Question

$\int\frac{\cos \theta}{2-\sin^{2}\theta}d \theta=$ ________.

• A. $\frac{1}{2}\log|\frac{\sqrt{2}-\sin \theta}{\sqrt{2}+\sin \theta}|+C$
• B. $\frac{1}{2}\log|\frac{\sqrt{2}+\sin \theta}{\sqrt{2}-\sin \theta}|+C$
• C. $\log|\frac{\sqrt{2}+\sin \theta}{\sqrt{2}-\sin \theta}|+C$
• D. $\frac{1}{\sqrt{2}}\log|\frac{\sqrt{2}+\sin \theta}{\sqrt{2}-\sin \theta}|+C$
• E. $\frac{1}{2\sqrt{2}}\log|\frac{\sqrt{2}+\sin \theta}{\sqrt{2}-\sin \theta}|+C$

Hint:

Recognize $du$ in the numerator to pick your $u$.

Answer 1

Answer: (E)

Step 1: Concept
Use substitution $u = \sin \theta$.

Step 2: Meaning
$du = \cos \theta d\theta$. Integral becomes $\int \frac{1}{2-u^2} du$.

Step 3: Analysis
Apply standard formula $\int \frac{1}{a^2-x^2}dx = \frac{1}{2a}\log|\frac{a+x}{a-x}|+C$.
Here $a = \sqrt{2}$.

Step 4: Conclusion
$\frac{1}{2\sqrt{2}}\log|\frac{\sqrt{2}+u}{\sqrt{2}-u}| + C = \frac{1}{2\sqrt{2}}\log|\frac{\sqrt{2}+\sin \theta}{\sqrt{2}-\sin \theta}| + C$. Final Answer: (E)