What is the integral of 2zcos(z^2) ?

The integral of 2zcos(z^2) is sin(z^2)+C

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integral of 2zcos(z^2)

\int 2 z cos (z^{2}) d z

sin (z^{2}) + C

Solution steps

\int 2 z cos (z^{2}) d z

Take the constant out:

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

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

Apply u-substitution

: \int \frac{cos ( u )}{2} d u

= 2 \cdot \int \frac{cos ( u )}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int cos (u) d u

Use the common integral:

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

= 2 \cdot \frac{1}{2} sin (u)

Substitute back

u = z^{2}

= 2 \cdot \frac{1}{2} sin (z^{2})

Simplify

2 \cdot \frac{1}{2} sin (z^{2}) : sin (z^{2})

= sin (z^{2})

Add a constant to the solution

= sin (z^{2}) + C

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