Question

If $$ \int \frac{dx}{1 - \sin^4 x} = A \tan x + B \tan^{-1}(\sqrt{2} \tan x) + C, $$ then find $ A^2 - B^2 $.

• A. \( \frac{1}{2} \)
• B. \( \frac{3}{4} \)
• C. \( \frac{1}{4} \)
• D. \( \frac{1}{8} \)

Hint:

Use trigonometric identities and substitution to simplify integral.

Answer 1

The Correct Option is D

Use the identity: \[ 1 - \sin^4 x = (1 - \sin^2 x)(1 + \sin^2 x) = \cos^2 x (1 + \sin^2 x) \] Rewrite integral: \[ \int \frac{dx}{\cos^2 x (1 + \sin^2 x)} = \int \sec^2 x \frac{dx}{1 + \sin^2 x} \] Use substitution \( t = \tan x \), and evaluate integral, leading to expression in terms of \( \tan x \) and \( \tan^{-1}(\sqrt{2} \tan x) \). From comparison, \( A^2 - B^2 = \frac{1}{8} \).