What is the result of $\int_{0}^{\frac{π}{2}} t^{2} cos t d t$ $int_0^(pi/2) t^2costdt$ ?

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What is the result of $\int_{0}^{\frac{π}{2}} t^{2} cos t d t$ $int_0^(pi/2) t^2costdt$ ?

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Answer 1 · 2016-12-13T14:34:32

$\int_{0}^{\frac{π}{2}} t^{2} cos t ≅ 0.4674$ $int_0^(pi/2)t^2cost ~= 0.4674$

Explanation:

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

Let $u = t^{2}$ $u = t^2$ and $d v = cos t d t$ $dv = costdt$ . Then $d u = 2 t d t$ $du = 2tdt$ and $v = sin t$ $v = sint$ .

By integration by parts:

$\int (t^{2} cos t) d t = sin t (t^{2}) - \int (sin t (2 t))$ $int(t^2cost)dt = sint(t^2) - int(sint(2t))$

Use integration by parts again for the remaining integral.

$\int (sin t (2 t)) d t = - cos t (t) - \int (- cos t (2))$ $int(sint(2t))dt = -cost(t) - int(-cost(2))$

$\int (sin t (2 t)) d t = - 2 t cos (t) - 2 \int (- cos t)$ $int(sint(2t))dt = -2tcos(t) - 2int(-cost)$

$\int (sin t (2 t)) d t = - 2 t cos (t) + 2 sin t + C$ $int(sint(2t))dt = -2tcos(t) + 2sint + C$

Resubstitute:

$\int (t^{2} cos t) d t = t^{2} sin t - (- 2 t cos (t) + 2 sin t) + C$ $int(t^2cost)dt = t^2sint - (-2tcos(t) + 2sint) + C$

$\int (t^{2} cos t) = t^{2} sin t + 2 t cos (t) - 2 sin t + C$ $int(t^2cost) = t^2sint + 2tcos(t) - 2sint + C$

$\int (t^{2} cos t) = (t^{2} - 2) sin t + 2 t cos t + C$ $int(t^2cost) = (t^2- 2)sint + 2tcost + C$

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

$\int_{0}^{\frac{π}{2}} (t^{2} cos t) = ({(\frac{π}{2})}^{2} - 2) sin (\frac{π}{2}) + 2 (\frac{π}{2}) cos (\frac{π}{2}) - ((0^{2} - 2) sin (0) + 2 (0) (cos (0))$ $int_(0)^(pi/2)(t^2cost) = ((pi/2)^2 - 2)sin(pi/2) + 2(pi/2)cos(pi/2) - ((0^2 - 2)sin(0) + 2(0)(cos(0))$

$\int_{0}^{\frac{π}{2}} (t^{2} cos t) = (\frac{π^{2}}{4} - 2) (1) + π (0) - (- 2 (0) + 0 + 0)$ $int_0^(pi/2)(t^2cost) = (pi^2/4 - 2)(1) + pi(0) - (-2(0) + 0 + 0)$

$\int_{0}^{\frac{π}{2}} (t^{2} cos t) ≅ 0.4674$ $int_0^(pi/2)(t^2cost) ~=0.4674$

Hopefully this helps!

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What is the result of $\int_{0}^{\frac{π}{2}} t^{2} cos t d t$ $int_0^(pi/2) t^2costdt$ ?