Question

Calculus (2) Master Needed to show steps of evaluating the integral b=0, a= pi/2 :
`\int\limits^a_b {(t^2-3t)sint} \, dt` I get stuck after applying integration by parts and simplified the negative, I get
`(t^2-3t)*-cost |(\pi /2)/(0) + \int\limits^a_b {cost * (t-3)} \, dt` what am I supposed to do at this stage? Or did I do it wrong? Step by step is needed!

Answer 1

`\displaystyle \int\limits^(\pi/2)_ {0} \,(t^2-3t)\sin t\, dt = \pi -5`

Explanation:

We want to evaluate the integral:

`\displaystyle \int\limits^(\pi/2)_ {0} \,(t^2-3t)\sin(t)\, dt`

Since this is a product of two functions, we can use Integration by Parts. Recall that:

`\displaystyle \int u\, dv = uv - \int v\, du`

Let u be t² - 3t and let dv be sin(t).

Thus:

`du = 2t - 3\, dt`

And:

`\displaystyle v = \int \sin t\, dt = - \cos t`

So:

`\displaystyle =\left(-\cos t\right)\left(t^2-3t\right)\Bigg|_0^(\pi/2)-\int_(0)^(\pi/2) \left(-\cos t\right)\left(2t-3\right)\, dt`

Simplify:

`\displaystyle = - \left(\cos t\right)(t^2-3t)\Bigg|_(0)^(\pi /2)+\int\limits^(\pi/2)_ {0} \,\cos(t)(2t-3)\, dt`

We can use IBP once more.

Let u be 2t - 3 and let dv be cos(t). Therefore:

`du = 2\, dt`

And:

`\displaystyle v = \int \cos t\, dt = \sin t`

So:

`\displaystyle = - \left(\cos t\right)(t^2-3t)\Bigg|_0^(\pi/2)+\left(\sin(t)(2t-3)\Bigg|_0^(\pi /2)-\int\limits^(\pi/2)_0 \, 2\sin(t)dt \right)`

Evaluate:

`\displaystyle =-\left(\cos t\right)(t^2-3t)+\sin(t)(2t-3)+2\cos(t)\Bigg|_0^(\pi /2)`

Evaluate the first section:

`\displaystyle \begin{aligned} &= -\cos\left((\pi)/(2)\right)\left(\left((\pi)/(2)\right)^2-3\left((\pi)/(2)\right)\right)+\sin\left((\pi)/(2)\right)\left(2\left((\pi)/(2)\right)-3\right)+2\cos \left((\pi)/(2)\right)\right) \\ \\ &= (0) + (1)(\pi -3) + (0) \\ \\ &= \pi -3 \end{aligned}`

And the second:

`\displaystyle \begin{aligned} &= -\cos(0)((0)^2-3(0))+(\sin(0))(2(0)-3)+2\cos(0 ) \\ \\&=(0)+(0)+(2) \\ \\ &= 2\end{aligned}`

Therefore:

`\displaystyle \int\limits^(\pi/2)_ {0} \,(t^2-3t)\sin(t)\, dt = (\pi -3) -(2) = \pi -5`