What is the integral from 0 to x of 10cos(x+y+z) ?

The integral from 0 to x of 10cos(x+y+z) is 10(sin(2x+y)-sin(x+y))

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integral from 0 to x of 10cos(x+y+z)

\int_{0}^{x} 10 cos (x + y + z) d z

10 (sin (2 x + y) - sin (x + y))

Solution steps

\int_{0}^{x} 10 cos (x + y + z) d z

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 \cdot \int_{0}^{x} cos (x + y + z) d z

Apply u-substitution

: \int_{x + y}^{2 x + y} cos (u) d u

= 10 \cdot \int_{x + y}^{2 x + y} cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= 10 [sin (u)]_{x + y}^{2 x + y}

Compute the boundaries:

sin (2 x + y) - sin (x + y)

= 10 (sin (2 x + y) - sin (x + y))

\frac{\partial}{\partial x} (x^{4} y^{2} + 4 y^{4} - x^{3})

\int 5 cos^{4} (x) d x

\frac{\partial}{\partial z} (- 3 z)

x \cdot cos (\frac{3 π x}{2}), 0

\int_{0}^{\frac{4}{5}} \frac{1}{16 + 25 x ^{2}} d x

What is the integral from 0 to x of 10cos(x+y+z) ?
The integral from 0 to x of 10cos(x+y+z) is 10(sin(2x+y)-sin(x+y))