Question

$\int \frac{1}{8 \sin^2 x + 1}\, dx$ is equal to:

• A. $\sin^{-1}(\tan x) + C$
• B. $\frac{1}{3} \sin^{-1}(\tan x) + C$
• C. $\frac{1}{3} \tan^{-1}(3 \tan x) + C$
• D. $\tan^{-1}(3 \tan x) + C$
• E. $\sin^{-1}(3 \tan x) + C$

Hint:

Dividing by $\cos^2 x$ is the "golden rule" for any rational function containing only even powers of $\sin x$ and $\cos x$. It immediately transforms trigonometric complexities into solvable algebraic forms using the $\tan^{-1}$ identity.

Answer 1

The Correct Option is C

Concept: For integrals of the form $\int \frac{1}{a \sin^2 x + b \cos^2 x} \, dx$, a standard technique is to divide both the numerator and denominator by $\cos^2 x$ to convert the expression into one involving $\tan x$ and $\sec^2 x$.

Step 1: Divide the numerator and denominator by $\cos^2 x$.
\[ \int \frac{\sec^2 x}{8 \tan^2 x + \sec^2 x} \, dx \] Substitute $\sec^2 x = 1 + \tan^2 x$ in the denominator: \[ \int \frac{\sec^2 x}{8 \tan^2 x + (1 + \tan^2 x)} \, dx = \int \frac{\sec^2 x}{9 \tan^2 x + 1} \, dx \]

Step 2: Use substitution.
Let $u = 3 \tan x$, then $du = 3 \sec^2 x \, dx \Rightarrow \frac{1}{3} du = \sec^2 x \, dx$. The integral becomes: \[ \frac{1}{3} \int \frac{1}{u^2 + 1} \, du = \frac{1}{3} \tan^{-1} u + C \]

Step 3: Substitute back.
\[ \frac{1}{3} \tan^{-1}(3 \tan x) + C \]