# What is the result of int_0^(pi/2) t^2costdt?

Dec 13, 2016

${\int}_{0}^{\frac{\pi}{2}} {t}^{2} \cos t \cong 0.4674$

#### Explanation:

We start by using the techniques we would use to find the indefinite integral. Use integration by parts.

Let $u = {t}^{2}$ and $\mathrm{dv} = \cos t \mathrm{dt}$. Then $\mathrm{du} = 2 t \mathrm{dt}$ and $v = \sin t$.

By integration by parts:

$\int \left({t}^{2} \cos t\right) \mathrm{dt} = \sin t \left({t}^{2}\right) - \int \left(\sin t \left(2 t\right)\right)$

Use integration by parts again for the remaining integral.

$\int \left(\sin t \left(2 t\right)\right) \mathrm{dt} = - \cos t \left(t\right) - \int \left(- \cos t \left(2\right)\right)$

$\int \left(\sin t \left(2 t\right)\right) \mathrm{dt} = - 2 t \cos \left(t\right) - 2 \int \left(- \cos t\right)$

$\int \left(\sin t \left(2 t\right)\right) \mathrm{dt} = - 2 t \cos \left(t\right) + 2 \sin t + C$

Resubstitute:

$\int \left({t}^{2} \cos t\right) \mathrm{dt} = {t}^{2} \sin t - \left(- 2 t \cos \left(t\right) + 2 \sin t\right) + C$

$\int \left({t}^{2} \cos t\right) = {t}^{2} \sin t + 2 t \cos \left(t\right) - 2 \sin t + C$

$\int \left({t}^{2} \cos t\right) = \left({t}^{2} - 2\right) \sin t + 2 t \cos t + C$

Since this is a definite integral, we can forget about the "$C$".

int_(0)^(pi/2)(t^2cost) = ((pi/2)^2 - 2)sin(pi/2) + 2(pi/2)cos(pi/2) - ((0^2 - 2)sin(0) + 2(0)(cos(0))

${\int}_{0}^{\frac{\pi}{2}} \left({t}^{2} \cos t\right) = \left({\pi}^{2} / 4 - 2\right) \left(1\right) + \pi \left(0\right) - \left(- 2 \left(0\right) + 0 + 0\right)$

${\int}_{0}^{\frac{\pi}{2}} \left({t}^{2} \cos t\right) \cong 0.4674$

Hopefully this helps!