What is the integral of (xsin^2(x))/2 ?

The integral of (xsin^2(x))/2 is 1/16 (2x^2-2xsin(2x)-cos(2x))+C

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integral of (xsin^2(x))/2

\int \frac{x sin ^{2} ( x )}{2} d x

\frac{1}{16} (2 x^{2} - 2 x sin (2 x) - cos (2 x)) + C

Solution steps

\int \frac{x sin ^{2} ( x )}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int x sin^{2} (x) d x

Apply Integration By Parts

: \frac{1}{2} x (x - \frac{1}{2} sin (2 x)) - \int \frac{1}{2} (x - \frac{1}{2} sin (2 x)) d x

= \frac{1}{2} (\frac{1}{2} x (x - \frac{1}{2} sin (2 x)) - \int \frac{1}{2} (x - \frac{1}{2} sin (2 x)) d x)

\int \frac{1}{2} (x - \frac{1}{2} sin (2 x)) d x = \frac{1}{2} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x))

= \frac{1}{2} (\frac{1}{2} x (x - \frac{1}{2} sin (2 x)) - \frac{1}{2} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x)))

Simplify

\frac{1}{2} (\frac{1}{2} x (x - \frac{1}{2} sin (2 x)) - \frac{1}{2} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x))) : \frac{1}{16} (2 x^{2} - 2 x sin (2 x) - cos (2 x))

= \frac{1}{16} (2 x^{2} - 2 x sin (2 x) - cos (2 x))

Add a constant to the solution

= \frac{1}{16} (2 x^{2} - 2 x sin (2 x) - cos (2 x)) + C

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What is the integral of (xsin^2(x))/2 ?
The integral of (xsin^2(x))/2 is 1/16 (2x^2-2xsin(2x)-cos(2x))+C