The given integral is:

$I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\cos(x)}{\sin^3(x)} \, dx$

To solve this, we can proceed with the following steps:

Step 1: Substitution

We notice that the numerator is $\cos(x)$ and the denominator contains $\sin(x)$. This suggests a substitution based on the sine function.

Let: $u = \sin(x)$ Then: $du = \cos(x) \, dx$

This substitution transforms the integral as follows:

• When $x = \frac{\pi}{4}$, $u = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
• When $x = \frac{\pi}{3}$, $u = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

The integral becomes: $I = \int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}} \frac{du}{u^3}$

Step 2: Simplify the integral

The integral $\int \frac{1}{u^3} \, du$ can be rewritten as: $\int u^{-3} \, du$

Using the power rule for integration: $\int u^{-3} \, du = \frac{u^{-2}}{-2} = -\frac{1}{2u^2}$

Step 3: Evaluate the limits

Now, evaluate the expression at the limits of integration: $I = -\frac{1}{2} \left[ \frac{1}{u^2} \right]_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}$ Substituting the limits: $I = -\frac{1}{2} \left( \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2} - \frac{1}{\left(\frac{\sqrt{2}}{2}\right)^2} \right)$ Simplify the expressions: $I = -\frac{1}{2} \left( \frac{1}{\frac{3}{4}} - \frac{1}{\frac{2}{4}} \right)$ $I = -\frac{1}{2} \left( \frac{4}{3} - 2 \right)$ $I = -\frac{1}{2} \left( \frac{4}{3} - \frac{6}{3} \right)$ $I = -\frac{1}{2} \left( \frac{-2}{3} \right)$ $I = \frac{1}{3}$

Thus, the value of the integral is: $\boxed{\frac{1}{3}}$

Would you like further details or have any questions?

Here are 5 related questions to explore further:

1. How would you solve this integral if the powers of sine were different?
2. What if the limits of integration were different?
3. How does substitution work for more complex trigonometric integrals?
4. What are the techniques for integrating powers of sine and cosine?
5. Can this integral be solved using other methods, like integration by parts?

Tip: Always check if a simple substitution can simplify trigonometric integrals involving powers of sine and cosine.