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

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

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integral of-x^2sin(x)

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

x^{2} cos (x) - 2 x sin (x) - 2 cos (x) + C

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

: - x^{2} cos (x) - \int - 2 x cos (x) d x

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

\int - 2 x cos (x) d x = - 2 (x sin (x) + cos (x))

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (\frac{e ^{7 - x}}{y ^{5}})

\frac{\partial}{\partial x} (2 y - x e^{x y})

n = 0 \sum \infty \frac{1}{n ^{\frac{2}{3}}}

\frac{d}{d x} (e^{(2 x)} (x^{3} + 6 x - 1) ln (x))

\frac{d}{d x} (\frac{sin ^{2} ( x ) + 1}{cos ^{2} ( x ) + 1})

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