What is the integral of 6x^2cos(2x^3+1) ?

The integral of 6x^2cos(2x^3+1) is sin(2x^3+1)+C

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integral of 6x^2cos(2x^3+1)

\int 6 x^{2} cos (2 x^{3} + 1) d x

sin (2 x^{3} + 1) + C

Solution steps

\int 6 x^{2} cos (2 x^{3} + 1) d x

Take the constant out:

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

= 6 \cdot \int x^{2} cos (2 x^{3} + 1) d x

Apply u-substitution

: \int \frac{u cos ( 2 u ^{\frac{3}{2}} + 1 )}{2 x} d u

= 6 \cdot \int \frac{cos ( 2 u + 1 )}{3} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int cos (2 u + 1) d u

Apply u-substitution

: \int cos (v) \frac{1}{2} d v

= 6 \cdot \frac{1}{3} \cdot \int cos (v) \frac{1}{2} d v

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \int cos (v) d v

Use the common integral:

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

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

Substitute back

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

Simplify

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

= sin (2 x^{3} + 1)

Add a constant to the solution

= sin (2 x^{3} + 1) + C

x + 3

\frac{d y}{d x} = y (1 - 0.0004 y)

(e^{x} - 2 x y) d x - x^{2} d y = 0

2 e^{- 3 x}

(y + 1) d x - x d y = 0

What is the integral of 6x^2cos(2x^3+1) ?
The integral of 6x^2cos(2x^3+1) is sin(2x^3+1)+C