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

1. amyna

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2. anonymous

You can use u-substitution. Let u = tan x

3. amyna

i havent learned u-sub yet, but thanks for your help

4. phi

do you know d/dx tan x ?

5. 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 $\int u \ du$

6. 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

7. DecentNabeel

$\mathrm{Apply\:Integral\:Substitution}:\quad \int\limits f\left(g\left(x\right)\right)\cdot g^'\left(x\right)dx=\int\limits f\left(u\right)du,\:\quad u=g\left(x\right)$ $u=\tan \left(x\right):\quad \quad du=\sec ^2\left(x\right)dx,\:\quad \:dx=\cos ^2\left(x\right)du$ $=\int\limits \:u\sec ^2\left(x\right)\cos ^2\left(x\right)du$ $=\int\limits \:udu$ $\mathrm{Apply\:the\:Power\:\Rule}:\quad \int\limits x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -$ $=\frac{u^{1+1}}{1+1}$ $\mathrm{Substitute\:back}\:u=\tan \left(x\right)$ $=\frac{\tan ^{1+1}\left(x\right)}{1+1}$ $=\frac{\tan ^2\left(x\right)}{2}+c$

8. amyna

Thank You!!