What is the integral of -costsint+cos^2tsint+4sin^3tcostdt from o to pi?

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What is the integral of -costsint+cos^2tsint+4sin^3tcostdt from o to pi?

$\int_{0}^{π} - cos (t) sin (t) + {cos}^{2} (t) sin (t) + 4 {sin}^{3} (t) cos (t) d t$ $\int_0^pi -cos(t)sin(t)+cos^2(t)sin(t)+4sin^3(t)cos(t)dt$
I assume one can substitue with u=cos(t), but i fail to get the right answer of 2/3, i get -2/3.

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Answer 1 · 2018-03-17T17:25:50

$\frac{2}{3}$ $2/3$

Explanation:

$I = \int_{0}^{π} - cos t sin t + {cos}^{2} t sin t + 4 {sin}^{3} t cos t d t$ $I=int_0^pi-costsint+cos^2tsint+4sin^3tcostdt$
$I = - \frac{1}{2} \int_{0}^{π} sin 2 t d t + \int_{0}^{π} {cos}^{2} t sin t d t + \int_{0}^{π} 4 {sin}^{3} t cos t d t$ $I=-1/2int_0^pisin2tdt+int_0^picos^2tsintdt+int_0^pi4sin^3tcostdt$

Let us take, $I = I_{1} + I_{2} + I_{3}$ $I=I_1+I_2+I_3$
$I_{1} = - \frac{1}{2} {[\frac{- cos 2 t}{2}]}_{0}^{π} = \frac{1}{4} [cos 2 π - cos 0] = \frac{1}{4} [1 - 1] = 0$ $I_1=-1/2[(-cos2t)/2]_0^pi=1/4[cos2pi-cos0]=1/4[1-1]=0$
$I_{2} = - \int_{0}^{π} {cos}^{2} t (- sin t) d t = - \int_{0}^{π} {[cos t]}^{2} \cdot \frac{d}{d t} (cos t) d t$ $I_2=-int_0^picos^2t(-sint)dt=-int_0^pi[cost]^2*d/(dt)(cost)dt$
$\Rightarrow I_{2} = - {[\frac{{(cos t)}^{3}}{3}]}_{0}^{π} = - \frac{1}{3} [{(cos π)}^{3} - {(cos 0)}^{3}]$ $=>I_2=-[(cost)^3/3]_0^pi=-1/3[(cospi)^3-(cos0)^3]$
$I_{2} = - \frac{1}{3} [{(- 1)}^{3} - {(1)}^{3}] = - \frac{1}{3} [- 1 - 1] = - \frac{1}{3} [- 2]$ $I_2=-1/3[(-1)^3-(1)^3]=-1/3[-1-1]=-1/3[-2]$
$I_{2} = \frac{2}{3}$ $I_2=2/3$
$I_{3} = 4 \int_{0}^{π} {[sin t]}^{3} \frac{d}{d t} (sin t) d t = 4 {[\frac{{(sin t)}^{4}}{4}]}_{0}^{π} = {[{(sin π)}^{4} - (] sin 0)}^{4}] = [0 - 0] = 0$ $I_3=4int_0^pi[sint]^3d/(dt)(sint)dt=4[(sint)^4/4]_0^pi=[(sinpi)^4-(]sin0)^4]=[0-0]=0$
$I = I_{1} + I_{2} + I_{3} = 0 + \frac{2}{3} + 0 = \frac{2}{3}$ $I=I_1+I_2+I_3=0+2/3+0=2/3$

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What is the integral of -costsint+cos^2tsint+4sin^3tcostdt from o to pi?

$\int_{0}^{π} - cos (t) sin (t) + {cos}^{2} (t) sin (t) + 4 {sin}^{3} (t) cos (t) d t$ $\int_0^pi -cos(t)sin(t)+cos^2(t)sin(t)+4sin^3(t)cos(t)dt$
I assume one can substitue with u=cos(t), but i fail to get the right answer of 2/3, i get -2/3.

1 Answer

Explanation:

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What is the integral of -costsint+cos^2tsint+4sin^3tcostdt from o to pi?

∫π0−cos(t)sin(t)+cos2(t)sin(t)+4sin3(t)cos(t)dt\int_0^pi -cos(t)sin(t)+cos^2(t)sin(t)+4sin^3(t)cos(t)dt I assume one can substitue with u=cos(t), but i fail to get the right answer of 2/3, i get -2/3.

1 Answer

Explanation:

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$\int_{0}^{π} - cos (t) sin (t) + {cos}^{2} (t) sin (t) + 4 {sin}^{3} (t) cos (t) d t$ $\int_0^pi -cos(t)sin(t)+cos^2(t)sin(t)+4sin^3(t)cos(t)dt$
I assume one can substitue with u=cos(t), but i fail to get the right answer of 2/3, i get -2/3.