What is the integral of 6xcos(3x^2) ?

The integral of 6xcos(3x^2) is sin(3x^2)+C

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integral of 6xcos(3x^2)

\int 6 x cos (3 x^{2}) d x

sin (3 x^{2}) + C

Solution steps

\int 6 x cos (3 x^{2}) d x

Take the constant out:

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

= 6 \cdot \int x cos (3 x^{2}) d x

Apply u-substitution

: \int \frac{cos ( u )}{6} d u

= 6 \cdot \int \frac{cos ( u )}{6} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int cos (u) d u

Use the common integral:

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

= 6 \cdot \frac{1}{6} sin (u)

Substitute back

u = 3 x^{2}

= 6 \cdot \frac{1}{6} sin (3 x^{2})

Simplify

6 \cdot \frac{1}{6} sin (3 x^{2}) : sin (3 x^{2})

= sin (3 x^{2})

Add a constant to the solution

= sin (3 x^{2}) + C

\frac{\partial}{\partial x} ((x + y)^{z})

x \to π lim (x)

\frac{1}{s ^{2} + 2 \cdot π \cdot s + π ^{2}}

n = 3 \sum \infty e^{5 - 6 n}

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = 0

What is the integral of 6xcos(3x^2) ?
The integral of 6xcos(3x^2) is sin(3x^2)+C