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Question

If I=2ππ4π4dx(1+esinx)(2cos2x),
then 27 I2 equals to

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Solution

I=2ππ4π4dx(1+esinx)(2cos2x)(i)
Using the property:
abf(x) dx=abf(a+bx) dx
I=2ππ4π4dx(1+esinx)(2cos2x)
I=2ππ4π4esinx dx(1+esinx)(2cos2x)(ii)
Adding equation (i) and (ii), we get-
2I=2ππ4π4dx(2cos2x)
aaf(x)dx=2a0f(x)dxEven function
2I=4ππ40dx(1+2sin2x)
I=2ππ40sec2x dx(sec2x+2tan2x)
I=2ππ40sec2x dx(1+tan2x+2tan2x)
Put tanx=usec2x dx=du
I=2π10du(1+3u2)=[23πtan1(3u)]10
I=23π×(π30)=233
27I2=27×(233)2=4

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