Evaluating ∫−π4π4(sin⁡x+2cos⁡x)3dx

veceriraby

veceriraby

Answered

2022-01-23

Evaluating π4π4(sinx+2cosx)3dx

Answer & Explanation

Damian Roberts

Damian Roberts

Expert

2022-01-24Added 14 answers

Just expand the binomial
(sinx+2cosx)3=sin3x+6sin2xcosx+12sinxcos2x+8cos3x
Now note that
π4π4sin3xdx=0=π4π412sinxcos2xdx
(because the functions inside the integrals are odd functions and since we are in a symmetric interval). The rest of the integrals are easy, since
π4π4(6sin2xcosx+8cos3x)dx=π4π4(2sin2x+8)cosxdx
and if we set u=sinx, the integral is much easier :
π4π4(2sin2x+8)cosxdx=1212(2u2+8)du=2323

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