What is the integral of 2t^2cos(t^3-2) ?

The integral of 2t^2cos(t^3-2) is 2/3 sin(t^3-2)+C

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integral of 2t^2cos(t^3-2)

\int 2 t^{2} cos (t^{3} - 2) d t

\frac{2}{3} sin (t^{3} - 2) + C

Solution steps

\int 2 t^{2} cos (t^{3} - 2) d t

Take the constant out:

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

= 2 \cdot \int t^{2} cos (t^{3} - 2) d t

Apply u-substitution

: \int \frac{cos ( u )}{3} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = t^{3} - 2

= 2 \cdot \frac{1}{3} sin (t^{3} - 2)

Simplify

2 \cdot \frac{1}{3} sin (t^{3} - 2) : \frac{2}{3} sin (t^{3} - 2)

= \frac{2}{3} sin (t^{3} - 2)

Add a constant to the solution

= \frac{2}{3} sin (t^{3} - 2) + C

\int cot (\frac{x}{a}) d x

\int x^{2} sec (x^{3} - 1) tan (x^{3} - 1) d x

\int (4 y + y) d y

\int \frac{sin ( 2 t )}{2} d t

What is the integral of 2t^2cos(t^3-2) ?
The integral of 2t^2cos(t^3-2) is 2/3 sin(t^3-2)+C