To solve the integral:

$\int \frac{\sin 2x}{1 + \sin^2 x} \, dx$

Step-by-Step Solution

1. Rewrite $\sin 2x$: Recall that $\sin 2x = 2 \sin x \cos x$. Substitute this identity: $\int \frac{2 \sin x \cos x}{1 + \sin^2 x} \, dx$

2. Substitute for $u$: Let $u = \sin x$. Then, $du = \cos x \, dx$.

   Substitute $u$ and $du$ into the integral: $\int \frac{2u \cos x}{1 + u^2} \cdot \frac{dx}{\cos x}$ Simplifying, the $\cos x$ terms cancel, leaving: $\int \frac{2u}{1 + u^2} \, du$

3. Simplify the Integral: Now, we have: $\int \frac{2u}{1 + u^2} \, du$ This can be separated as: $2 \int \frac{u}{1 + u^2} \, du$

4. Integrate: Notice that $\frac{u}{1 + u^2}$ has a standard form where the integral of $\frac{u}{1 + u^2}$ is $\frac{1}{2} \ln |1 + u^2|$.

   So, $2 \int \frac{u}{1 + u^2} \, du = 2 \cdot \frac{1}{2} \ln |1 + u^2| = \ln |1 + u^2|$

5. Back-substitute $u = \sin x$: Replace $u$ with $\sin x$ to get the final answer: $\int \frac{\sin 2x}{1 + \sin^2 x} \, dx = \ln |1 + \sin^2 x| + C$

Final Answer

$\int \frac{\sin 2x}{1 + \sin^2 x} \, dx = \ln |1 + \sin^2 x| + C$

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

Related Questions

1. How do we handle integrals with trigonometric identities?
2. What substitution techniques are helpful for integrals involving trigonometric functions?
3. Can you explain the integration of rational functions with trigonometric terms?
4. How does substitution simplify integrals in general?
5. What is the process to verify the result of an indefinite integral?

Tip

When you encounter a trigonometric integral, always consider substituting using identities or common functions to simplify the integral before solving.