Mathematics

Evaluate the following definite integrals :
$$\displaystyle \int _{0}^{\pi /2} \cos^2 x\ dx$$


ANSWER

$$\frac { \pi }{ 4 } $$


SOLUTION

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

$$=\displaystyle \int _{0}^{\pi /2} \dfrac {1+\cos 2x}{2} dx$$........$$using \ (\cos2 x=2cos^2 x-1)$$

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

$$=\dfrac {1}{2}\left [\left (\dfrac {\pi}{2}+\dfrac {\sin \pi}{2} \right) -\left (0+\dfrac {\sin 0}{2}\right) \right] =\dfrac {\pi}{4}$$

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