Finding value of ∫0∞xsin⁡x(x2+1)3dx

Berasiniz1

Berasiniz1

Answered

2022-01-24

Finding value of 0xsinx(x2+1)3dx

Answer & Explanation

enveradapb

enveradapb

Expert

2022-01-25Added 13 answers

I=0xsinx(1+x2)3dx=14sinx(1+x2)20=0+140cosx(1+x2)2dx
Now we have that I=14J(1) because we can take:
J(a)=0cosxa+x2dxJ(a)=0cosx(a+x2)2dx
Thus all we need to do is to find J(a) then take a derivate and set a=1.
J(a)=π2aeaJ(a)=π2(12a32ea1aea12a)
I=14J(1)=π8(12e1+12e1)=π8e

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