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Question

2ex1+sin2x1+cos2xdx=

A
exsecx+c
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B
ex tan x+c
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C
ex secx tan x+c
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D
ex cosec x+c
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Solution

The correct option is A exsecx+c
2ex1+sin2x1+cos2xdx=2exsinx+cosx2cos2xdx
ex(tanx secx+secx)dx
it is in the form of
ex(f(x)+f1(x))dx
=exf(x)+c
so
ex(tanx secx+secx)dx=exsecx+c
so,2ex1+sin2x1+cos2xdx=exsecx+c
while x ϵ(π/4,3π/4)

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