Mathematics

Evaluate the given integral.
$$\displaystyle\int{\dfrac{\sqrt{1+\cos{2x}}}{2}dx}$$


SOLUTION
$$I=\displaystyle\int{\dfrac{\sqrt{1+\cos{2x}}}{2}dx}$$

$$=\displaystyle\int{\dfrac{\sqrt{1+2{\cos}^{2}{x}-1}}{2}dx}$$            ........$$(cos2x=2cos^2x-1)$$

$$=\displaystyle\int{\dfrac{\sqrt{2{\cos}^{2}{x}}}{2}dx}$$

$$=\dfrac{\sqrt{2}}{2}\displaystyle\int{\cos{x}\,dx}$$

$$=\dfrac{\sqrt{2}\times \sqrt{2}}{2\times \sqrt{2}}\sin{x}+c$$

$$=\dfrac{2}{2\times \sqrt{2}}\sin{x}+c$$

$$=\dfrac{1}{\sqrt{2}}\sin{x}+c$$
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Subjective Medium Published on 17th 09, 2020
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