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∫π0−cos(t)sin(t)+cos2(t)sin(t)+4sin3(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

2/323

Explanation:

I=int_0^pi-costsint+cos^2tsint+4sin^3tcostdtI=∫π0−costsint+cos2tsint+4sin3tcostdt
I=-1/2int_0^pisin2tdt+int_0^picos^2tsintdt+int_0^pi4sin^3tcostdtI=−12∫π0sin2tdt+∫π0cos2tsintdt+∫π04sin3tcostdt

Let us take, I=I_1+I_2+I_3I=I1+I2+I3
I_1=-1/2[(-cos2t)/2]_0^pi=1/4[cos2pi-cos0]=1/4[1-1]=0I1=−12[−cos2t2]π0=14[cos2π−cos0]=14[1−1]=0
I_2=-int_0^picos^2t(-sint)dt=-int_0^pi[cost]^2*d/(dt)(cost)dtI2=−∫π0cos2t(−sint)dt=−∫π0[cost]2⋅ddt(cost)dt
=>I_2=-[(cost)^3/3]_0^pi=-1/3[(cospi)^3-(cos0)^3]⇒I2=−[(cost)33]π0=−13[(cosπ)3−(cos0)3]
I_2=-1/3[(-1)^3-(1)^3]=-1/3[-1-1]=-1/3[-2]I2=−13[(−1)3−(1)3]=−13[−1−1]=−13[−2]
I_2=2/3I2=23
I_3=4int_0^pi[sint]^3d/(dt)(sint)dt=4[(sint)^4/4]_0^pi=[(sinpi)^4-(]sin0)^4]=[0-0]=0I3=4∫π0[sint]3ddt(sint)dt=4[(sint)44]π0=[(sinπ)4−(]sin0)4]=[0−0]=0
I=I_1+I_2+I_3=0+2/3+0=2/3I=I1+I2+I3=0+23+0=23

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