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Question

If a1,a2 and a3 are the three values of a which satisfy the equation π/20(sinx+acosx)3 dx4aπ2π/20xcosx dx=2, then the value of 1000(a21+a22+a23) is

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Solution

I=π/20(sinx+acosx)3dx4aπ2π/20xcosxdx=2

Let I1=π/20(sinx+acosx)3dx
=π/20((sinx+acosx)1+a21+a2)3dx
=(1+a2)3/2π/20(sin(x+α))3dx
where cosα=11+a2 and sinα=a1+a2
I1=(1+a2)3/2π/20[1cos2(x+α)]sin(x+α) dx
Put cos(x+α)=t
sin(x+α)dx=dt
I1=(1+a2)3/2sinαcosα(1t2)dt
=(1+a2)3/2[t33t]sinαcosα
=(1+a2)3/2[13(sin3αcos3α)(sinαcosα)]
=(1+a2)3/2[(sinα+cosα)13(sinα+cosα)(1sinαcosα)]
=(1+a2)3/2(sinα+cosα)3[31+sinαcosα]
=(1+a2)3/23(1+a1+a2)(2+a1+a2)
=1+a3(2+2a2+a)


I2=π/20xI cosxII dx
=xsinxπ/20π/20sinx dx
=π21=π22

I=1+a3(2+2a2+a)4aπ2π22=2
2a3+3a23a4=0

a1+a2+a3=32
a1a2=32
a21=(32)22(32)
=94+3=214
1000a21=1000×214
=250×21=5250

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