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secx(secx+secxtanx)dx
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 IF U FIND MY ANSWER HELPFULL PLEASE APPROVE IT BY CLICKING YES
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
IF U FIND MY ANSWER HELPFULL PLEASE APPROVE IT BY CLICKING YES
I = ∫sec^2x(1+tanx)dx Put 1+tanx = t => sec^2xdx = dt I = ∫t.dt = t^2/2 = (1+tanx)^2/2 If u like my answer then do approve it.......
I = ∫sec^2x(1+tanx)dx
Put 1+tanx = t => sec^2xdx = dt
I = ∫t.dt = t^2/2 = (1+tanx)^2/2
If u like my answer then do approve it.......
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