Mathematics

Evaluate the following definite integral:

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


SOLUTION

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

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

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

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

$$=\dfrac {\pi}{4}$$
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Subjective Medium Published on 17th 09, 2020
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