What is the integral of 12xzcos(xz^2) ?

The integral of 12xzcos(xz^2) is 6sin(xz^2)+C

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integral of 12xzcos(xz^2)

\int 12 x z cos (x z^{2}) d z

6 sin (x z^{2}) + C

Solution steps

\int 12 x z cos (x z^{2}) d z

Take the constant out:

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

= 12 x \cdot \int cos (x z^{2}) z d z

Apply u-substitution

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

= 12 x \cdot \int \frac{cos ( xu )}{2} d u

Take the constant out:

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

= 12 x \frac{1}{2} \cdot \int cos (xu) d u

Apply u-substitution

: \int \frac{cos ( v )}{x} d v

= 12 x \frac{1}{2} \cdot \int \frac{cos ( v )}{x} d v

Take the constant out:

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

= 12 x \frac{1}{2} \cdot \frac{1}{x} \cdot \int cos (v) d v

Use the common integral:

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

= 12 x \frac{1}{2} \cdot \frac{1}{x} sin (v)

Substitute back

= 12 x \frac{1}{2} \cdot \frac{1}{x} sin (x z^{2})

Simplify

12 x \frac{1}{2} \cdot \frac{1}{x} sin (x z^{2}) : 6 sin (x z^{2})

= 6 sin (x z^{2})

Add a constant to the solution

= 6 sin (x z^{2}) + C

f (x) = ln (1 + sin^{2} (x))

x \to - 0 lim (x^{2})

\int e^{x} (x^{2} - 2 x - 1) d x

\frac{\partial}{\partial x} (x ln (2 y - x))

\frac{s - 3}{( s - 2 ) ^{2}}

What is the integral of 12xzcos(xz^2) ?
The integral of 12xzcos(xz^2) is 6sin(xz^2)+C