helpful to use reduction formulas or identities to break the powers down into simpler integrals.