What is the integral of x^2ycos(x)+2xysin(x)-y^2e^x ?

The integral of x^2ycos(x)+2xysin(x)-y^2e^x is yx^2sin(x)-y^2e^x+C

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integral of x^2ycos(x)+2xysin(x)-y^2e^x

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

y x^{2} sin (x) - y^{2} e^{x} + C

Solution steps

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

\int y^{2} e^{x} d x = y^{2} e^{x}

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

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

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

Simplify

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

= y x^{2} sin (x) - y^{2} e^{x}

Add a constant to the solution

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

\int (\frac{1}{( 25 - x ^{2} ) ^{\frac{3}{2}}}) d x

\int \frac{5 + x}{5 - x} d x

f (x) = \frac{2 x ^{2} - 5 x + 3}{5 x - 3}

\int \frac{e ^{x} + 1}{x} d x

f (x) = \frac{4}{x + 1}, at x = 1

What is the integral of x^2ycos(x)+2xysin(x)-y^2e^x ?
The integral of x^2ycos(x)+2xysin(x)-y^2e^x is yx^2sin(x)-y^2e^x+C