First slide

Introduction to integration

Question

$\int_{0}^{π / 2} \sqrt{1 + S i n 2 x} d x =$

Very Easy

A

1
B

-2
C

$\sqrt{2}$
D

2

Solution

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

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