If I=2ππ4∫−π4dx(1+esinx)(2−cos2x),
then 27I2 equals to
A
4.00
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B
4
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C
4.0
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Solution
I=2ππ4∫−π4dx(1+esinx)(2−cos2x)⋯(i)
Using the property: a∫bf(x)dx=a∫bf(a+b−x)dx I=2ππ4∫−π4dx(1+e−sinx)(2−cos2x) I=2ππ4∫−π4esinxdx(1+esinx)(2−cos2x)⋯(ii)
Adding equation (i) and (ii), we get- 2I=2ππ4∫−π4dx(2−cos2x) ∵a∫−af(x)dx=2a∫0f(x)dx⇒Even function 2I=4ππ4∫0dx(1+2sin2x) I=2ππ4∫0sec2xdx(sec2x+2tan2x) I=2ππ4∫0sec2xdx(1+tan2x+2tan2x)
Put tanx=u⇒sec2xdx=du I=2π1∫0du(1+3u2)=[2√3πtan−1(√3u)]10 I=2√3π×(π3−0)=23√3 ∴27I2=27×(23√3)2=4