For a whole number n, let \(I_{n} = \int_{0}^{\frac{\pi}{4}}\displaystyle\tan^{n}x\,dx\). Which of the following statements are correct?
(A) \(I_{2} + I_{0} = 1\)
(B) \(I_{3} + I_{1} = \displaystyle\frac{1}{2}\)
(C) \(I_{1} = \log_e 2\)
(D) \(I_{4} = \displaystyle\frac{\pi}{4} - \frac{2}{3}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
For \(n \ge 2\), write \(\tan^{n}x = \tan^{n-2}x\,(\sec^{2}x - 1)\). Then
\( \displaystyle I_{n} + I_{n-2} = \int_{0}^{\pi/4}\tan^{n-2}x\,\sec^{2}x\,dx \)
With \(t = \tan x\), \(dt = \sec^{2}x\,dx\), limits 0 to 1:
\( \displaystyle I_{n} + I_{n-2} = \int_{0}^{1}t^{n-2}\,dt = \frac{1}{n-1} \)
Also
\( \displaystyle I_{0} = \frac{\pi}{4},\qquad I_{1} = \left[\log_e\sec x\right]_{0}^{\pi/4} = \log_e\sqrt{2} = \frac{1}{2}\log_e 2 \)
From the relation with \(n = 2\) and \(n = 4\):
\( \displaystyle I_{2} = 1 - \frac{\pi}{4} \)
\( \displaystyle I_{4} = \frac{1}{3} - I_{2} = \frac{\pi}{4} - \frac{2}{3} \)
So \(I_{n} + I_{n-2} = \frac{1}{n-1}\), \(I_{1} = \frac{1}{2}\log_e 2\) and \(I_{4} = \frac{\pi}{4} - \frac{2}{3}\).
Checking the statements
(A) \(I_{2} + I_{0} = 1\) — Correct. With n = 2, \(\frac{1}{2-1} = 1\).
(B) \(I_{3} + I_{1} = \frac{1}{2}\) — Correct. With n = 3, \(\frac{1}{3-1} = \frac{1}{2}\).
(C) \(I_{1} = \log_e 2\) — Incorrect. \(I_{1} = \log_e\sec\frac{\pi}{4} = \log_e\sqrt{2} = \frac{1}{2}\log_e 2\).
(D) \(I_{4} = \frac{\pi}{4} - \frac{2}{3}\) — Correct. \(I_{4} = \frac{1}{3} - \left(1 - \frac{\pi}{4}\right) = \frac{\pi}{4} - \frac{2}{3}\).