1. ## help with integration

$\displaystyle \int_{0}^{2} t^2sin{t^3} + 2.5dt$
So I started with integration by parts.
$\displaystyle \int_{0}^{2} t^2sint^3dt + \int_{0}^{2} 2.5dt$
$\displaystyle u = t^2$
$\displaystyle du = 2tdt$
$\displaystyle dv = sint^3dt$
$\displaystyle v = \frac{-1}{3t^2}cos(t^3)$
but then I realized this was gonna go on for a long time. Is there an easier way to take this integral?

2. Originally Posted by steveo0
$\displaystyle \int_{0}^{2} t^2sin{t^3} + 2.5dt$
So I started with integration by parts.
$\displaystyle \int_{0}^{2} t^2sint^3dt + \int_{0}^{2} 2.5dt$
$\displaystyle u = t^2$
$\displaystyle du = 2tdt$
$\displaystyle dv = sint^3dt$
$\displaystyle v = \frac{-1}{3t^2}cos(t^3)$
but then I realized this was gonna go on for a long time. Is there an easier way to take this integral?
If you mean $\displaystyle \int_0^2t^2\sin\left(t^3\right)+2.5\,dt$, then we see that we have $\displaystyle \int_0^2t^2\sin\left(t^3\right)\,dt+2.5\int_0^2\,d t$

To solve the first integral, make the substitution $\displaystyle u=t^3$
The second one is very straightforward.

Can you continue from here?

3. oh! yeah. thanks

4. Hello, steveo0!

By-parts is not necessary . . .

We are concerned with: .$\displaystyle \int_{0}^{2} t^2\sin(t^3)\,dt$

Let: .$\displaystyle u \,=\,t^3\quad\Rightarrow\quad du \,=\,3t^2\,dt \quad\Rightarrow\quad t^2\,dt \,=\,\tfrac{1}{3}du$

We have: . $\displaystyle \int \underbrace{\sin(t^3)}_{\downarrow}\,\underbrace{( t^2\,dt)}_{\downarrow}$
Substitute: .$\displaystyle \int \sin u\,\left(\tfrac{1}{3}du \right) \;=\;\tfrac{1}{3}\int \sin u\,du$ . . . Got it?