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

The integral of 2xe^z-y^2sin(x) is y^2cos(x)+e^zx^2+C

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integral of 2xe^z-y^2sin(x)

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

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

Solution steps

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

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

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

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

Simplify

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

Add a constant to the solution

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

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What is the integral of 2xe^z-y^2sin(x) ?
The integral of 2xe^z-y^2sin(x) is y^2cos(x)+e^zx^2+C