Question

If \(\int \frac{sin2x}{sin^4x+cos^4x}\,dx = f(x)+c\) where c is constant of integration, then the value of \(tan(f(x))\) is

• A. \(tanx\)
• B. \(tan^2x\)
• C. \(tan^4x\)
• D. \(cot^4x\)

Hint:

Divide the numerator and denominator by cos^4 x and substitute u = tan^2 x.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Integrals with \(\sin^4 x + \cos^4 x\) in the denominator are usually done by dividing by \(\cos^4 x\), which turns everything into \(\tan x\) and \(\sec^2 x\).

Step 2: Rewrite the integrand.
\(\sin 2x = 2\sin x\cos x = 2\tan x\cos^2 x\). Divide the numerator and denominator by \(\cos^4 x\):
\[ \frac{\sin 2x}{\sin^4 x + \cos^4 x} = \frac{2\tan x\sec^2 x}{\tan^4 x + 1} \]

Step 3: Substitute.
Let \(u = \tan^2 x\), so \(du = 2\tan x\sec^2 x\,dx\). Also \(\tan^4 x = u^2\).
\[ \int\frac{du}{u^2 + 1} = \tan^{-1}u + c = \tan^{-1}(\tan^2 x) + c \]

Step 4: Find tan f(x).
So \(f(x) = \tan^{-1}(\tan^2 x)\) and \(\tan f(x) = \tan^2 x\).

Step 5: Check the options.
Option (B) matches. The others do not come from this substitution.

Final Answer:
\(\tan(f(x)) = \tan^2 x\), option (B). \[ \boxed{\tan^2 x} \]