First slide

Methods of integration

Question

$\int_{0}^{1} \frac{dx}{e^{x} + e^{- x}} dx$ is equal to

Easy

A

$\frac{π}{4}$
B

$\tan^{- 1} e - \frac{π}{4}$
C

$\tan^{- 1} e$
D

$\frac{π}{4} \tan^{- 1} e$

Solution

$\begin{array}{l} \int_{0}^{1} \frac{dx}{e^{x} + e^{- x}} = \int_{0}^{1} \frac{e^{x}}{e^{2 x} + 1} dx \\ Let t = e^{x} \Rightarrow dt = e^{x} dx \\ = \int_{1}^{e} \frac{dt}{t^{2} + 1} = {[\tan^{- 1} t]}_{1}^{e} \\ = \tan^{- 1} e - \tan^{- 1} (1) = \tan^{- 1} e - \frac{π}{4} \end{array}$

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