What is the integral of (y^2cos(z)-2ze^x) ?

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

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integral of (y^2cos(z)-2ze^x)

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

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

Solution steps

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

Apply the Sum Rule:

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

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

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

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

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

Add a constant to the solution

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

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