Mathematics

# $\displaystyle \int _{0}^{\pi /2} \cos^2 x\ dx$

##### SOLUTION

$I=\displaystyle \int _{0}^{\pi /2} \cos^2 x\ dx$

$=\displaystyle \int _{0}^{\pi /2} \dfrac {1+\cos 2x}{2} dx$

$=\dfrac {1}{2} \left [x +\dfrac {\sin 2x}{2} \right]_0^{\pi /2}$

$=\dfrac {1}{2}\left [\left (\dfrac {\pi}{2}+\dfrac {\sin \pi}{2} \right) -\left (0+\dfrac {\sin 0}{2}\right) \right]$

$=\dfrac {\pi}{4}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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