What is the antiderivative of $x {(sin x)}^{2}$ $x(sinx)^2$ ?

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What is the antiderivative of $x {(sin x)}^{2}$ $x(sinx)^2$ ?

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Answer 1 · 2018-01-04T21:13:32

$\frac{1}{4} x^{2} - \frac{1}{4} x sin 2 x + \frac{1}{8} cos 2 x$ $1/4x^2-1/4xsin2x+1/8cos2x$

Explanation:

$I = \int x {sin}^{2} x d x$ $I = intxsin^2xdx$

$cos 2 x = {cos}^{2} - {sin}^{2} x, \Rightarrow {sin}^{2} x = \frac{1 - cos 2 x}{2}$ $cos2x= cos^2-sin^2x , rArr sin^2x= (1-cos2x)/2$

$I = \int x {sin}^{2} x d x = \int \frac{x (1 - cos 2 x)}{2} d x$ $I = intxsin^2xdx = int(x(1-cos2x))/2dx$
$I = \frac{1}{2} \int x d x - \frac{1}{2} \int x cos 2 x d x$ $I = 1/2intxdx- 1/2intxcos2xdx$

Let evaluate the second part first. Let

$u = 2 x \Rightarrow d u = 2 d x$ $u = 2x rArr du = 2dx$ and perform integration by parts

$\frac{1}{2} \int x cos 2 x d x = \frac{1}{8} \int u cos u d u = \frac{1}{8} (u sin u - \int sin u d u)$ $1/2intxcos2xdx = 1/8 intucosudu= 1/8 (usinu-intsinudu)$
$= \frac{1}{8} (u sin u + cos u)$ $=1/8 (usinu+cosu)$

$I = \frac{1}{4} x^{2} - \frac{1}{8} (2 x sin 2 x + cos 2 x)$ $I = 1/4x^2-1/8(2xsin2x+cos2x)$
$I = \frac{1}{4} x^{2} - \frac{1}{4} x sin 2 x + \frac{1}{8} cos 2 x$ $I= 1/4x^2-1/4xsin2x+1/8cos2x$

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