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

The integral of-xsin^2(x) is -1/2 x^2+1/4 xsin(2x)+(x^2)/4+1/8 cos(2x)+C

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integral of-xsin^2(x)

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

- \frac{1}{2} x^{2} + \frac{1}{4} x sin (2 x) + \frac{x ^{2}}{4} + \frac{1}{8} cos (2 x) + C

Solution steps

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

Take the constant out:

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

= - \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} 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} 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} x (x - \frac{1}{2} sin (2 x)) - \frac{1}{2} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x))) : - \frac{1}{2} x^{2} + \frac{1}{4} x sin (2 x) + \frac{x ^{2}}{4} + \frac{1}{8} cos (2 x)

= - \frac{1}{2} x^{2} + \frac{1}{4} x sin (2 x) + \frac{x ^{2}}{4} + \frac{1}{8} cos (2 x)

Add a constant to the solution

= - \frac{1}{2} x^{2} + \frac{1}{4} x sin (2 x) + \frac{x ^{2}}{4} + \frac{1}{8} cos (2 x) + C

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What is the integral of-xsin^2(x) ?
The integral of-xsin^2(x) is -1/2 x^2+1/4 xsin(2x)+(x^2)/4+1/8 cos(2x)+C