Let \(f(x) = \int\displaystyle\frac{\tan^3x}{\cos^2x}\,dx\) and \(f\left(\displaystyle\frac{\pi}{4}\right) = 0\). Then \(f(0)\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle -\frac{1}{4}\)
Since \(\frac{1}{\cos^2x} = \sec^2x\) is the derivative of \(\tan x\), substitute
\(t = \tan x,\quad dt = \sec^2x\,dx\)
Then
\( \displaystyle f(x) = \int t^3\,dt = \frac{\tan^4x}{4}+C \)
The condition \(f\left(\frac{\pi}{4}\right) = 0\), with \(\tan\frac{\pi}{4} = 1\), fixes the constant:
\( \displaystyle \frac{1}{4}+C = 0 \)
\( \displaystyle C = -\frac{1}{4} \)
Since \(\tan 0 = 0\):
\( \displaystyle f(0) = 0-\frac{1}{4} = -\frac{1}{4} \)