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

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

Mathematics
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|dw:1437252577947:dw|
You can use u-substitution. Let u = tan x
i havent learned u-sub yet, but thanks for your help

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Other answers:

  • phi
do you know d/dx tan x ?
  • phi
\[ d \tan x = \sec^2 x \ dx \] so your problem is \[ \int \tan x\ \ d \tan x \] or if you rename tan x = u \[ \int u \ du \]
  • phi
peachpi 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
\[\mathrm{Apply\:Integral\:Substitution}:\quad \int\limits f\left(g\left(x\right)\right)\cdot g^'\left(x\right)dx=\int\limits f\left(u\right)du,\:\quad u=g\left(x\right)\] \[u=\tan \left(x\right):\quad \quad du=\sec ^2\left(x\right)dx,\:\quad \:dx=\cos ^2\left(x\right)du\] \[=\int\limits \:u\sec ^2\left(x\right)\cos ^2\left(x\right)du\] \[=\int\limits \:udu\] \[\mathrm{Apply\:the\:Power\:\Rule}:\quad \int\limits x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -\] \[=\frac{u^{1+1}}{1+1}\] \[\mathrm{Substitute\:back}\:u=\tan \left(x\right)\] \[=\frac{\tan ^{1+1}\left(x\right)}{1+1}\] \[=\frac{\tan ^2\left(x\right)}{2}+c\]
Thank You!!

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