If In=∫π/4π/2 cotn⁡xdx, then I1+I3,I2+I4,I3+I5,I4+I6,… are in

A
A.P
B
G.P
C
H.P
D
none of these

Solution:

We have,

$\begin{array}{l} I_{n} = \int_{π / 4}^{π / 2} \cot^{n} x d x \Rightarrow I_{n + 2} = \int_{π / 4}^{π / 2} \cot^{n + 2} x d x \\ ∴ I_{n} + I_{n + 2} = \int_{π / 4}^{π / 2} (\cot^{n} x + \cot^{n + 2} x) d x = \int_{π / 4}^{π / 2} \cot^{n} x {cosec}^{2} x d x \end{array}$

$\Rightarrow I_{n} + I_{n + 2} = - \int_{1}^{0} t^{11} d t,$ where $t = \cot x$

$\begin{array}{l} \Rightarrow I_{n} + I_{n + 2} = - {[\frac{t^{n - 1}}{n + 1}]}_{1}^{0} = \frac{1}{n + 1} \\ \Rightarrow \frac{1}{I_{n} + I_{n + 2}} = n + 1 \end{array}$

$\Rightarrow \frac{1}{I_{1} + I_{3}}, \frac{1}{I_{2} + I_{4}}, \frac{1}{I_{3} + I_{5}}, \dots$ are in A.P.

$\Rightarrow I_{1} + I_{3}, I_{2} + I_{4}, I_{3} + I_{5}, \dots$ are in H.P.

If $I_{n} = \int_{π / 4}^{π / 2} \cot^{n} x d x,$ then $I_{1} + I_{3}, I_{2} + I_{4}, I_{3} + I_{5}, I_{4} + I_{6}, \dots$ are in

Solution:

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