To evaluate the integral

$\int \sin^6(x) \cos^3(x) \, dx,$

we can use a substitution approach. Since we have a power of $\cos(x)$, we can use the identity:

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

to express $\sin^6(x)$ in terms of $\cos(x)$, or alternatively use a substitution if we simplify with trigonometric identities.

Here’s a step-by-step approach:

Step 1: Rewrite the Integral

Since $\cos^3(x)$ is present, we can let $u = \sin(x)$. Then, $du = \cos(x) \, dx$, and $\cos(x) \, dx = du$.

Notice that we have $\cos^3(x) \, dx$ in the integral, which can be split as: $\cos^2(x) \cos(x) \, dx = (1 - \sin^2(x)) \cos(x) \, dx.$

Letting $u = \sin(x)$, then $\sin^6(x) = u^6$, $\cos(x) \, dx = du$, and $\cos^2(x) = 1 - u^2$.

Step 2: Substitute and Simplify

The integral now becomes: $\int u^6 (1 - u^2) u \, du.$ Expanding this gives: $\int (u^6 - u^8) u \, du = \int u^9 - u^{11} \, du.$

Step 3: Integrate Term by Term

Now, integrate each term: $\int u^9 \, du - \int u^{11} \, du = \frac{u^{10}}{10} - \frac{u^{12}}{12} + C.$

Step 4: Substitute Back

Since $u = \sin(x)$, we substitute back: $\frac{\sin^{10}(x)}{10} - \frac{\sin^{12}(x)}{12} + C.$

Final Answer:

$\int \sin^6(x) \cos^3(x) \, dx = \frac{\sin^{10}(x)}{10} - \frac{\sin^{12}(x)}{12} + C.$

Would you like more details on any specific step or have any questions?

Here are some related questions to explore further:

1. How would you approach this integral if the powers of $\sin(x)$ and $\cos(x)$ were different?
2. What other trigonometric identities could simplify integrals involving powers of $\sin(x)$ and $\cos(x)$?
3. How does substitution work for integrals involving trigonometric functions?
4. Could integration by parts be useful in similar trigonometric integrals?
5. How would the approach change if the integral involved $\tan(x)$ and $\sec(x)$?

Tip: When working with trigonometric integrals, always consider using identities or substitutions that reduce the powers of the trigonometric functions. This often simplifies the integration process.