Question

\(\int\frac{1}{1+3\sin^2 x+8 \cos^2 x}\ dx=\)

• A. \(\frac{1}{6}\tan^{-1}(\frac{2\tan x}{3})+C\)
• B. \(\frac{1}{6}\tan^{-1}(2\tan x)+C\)
• C. \(6\tan^{-1}(\frac{2\tan x}{3})+C\)
• D. \(\tan^{-1}(\frac{2\tan x}{3})+C\)

Hint:

When solving integrals involving trigonometric functions, consider using substitutions to simplify the expression. Using the substitution \( t = \tan x \) often helps transform the integral into a simpler form, as shown in this example.

Answer 1

The Correct Option is A

The correct answer is: (A): \(\frac{1}{6}\tan^{-1}\left(\frac{2\tan x}{3}\right) + C\)

We are tasked with evaluating the integral:

\(\int \frac{1}{1 + 3\sin^2 x + 8 \cos^2 x} \, dx\)

Step 1: Simplify the expression

The given expression contains both sine and cosine terms. We begin by rewriting the denominator to combine like terms:

\( 1 + 3\sin^2 x + 8\cos^2 x = 1 + 3\sin^2 x + 8(1 - \sin^2 x) = 1 + 3\sin^2 x + 8 - 8\sin^2 x \)

After simplifying, we get:

\( 9 - 5\sin^2 x \)

Step 2: Use a substitution to simplify further

Next, we use the substitution \( t = \tan x \), which implies that \( \sec^2 x \, dx = dt \). Since \( \sin^2 x = \frac{t^2}{1 + t^2} \), we can rewrite the denominator in terms of \( t \). The integral becomes:

\( \int \frac{1}{9 - 5\frac{t^2}{1 + t^2}} \, dt \)

Step 3: Simplify the expression

After simplifying the denominator, we get a rational function of \( t \), which can be integrated using standard methods. The resulting integral simplifies to:

\( \frac{1}{6}\tan^{-1}\left(\frac{2t}{3}\right) + C \)

Step 4: Substitute back to original variable

Finally, substitute \( t = \tan x \) back into the expression to get the final result:

\( \frac{1}{6}\tan^{-1}\left(\frac{2\tan x}{3}\right) + C \)

Conclusion:
The correct answer is (A): \(\frac{1}{6}\tan^{-1}\left(\frac{2\tan x}{3}\right) + C\)