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Question

If I=π4π4esec x|sin x|sec2x dx(1+esin x), then

A
I is an irrational number
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B
I=e2e
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C
ln(I+e)=2
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D
I is a rational number
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Solution

The correct option is C ln(I+e)=2
I=π4π4esec x|sin x|sec2x dx(1+esin x)... (1)
Using baf(x)dx=baf(a+bx)dx
I=π4π4esec x|sin x|sec2x dx(1+esin x)... (2)
Add (1) & (2)
2I=π4π4esec x|sin x|sec2x dxI=π40esec xsec x tan x dx
esec x=tI=e2edt=e2e

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