Q81 / 150 · MHT CET 2026 (13 April Shift 1) · Mathematics
If $\displaystyle\int \dfrac{\sin 2x}{\sin^4 x + \cos^4 x} \, dx = f(x) + c$ where c is constant of integration, then the value of $\tan(f(x))$ is
- A$\tan x$
- Correct answerB$\tan^2 x$
- C$\tan^4 x$
- D$\cot^4 x$
Solution
The given integral is $\displaystyle\int \dfrac{\sin 2x}{\sin^4 x + \cos^4 x} \, dx$.
Dividing the numerator and the denominator by $\cos^4 x$, we get
$\displaystyle\int \dfrac{\dfrac{2\sin x \cos x}{\cos^4 x}}{\dfrac{\sin^4 x + \cos^4 x}{\cos^4 x}} \, dx$
$= \displaystyle\int \dfrac{2\tan x \sec^2 x}{\tan^4 x + 1} \, dx$
Substituting $\tan^2 x = t$, we get $2\tan x \sec^2 x \, dx = dt$.
The integral becomes
$\displaystyle\int \dfrac{dt}{t^2 + 1} = \tan^{-1}(t) + c$
Substituting back $t = \tan^2 x$, we get
$f(x) + c = \tan^{-1}(\tan^2 x) + c$
Comparing both sides, $f(x) = \tan^{-1}(\tan^2 x)$.
Therefore, $\tan(f(x)) = \tan(\tan^{-1}(\tan^2 x)) = \tan^2 x$.
Answer: $\tan^2 x$
Dividing the numerator and the denominator by $\cos^4 x$, we get
$\displaystyle\int \dfrac{\dfrac{2\sin x \cos x}{\cos^4 x}}{\dfrac{\sin^4 x + \cos^4 x}{\cos^4 x}} \, dx$
$= \displaystyle\int \dfrac{2\tan x \sec^2 x}{\tan^4 x + 1} \, dx$
Substituting $\tan^2 x = t$, we get $2\tan x \sec^2 x \, dx = dt$.
The integral becomes
$\displaystyle\int \dfrac{dt}{t^2 + 1} = \tan^{-1}(t) + c$
Substituting back $t = \tan^2 x$, we get
$f(x) + c = \tan^{-1}(\tan^2 x) + c$
Comparing both sides, $f(x) = \tan^{-1}(\tan^2 x)$.
Therefore, $\tan(f(x)) = \tan(\tan^{-1}(\tan^2 x)) = \tan^2 x$.
Answer: $\tan^2 x$
Hence, the correct answer is option: 'B'.