What is the integral of (cos(x+4y)-2xz) ?

The integral of (cos(x+4y)-2xz) is -zx^2+sin(4y+x)+C

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integral of (cos(x+4y)-2xz)

\int (cos (x + 4 y) - 2 x z) d x

- z x^{2} + sin (4 y + x) + C

Solution steps

\int cos (x + 4 y) - 2 x z d x

Rewrite using trig identities

= \int cos (x) cos (4 y) - sin (x) sin (4 y) - 2 z 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 cos (x) cos (4 y) d x - \int sin (x) sin (4 y) d x - \int 2 z x d x

\int cos (x) cos (4 y) d x = cos (4 y) sin (x)

\int sin (x) sin (4 y) d x = - sin (4 y) cos (x)

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

= cos (4 y) sin (x) - (- sin (4 y) cos (x)) - z x^{2}

Simplify

cos (4 y) sin (x) - (- sin (4 y) cos (x)) - z x^{2} : - z x^{2} + sin (4 y + x)

= - z x^{2} + sin (4 y + x)

Add a constant to the solution

= - z x^{2} + sin (4 y + x) + C

\int \frac{8}{x ^{3} - 4 x} d x

\int \frac{2 y + 6}{y + 3} d y

\int x^{2} e^{\frac{- x ^{2}}{2}} d x

\int e^{x} + 5 sin (x) d x

What is the integral of (cos(x+4y)-2xz) ?
The integral of (cos(x+4y)-2xz) is -zx^2+sin(4y+x)+C