First slide

Evaluation of definite integrals

Question

$If \int_{0}^{k} \frac{\cos x}{1 + \sin^{2} x} d x = \frac{π}{4} then k =$

Easy

A

0
B

$\frac{π}{4}$
C

$\frac{π}{2}$
D

None

Solution

$\begin{array}{l} \int_{0}^{k} \frac{d t}{1 + t^{2}} = \frac{π}{4} put \sin x = t \\ \Rightarrow {(\tan^{- 1} t)}_{0}^{k} = \frac{π}{4} \cos x \cdot d x = d t \\ \Rightarrow {(\tan^{- 1} (\sin x))}_{0}^{k} = \frac{π}{4} \\ \tan^{1} (\sin k) - \tan^{- 1} (\sin 0) = \frac{π}{4} \\ \tan^{- 1} (\sin k) - 0 = \frac{π}{4} \Rightarrow \sin k = 1 \\ k = \frac{π}{2} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App