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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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