What is the integral of (x^2sin(x)cos(x)+xsin^2(x)) ?

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

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integral of (x^2sin(x)cos(x)+xsin^2(x))

\int (x^{2} sin (x) cos (x) + x sin^{2} (x)) d x

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

Solution steps

\int x^{2} sin (x) cos (x) + x sin^{2} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{2} sin (x) cos (x) d x + \int x sin^{2} (x) d x

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

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

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

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

Add a constant to the solution

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

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