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

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

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integral of (yz^2-sin(x)sin(pi-y))

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

sin (- y + π) cos (x) + y z^{2} x + C

Solution steps

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

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

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

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

Simplify

= sin (- y + π) cos (x) + y z^{2} x

Add a constant to the solution

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

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