What is the integral of (-x^2y^2sin(x)+2xy^2cos(x)) ?

The integral of (-x^2y^2sin(x)+2xy^2cos(x)) is y^2x^2cos(x)+C

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integral of (-x^2y^2sin(x)+2xy^2cos(x))

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

y^{2} x^{2} cos (x) + C

Solution steps

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

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

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

= - y^{2} (- x^{2} cos (x) + 2 (x sin (x) + cos (x))) + 2 y^{2} (x sin (x) + cos (x))

Simplify

- y^{2} (- x^{2} cos (x) + 2 (x sin (x) + cos (x))) + 2 y^{2} (x sin (x) + cos (x)) : y^{2} x^{2} cos (x)

= y^{2} x^{2} cos (x)

Add a constant to the solution

= y^{2} x^{2} cos (x) + C

\int \frac{e ^{\frac{1}{x ^{7}}}}{x ^{8}} d x

\frac{d}{d x} ((1 - x^{2})^{\frac{3}{2}})

\frac{d}{d x} (- 2 x^{- 3} \cdot ln (x + x^{- 1}))

x \to - 9 lim (\frac{x ^{2} - 7}{9 - x})

\frac{\partial}{\partial x} (4 x^{2} y^{3} + 7 x y^{2} - 7 x^{2})

What is the integral of (-x^2y^2sin(x)+2xy^2cos(x)) ?
The integral of (-x^2y^2sin(x)+2xy^2cos(x)) is y^2x^2cos(x)+C