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

#\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
maganbhai P.
Mar 17, 2018

#2/3#

Explanation:

#I=int_0^pi-costsint+cos^2tsint+4sin^3tcostdt#
#I=-1/2int_0^pisin2tdt+int_0^picos^2tsintdt+int_0^pi4sin^3tcostdt#

Let us take, #I=I_1+I_2+I_3#
#I_1=-1/2[(-cos2t)/2]_0^pi=1/4[cos2pi-cos0]=1/4[1-1]=0#
#I_2=-int_0^picos^2t(-sint)dt=-int_0^pi[cost]^2*d/(dt)(cost)dt#
#=>I_2=-[(cost)^3/3]_0^pi=-1/3[(cospi)^3-(cos0)^3]#
#I_2=-1/3[(-1)^3-(1)^3]=-1/3[-1-1]=-1/3[-2]#
#I_2=2/3#
#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+2/3+0=2/3#

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