First slide

Introduction to integration

Question

$\int \frac{1}{1 + \sin x} d x .$

Easy

A

$S e c x + T a n x + C$
B

$S e c x - T a n x + C$
C

$T a n x - S e c x + C$
D

$T a n x - S e c x + C$

Solution

$\begin{array}{l} \int \frac{1}{1 + \sin x} d x = \int \frac{1}{(1 + \sin x)} \cdot \frac{(1 - \sin x)}{(1 - \sin x)} d x \\ = \int \frac{1 - \sin x}{1 - \sin^{2} x} d x \\ == \int \frac{1 - \sin x}{\cos^{2} x} d x \\ = \int \frac{1}{\cos^{2} x} d x - \int \frac{\sin x}{\cos^{2} x} d x \\ = \int \sec^{2} x d x - \int \tan x \sec x d x \\ = \tan x - \sec x + C \end{array}$

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