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Special Integrals - 1

Evaluate : ...

Question

Evaluate : $\int_{0}^{π} \frac{sin θ + cos θ}{\sqrt{1 + sin 2 θ}} d θ$

A

π

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B

2 π

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C

π / 2

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D

π / 3

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Solution

The correct option is A $π$
$\int_{0}^{π} \frac{s i n θ + c o s θ}{\sqrt{1 + s i n 2 θ}} \cdot d θ$

Let $I = \int \frac{s i n θ + c o s θ}{\sqrt{1 + s i n 2 θ}} \cdot d θ = \int \frac{sin θ + cos θ}{\sqrt{{sin}^{2} θ + {cos}^{2} θ + 2 sin θ cos θ}}$
$= \int \frac{s i n θ + c o s θ}{\sqrt{(s i n θ + c o s θ)^{2}}} d θ = \int \frac{s i n θ + c o s θ}{(s i n θ + c o s θ)} \cdot d θ$
$= \int 1 \cdot d θ = θ + c$

$\int_{0}^{π} \frac{s i n θ + c o s θ}{\sqrt{1 + s i n 2 θ}} \cdot d θ = \int_{0}^{π} d θ$
$= [θ + c]_{0}^{π}$
$= (π + c) - (0 + c)$
$= π$

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