Mathematics

$\int { \sin ^{ 2 }{ b } } xdx\quad \quad$

SOLUTION
$\displaystyle\int{{\sin}^{2}{bx}dx}$

$=\dfrac{1}{2}\displaystyle\int{2{\sin}^{2}{bx}dx}$

$=\dfrac{1}{2}\displaystyle\int{\left(1-\cos{2bx}\right)dx}$

$=\dfrac{1}{2}\left[x-\dfrac{\sin{2bx}}{2b}\right]+c$

$=\dfrac{x}{2}-\dfrac{\sin{2bx}}{4b}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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