Question

$\int \frac{\sin x}{3 + 4\cos^2 x} \, dx =$

• A. $\frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{2\cos x}{\sqrt{3}}\right) + C$
• B. $\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{\cos x}{3}\right) + C$
• C. $\frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{\cos x}{3}\right) + C$
• D. $-\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{2\cos x}{\sqrt{3}}\right) + C$

Answer 1

The Correct Option is A

1. Understand the integral:

We need to evaluate \( \int \frac{\sin x}{3 + 4 \cos^2 x} dx \).

2. Use substitution:

Let \( u = \cos x \), then \( du = -\sin x \, dx \) or \( -du = \sin x \, dx \).

The integral becomes:

\[ \int \frac{\sin x}{3 + 4 \cos^2 x} dx = -\int \frac{du}{3 + 4u^2} \]

3. Rewrite the denominator:

\[ 3 + 4u^2 = 4\left(\frac{3}{4} + u^2\right) = 4\left(\left(\frac{\sqrt{3}}{2}\right)^2 + u^2\right) \]

4. Apply the standard integral formula:

\[ \int \frac{1}{a^2 + t^2} dt = \frac{1}{a} \tan^{-1}\left(\frac{t}{a}\right) + C \]

Here, \( a = \frac{\sqrt{3}}{2} \) and \( t = u \), so:

\[ -\int \frac{du}{4\left(\left(\frac{\sqrt{3}}{2}\right)^2 + u^2\right)} = -\frac{1}{4} \cdot \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{u}{\frac{\sqrt{3}}{2}}\right) + C = -\frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{2u}{\sqrt{3}}\right) + C \]

5. Substitute back \( u = \cos x \):

\[ -\frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right) + C \]

Correct Answer: (A) \( -\frac{1}{2\sqrt{3}} \tan^{-1} \left( \frac{2 \cos x}{\sqrt{3}} \right) + C \)

Answer 2

Let $I = \int \frac{\sin x}{3 + 4\cos^2 x} \, dx$. Substitute $u = \cos x$, so $du = -\sin x \, dx$.

The integral becomes: \[ I = -\int \frac{1}{3 + 4u^2} \, du. \] Write $3 + 4u^2$ as $a^2 + u^2$ with $a^2 = 3$: \[ I = -\frac{1}{\sqrt{3}} \int \frac{1}{1 + \left(\frac{2u}{\sqrt{3}}\right)^2} \, du. \] Substitute back, and the integral simplifies to: \[ I = \frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{2\cos x}{\sqrt{3}}\right) + C. \]