https://file.aiquickdraw.com/mathbot/file/1730995097868f7l51e3x.jpg

To solve the integral:

\[
\int \frac{1 + \sin^2 x}{1 - \sin^2 x} \, dx
\]

we can simplify the integrand by using trigonometric identities. Notice that:

\[
1 - \sin^2 x = \cos^2 x
\]

Thus, we can rewrite the integral as:

\[
\int \frac{1 + \sin^2 x}{\cos^2 x} \, dx
\]

Next, separate the integrand:

\[
= \int \left( \frac{1}{\cos^2 x} + \frac{\sin^2 x}{\cos^2 x} \right) \, dx
\]

\[
= \int \sec^2 x \, dx + \int \tan^2 x \, dx
\]

Now, we can solve each part separately.

1. **First Integral:** \(\int \sec^2 x \, dx\)
   \[
   \int \sec^2 x \, dx = \tan x
   \]

2. **Second Integral:** \(\int \tan^2 x \, dx\)
   
   Recall that \(\tan^2 x = \sec^2 x - 1\), so we have:
   \[
   \int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx
   \]
   \[
   = \int \sec^2 x \, dx - \int 1 \, dx
   \]
   \[
   = \tan x - x
   \]

Putting it all together:

\[
\int \frac{1 + \sin^2 x}{1 - \sin^2 x} \, dx = \tan x + (\tan x - x) + C
\]

Simplifying, we get:

\[
= 2 \tan x - x + C
\]

Thus, the answer is:

\[
\int \frac{1 + \sin^2 x}{1 - \sin^2 x} \, dx = 2 \tan x - x + C
\]

Would you like more details on any part of the solution, or have any questions?

Here are 5 related questions you might find interesting:
1. How would the solution change if the integrand were \(\frac{1 - \sin^2 x}{1 + \sin^2 x}\)?
2. What is the result of integrating \(\frac{\sin^2 x}{1 - \sin^2 x}\) instead?
3. Can you solve \(\int \frac{1 + \cos^2 x}{1 - \cos^2 x} \, dx\)?
4. How does knowing the identity \(\sin^2 x + \cos^2 x = 1\) help in solving integrals like this?
5. What are some practical applications of integrals involving trigonometric functions?

**Tip:** Familiarity with trigonometric identities can greatly simplify solving integrals like these!

Evaluate the integral \( \int \frac{1 + \sin^2 x}{1 - \sin^2 x} \, dx \)

Learn how to evaluate the integral \( \int \frac{1 + \sin^2 x}{1 - \sin^2 x} \, dx \) using trigonometric identities and integral properties. This calculus problem explores trigonometric integration techniques and is suitable for students in undergraduate calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation