If a1,a2 and a3 are the three values of a which satisfy the equation π/2∫0(sinx+acosx)3dx−4aπ−2π/2∫0xcosxdx=2, then the value of 1000(a21+a22+a23) is
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Solution
I=π/2∫0(sinx+acosx)3dx−4aπ−2π/2∫0xcosxdx=2
Let I1=π/2∫0(sinx+acosx)3dx =π/2∫0((sinx+acosx)√1+a2√1+a2)3dx =(1+a2)3/2π/2∫0(sin(x+α))3dx where cosα=1√1+a2 and sinα=a√1+a2 I1=(1+a2)3/2π/2∫0[1−cos2(x+α)]sin(x+α)dx Put cos(x+α)=t ⇒−sin(x+α)dx=dt I1=−(1+a2)3/2−sinα∫cosα(1−t2)dt =(1+a2)3/2[t33−t]−sinαcosα =(1+a2)3/2[13(−sin3α−cos3α)−(−sinα−cosα)] =(1+a2)3/2[(sinα+cosα)−13(sinα+cosα)(1−sinαcosα)] =(1+a2)3/2(sinα+cosα)3[3−1+sinαcosα] =(1+a2)3/23(1+a√1+a2)(2+a1+a2) =1+a3(2+2a2+a)