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secx(secx+secxtanx)dx

secx(secx+secxtanx)dx

Grade:12

2 Answers

Vidyasagar Mohanraj
35 Points
8 years ago

we get on opening brackets as integral (sec^2x + sec^2x.tanx)dx

take sec62x common we get integral sec^2x(1+tanx)dx

let t = tanx

dt = sec^2x.dx

on substitution we get the previous integral to be simplified as

integral (1+t)dt

on integrating we get

t + t^2/2 + c

sub t= tanx 

tanx + tan^2x +c is the required answer

 

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Yogita Bang
39 Points
8 years ago

I = ∫sec^2x(1+tanx)dx

Put 1+tanx = t    => sec^2xdx = dt

I = ∫t.dt = t^2/2 = (1+tanx)^2/2

 

 

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