Mathematics

# $\displaystyle\int \sin^2(2x+5) dx$.

##### SOLUTION
We are given $I=\displaystyle\int \sin^{2}(2x+5) dx$
Now, we know that
$\cos 2\theta=1-2\sin^{2}\theta$
So, $1-\cos 2\theta=2\sin^{2}\theta$
$\therefore \sin^{2}\theta=\dfrac{1-\cos 2\theta}{2}$
$I=\displaystyle\int \left(\dfrac{1-\cos 2(2x+5)}{2}\right)dx$
$=\displaystyle\int \left(\dfrac{1}{2}-\dfrac{\cos(4x+10}{2}\right)dx$
$=\displaystyle\int \dfrac{dx}{2}-\dfrac{1}{2} \int \cos (4x+10) dx$
$=\dfrac{x}{2}-\dfrac{1}{2}\dfrac{\sin (4x+10)}{4}+C$
$I=\dfrac{x}{2}-\dfrac{\sin (4x+10)}{8}+C$
Where $C$ is any arbitrary constant

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

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