First slide

Introduction to definite integration

Question

$\int_{0}^{π} \frac{1}{1 + \sin x} d x =$

Easy

A

1
B

2
C

–1
D

–2

Solution

$\int_{0}^{π} \frac{1}{1 + \sin x} d x = 2 \int_{0}^{π / 2} \frac{1}{1 + \sin x} d x = 2 \int_{0}^{π / 2} \frac{1 - \sin x}{\cos^{2} x} d x = 2 \int_{0}^{π / 2} (\sec^{2} x - \sec x \tan x) d x$
$= 2 [\tan x - \sec {x]}_{0}^{π / 2} = - 2 {[\frac{1 - \sin x}{\cos x}]}_{0}^{π / 2} = - 2 {[\frac{\cos x}{1 + \sin x}]}_{0}^{π / 2} = - 2 [0 - 1] = 2$

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