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amyna
 one year ago
intergate (opposite of differentiation)
(tanxsec^2x) dx
i dont understand how to solve it!!
Thank for your help!
amyna
 one year ago
intergate (opposite of differentiation) (tanxsec^2x) dx i dont understand how to solve it!! Thank for your help!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can use usubstitution. Let u = tan x

amyna
 one year ago
Best ResponseYou've already chosen the best response.1i havent learned usub yet, but thanks for your help

phi
 one year ago
Best ResponseYou've already chosen the best response.0\[ d \tan x = \sec^2 x \ dx \] so your problem is \[ \int \tan x\ \ d \tan x \] or if you rename tan x = u \[ \int u \ du \]

phi
 one year ago
Best ResponseYou've already chosen the best response.0peachpi was suggesting that you rename tan x u = tan x then you find du du = sec^2 x dx and solving for dx dx= du/sec^2 x replace tan x with u and dx with du/sec^2 x and you get \[ \int u \ du \] which you integrate (to get (1/2) u^2 +C) then replace u with tan x to get the final answer

DecentNabeel
 one year ago
Best ResponseYou've already chosen the best response.0\[\mathrm{Apply\:Integral\:Substitution}:\quad \int\limits f\left(g\left(x\right)\right)\cdot g^'\left(x\right)dx=\int\limits f\left(u\right)du,\:\quad u=g\left(x\right)\] \[u=\tan \left(x\right):\quad \quad du=\sec ^2\left(x\right)dx,\:\quad \:dx=\cos ^2\left(x\right)du\] \[=\int\limits \:u\sec ^2\left(x\right)\cos ^2\left(x\right)du\] \[=\int\limits \:udu\] \[\mathrm{Apply\:the\:Power\:\Rule}:\quad \int\limits x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne \] \[=\frac{u^{1+1}}{1+1}\] \[\mathrm{Substitute\:back}\:u=\tan \left(x\right)\] \[=\frac{\tan ^{1+1}\left(x\right)}{1+1}\] \[=\frac{\tan ^2\left(x\right)}{2}+c\]
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