Question

If I$_n$ = $\int_{\pi/4}^{\pi/2} \cot^n x \,dx$, then :

• A. $\frac{1}{I_2+I_4}, \frac{1}{I_3+I_5}, \frac{1}{I_4+I_6}$ are in A.P.
• B. $I_2+I_4, I_3+I_5, I_4+I_6$ are in A.P.
• C. $\frac{1}{I_2+I_4}, \frac{1}{I_3+I_5}, \frac{1}{I_4+I_6}$ are in G.P.
• D. $I_2+I_4, (I_3+I_5)^2, I_4+I_6$ are in G.P.

Hint:

For definite integrals involving powers of trigonometric functions like $\int \sin^n x \,dx$, $\int \cos^n x \,dx$, $\int \tan^n x \,dx$, or $\int \cot^n x \,dx$, developing a reduction formula or a recurrence relation is a very powerful technique.

Answer 1

The Correct Option is A

Consider, \[ I_n + I_{n-2} = \int_{\pi/4}^{\pi/2}\left(\cot^n x+\cot^{\,n-2}x\right)dx = \int_{\pi/4}^{\pi/2}\cot^{\,n-2}x(\cot^2x+1)\,dx \] Using the identity \(1+\cot^2x=\csc^2x\), \[ I_n + I_{n-2}=\int_{\pi/4}^{\pi/2}\cot^{\,n-2}x\,\csc^2x\,dx \] Put \(u=\cot x\), then \(du=-\csc^2x\,dx\) Limits change as: \[ x=\pi/4 \Rightarrow u=1,\qquad x=\pi/2 \Rightarrow u=0 \] \[ I_n + I_{n-2} = \int_{1}^{0} u^{n-2}(-du) = \int_{0}^{1}u^{n-2}\,du = \left[\frac{u^{n-1}}{n-1}\right]_0^1 = \frac{1}{n-1} \] Now, \[ I_2+I_4=\frac{1}{3},\quad I_3+I_5=\frac{1}{4},\quad I_4+I_6=\frac{1}{5} \] Taking reciprocals, \[ 3,\;4,\;5 \] which form an arithmetic progression. Correct option: (A)