First slide

Evaluation of definite integrals

Question

$\int \frac{1 + \tan^{2} x}{1 - \tan^{2} x} d x =$

Moderate

A

$\frac{1}{2} \ln | \frac{1 + \tan^{2} x}{1 - \tan^{2} x} | + c$
B

$\frac{1}{2} \ln | \frac{1 - \tan^{2} x}{1 + \tan^{2} x} | + c$
C

$\frac{1}{2} \ln | \frac{1 + \tan x}{1 - \tan x} | + c$
D

$\frac{1}{2} \ln | \frac{1 - \tan x}{1 + \tan x} | + c$

Solution

$\int \frac{1 + \tan^{2} x}{1 - \tan^{2} x} d x =$ $\int \frac{\sec^{2} x}{1 - \tan^{2} x} d x = \int \frac{d t}{1 - t^{2}}$ where t = tanx
= $\frac{1}{2} \ln | \frac{1 + t}{1 - t} | + c$

$= \frac{1}{2} \ln | \frac{1 + \tan x}{1 - \tan x} | + c$

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