x ^3 x d x - Vidyadip

Question

$\int x \cos ^3 x d x$

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Answer

$\text { Let } I =\int x \cos ^3 x d x$
$\cos ^3 x =4 \cos ^3 x -3 \cos x$
$\therefore 4 \cos ^3 x =3 \cos x +\cos 3 x$
$\therefore \cos ^3 x =\frac{1}{4}(3 \cos x+\cos 3 x)$
$\therefore I =\frac{1}{4} \int x(3 \cos x+\cos 3 x) d x$
$=\frac{1}{4}\left[x \int(3 \cos x+\cos 3 x) d x-\int\left\{\frac{ d }{ d x}(x) \int(3 \cos x+\cos 3 x) d x\right\} d x\right]$
$=\frac{1}{4}\left[x\left(3 \sin x+\frac{\sin 3 x}{3}\right)-\int 1\left(3 \sin x+\frac{\sin 3 x}{3}\right) d x\right]$
$=\frac{1}{4}\left[3 x \sin x+\frac{x}{3} \sin 3 x-\left(-3 \cos x-\frac{1}{3} \cdot \frac{\cos 3 x}{3}\right)\right]+ c$
$\therefore I =\frac{1}{4}\left(3 x \sin x+\frac{x}{3} \sin 3 x+3 \cos x+\frac{1}{9} \cos 3 x\right)+ c $

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