How do you find the integral of $x {(sin x)}^{2}$ $x(sinx)^2$ ?

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How do you find the integral of $x {(sin x)}^{2}$ $x(sinx)^2$ ?

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Answer 1 · 2016-09-06T19:12:30

$\frac{1}{8} (2 x^{2} - 2 x sin 2 x - cos 2 x) + C$ $1/8(2x^2-2xsin2x-cos2x)+C$ .

Explanation:

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

Then, $I = \int {\frac{x (1 - cos 2 x)}{2}} d x = \frac{1}{2} \int x d x - \frac{1}{2} \int x sin 2 x d x = \frac{1}{4} x^{2} - \frac{1}{2} J,$ $I=int{(x(1-cos2x))/2}dx=1/2intxdx-1/2intxsin2xdx=1/4x^2-1/2J,$

where, $J = \int x cos 2 x d x$ $J=intxcos2xdx$

To find $J$ $J$ , we use the Rule of Integration by Parts (IBP) :

(IBP) : $\int u v d x = u \int v d x - \int (\frac{d u}{d x} \cdot \int v d x) d x$ $intuvdx=uintvdx-int((du)/dx*intvdx)dx$ .

Taking, $u = x, so that, \frac{d u}{d x} = 1, &, v = cos 2 x, so \int v d x = \frac{sin (2 x)}{2}$ $u=x", so that, "(du)/dx=1, &, v=cos2x", so "intvdx=sin(2x)/2$ .

$:. J=x/2sin2x-1/2intsin2xdx$

$=x/2sin2x-1/2(-1/2cos2x)$

$=x/2sin2x+1/4cos2x$

Hence, $I=x^2/4-x/4sin2x-1/8cos2x$

$=1/8(2x^2-2xsin2x-cos2x)+C$ .

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