Mathematics

Evaluate:$$\displaystyle\int_{0}^{\frac{\pi}{2}}{{\cos}^{2}{x}\,dx}$$


SOLUTION
$$\displaystyle\int_{0}^{\frac{\pi}{2}}{{\cos}^{2}{x}\,dx}$$
$$=\dfrac{1}{2}\displaystyle\int_{0}^{\frac{\pi}{2}}{2{\cos}^{2}{x}\,dx}$$
We know that $$\cos{2x}=2{\cos}^{2}{x}-1\Rightarrow\,2{\cos}^{2}{x}=1+\cos{2x}$$
$$=\dfrac{1}{2}\displaystyle\int_{0}^{\frac{\pi}{2}}{\left(1+\cos{2x}\right)\,dx}$$
$$=\dfrac{1}{2}\displaystyle\int_{0}^{\frac{\pi}{2}}{dx}+\dfrac{1}{2}\displaystyle\int_{0}^{\frac{\pi}{2}}{\cos{2x}\,dx}$$
$$=\dfrac{1}{2}\left[x\right]_{0}^{\frac{\pi}{2}}+\dfrac{1}{2}\left[\dfrac{\sin{2x}}{2}\right]_{0}^{\frac{\pi}{2}}$$
$$=\dfrac{1}{2}\left[x\right]_{0}^{\frac{\pi}{2}}+\dfrac{1}{4}\left[\sin{2x}\right]_{0}^{\frac{\pi}{2}}$$
$$=\dfrac{1}{2}\left[\dfrac{\pi}{2}-0\right]-\dfrac{1}{4}\left[\sin{\pi}-\sin{0}\right]$$
$$=\dfrac{\pi}{4}-0$$
$$=\dfrac{\pi}{4}$$
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Subjective Medium Published on 17th 09, 2020
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