∫ sin 𝑥

(cos 𝑥)5 𝑑𝑥 use substitution

To integrate \(\int \frac{\sin x}{(\cos x)^5} \, dx\) using substitution, follow these steps:

1. **Substitute a new variable:**
   Let \(u = \cos x\). Then we need to find \(du\).

2. **Find \(du\):**
   \[
   u = \cos x \implies \frac{du}{dx} = -\sin x
   \]
   Therefore,
   \[
   du = -\sin x \, dx \implies \sin x \, dx = -du
   \]

3. **Rewrite the integral:**
   Substitute \(u\) and \(dx\) into the integral:
   \[
   \int \frac{\sin x}{(\cos x)^5} \, dx = \int \frac{\sin x \, dx}{u^5}
   \]
   Using \(\sin x \, dx = -du\), we get:
   \[
   \int \frac{\sin x}{(\cos x)^5} \, dx = \int \frac{-du}{u^5} = -\int u^{-5} \, du
   \]

4. **Integrate with respect to \(u\):**
   \[
   -\int u^{-5} \, du = -\frac{u^{-4}}{-4} + C = \frac{u^{-4}}{4} + C
   \]
   Simplifying, we get:
   \[
   \frac{1}{4u^4} + C
   \]

5. **Substitute back in terms of \(x\):**
   Recall that \(u = \cos x\), so:
   \[
   \frac{1}{4u^4} + C = \frac{1}{4(\cos x)^4} + C = \frac{1}{4 \cos^4 x} + C
   \]

So, the final answer is:
\[
\int \frac{\sin x}{(\cos x)^5} \, dx = \frac{1}{4 \cos^4 x} + C
\]

Would you like more details or have any questions?

Here are 5 questions you might ask next:
1. How do you solve integrals involving products of trigonometric functions?
2. Can you explain how to use trigonometric identities to simplify integrals?
3. How do you solve integrals using partial fractions?
4. What is the method for solving integrals with exponential functions?
5. How do you evaluate definite integrals with trigonometric substitutions?

**Tip:** When using substitution, always make sure to correctly replace both the integrand and the differential to ensure a smooth transformation and accurate integration.

Learn how to integrate sin(x) / (cos(x))^5 using trigonometric substitution with a step-by-step solution. This problem involves advanced integration techniques and is suitable for advanced high school and college students.

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