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: Understanding the Concept
This integral can be solved using u-substitution. The presence of \(\sin\theta\) in the denominator and \(\cos\theta\) in the numerator suggests substituting for \(\sin\theta\). After substitution, the integral will be in a standard form.
Step 2: Key Formula or Approach
1. Use the substitution \(u = \sin\theta\).
2. Find the differential \(du\).
3. The integral will be transformed into the standard form \(\int \frac{dx}{a^2 - x^2}\).
4. Use the standard integration formula: \(\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln\left|\frac{a+x}{a-x}\right| + C\).
Step 3: Detailed Explanation
1. Perform the substitution.
Let \(u = \sin\theta\).
Then, differentiate with respect to \(\theta\):
\[ \frac{du}{d\theta} = \cos\theta \] \[ du = \cos\theta \, d\theta \] 2. Substitute into the integral.
The original integral is \(\int \frac{\cos\theta}{2 - \sin^2\theta} d\theta\).
Substituting \(u\) and \(du\), we get:
\[ \int \frac{1}{2 - u^2} du \] 3. Apply the standard integration formula.
This integral is in the form \(\int \frac{du}{a^2 - u^2}\), where \(a^2 = 2\), so \(a = \sqrt{2}\).
Using the formula \(\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln\left|\frac{a+x}{a-x}\right| + C\):
\[ \int \frac{du}{(\sqrt{2})^2 - u^2} = \frac{1}{2\sqrt{2}} \ln\left|\frac{\sqrt{2}+u}{\sqrt{2}-u}\right| + C \] 4. Substitute back for \(\theta\).
Replace \(u\) with \(\sin\theta\):
\[ \frac{1}{2\sqrt{2}} \ln\left|\frac{\sqrt{2}+\sin\theta}{\sqrt{2}-\sin\theta}\right| + C \] Since \(\sin\theta\) is always between -1 and 1, \(\sqrt{2}+\sin\theta\) and \(\sqrt{2}-\sin\theta\) are always positive, so we can use parentheses instead of the absolute value.
\[ \frac{1}{2\sqrt{2}}\log\left(\frac{\sqrt{2}+\sin\theta}{\sqrt{2}-\sin\theta}\right) + C \] This matches option (E).
Step 4: Final Answer
The integral is \(\frac{1}{2\sqrt{2}}\log\left(\frac{\sqrt{2}+\sin\theta}{\sqrt{2}-\sin\theta}\right) + C\).