Q.

If ∫-π/2π/282cosx dx(1+esinx)(1+sin4x)=απ+βloge(3+22), where α,β are integers, then α2+β2 equals ____ .         [2024]


Ans.

(8)

Let I=∫-π/2π/282cosx dx(1+esinx)(1+sin4x)

Using ∫-aaf(a+b-x) dx=∫-aaf(x) dx

⇒2I=∫-π/2π/2(82cosx(1+esinx)(1+sin4x)+esinx82cosx(esinx+1)(1+sin4x))dx

=∫-π/2π/282cosx 1+sin4xdx=2∫0π/282cosx 1+sin4xdx      [∵  ∫-aaf(x) dx=∫0af(x) dx, if f(-x)=f(x)]

Put t=sinx ⇒ dt=cosx dx 

∴  I=82∫01dt1+t4=42∫012dt1+t4

=42∫01(1+1/t2t2+1/t2)dt-42∫011-1t2t2+1t2dt

=42∫011+1t2(t-1t)2+2dt-42122[ln(1+1t-2)t+1t+2]01

=42·12tan-1(t-1t2)01-4222[ln(t+1t-2)(t+1t+2)]01

=2π+2ln(3+22)=απ+βloge(3+22)

⇒α=β=2

Then α2+β2=4+4=8