First slide

Evaluation of definite integrals

Question

$I f a_{n} = \int_{0}^{π / 2} \frac{\sin^{2} n x}{\sin x} d x$ then $a_{2} - a_{1}, a_{3} - a_{2},$
$a_{4} - a_{3}$ are in

Moderate

A

$A . P .$
B

$G . P .$
C

$H . P .$
D

Arith – Geo progression

Solution

$n > 1, a_{n} - a_{n - 1} = \int_{0}^{π / 2} \frac{\sin^{2} n x - \sin^{2} (n - 1) x}{\sin x} d x$
$\begin{array}{l} = \int_{0}^{π / 2} \frac{\sin (2 n - 1) x \sin x}{\sin x} d x \\ = \int_{0}^{π / 2} \sin (2 n - 1) x d x = - {\frac{\cos (2 n - 1) x}{2 n - 1}|}_{0}^{π / 2} \\ = \frac{1}{2 n - 1} \end{array}$
which is a Harmonic Progression.

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