Mathematics

Evaluate $\displaystyle \int \sin^3 x \cos^3 x \ dx$

SOLUTION
$I=\displaystyle \int \sin^3 x \cos^3 x \ dx$

$I=\displaystyle \dfrac{1}{8} \int (2\sin x \cos x)^3 \ dx$

$I=\displaystyle \dfrac{1}{8} \int \sin^3 2x \ dx$

$I=\displaystyle \dfrac{1}{8}\int \dfrac{3\sin 2x -\sin 6x}{4} \ dx$   $[\because \sin 3\theta=3\sin \theta-4\sin^3 \theta]$

$I=\displaystyle \dfrac{1}{32}\int (3\sin 2x -\sin 6x) \ dx$

$I=\displaystyle \dfrac{1}{32}\left[-\dfrac{3}{2} \cos 2x +\dfrac{1}{6} \cos 6x\right]+C$

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

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