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AmTran_Bus
 one year ago
Help me integrate tan^3x
AmTran_Bus
 one year ago
Help me integrate tan^3x

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AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1dw:1434148868513:dw

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1dw:1434148890008:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sec^2 x  1 = tan^2 x

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1Didn't I do that in my step above?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So now distribute. Do you know the antiderivative of \(\tan x\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so from distributing you should get \[\int \tan x\sec^2x\,dx\int\tan x\,dx=\int\tan x\sec^2x\,dx\ln\sec x+C\] You mentioned a substitution, that's a good idea. What do you think might work?

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1I would say u= tanx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep, that should work. Another one you can use: write \(\tan x\sec x\sec x\), then \(u=\sec x\) will work too.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits \tan x ~dx=\int\limits \frac{ \sin x }{ \cos x }dx=\ln \left \cos x \right\]

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1dw:1434149575520:dw

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1Well, but I need to consider my du

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, if \(u=\tan x\), then \(du=\sec^2x\,dx\), yes?

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1So it should say integral ( u du  rest) right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah that's \[\int u\,du\ln\sec x+C\] and so on.

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1Nice. Let me evaluate that and plug in u and see if it is right.

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1Well, guess im stuck

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1I know part of the answer has to bedw:1434149823843:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If I had to guess what the problem was... Hint: Identities.

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1Hum. I gave up and looked at the book. The answer is dw:1434149984333:dw

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1But I just dont see it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int u\,du=\frac{1}{2}u^2+C\] by the power rule. You set \(u=\tan x\), so this becomes \(\dfrac{\tan ^2x}{2}+C\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was thinking: \[\begin{split} \int\tan^3(x)~dx &= \int \frac{\sin^3(x)}{\cos^3(x)}dx \\ &= \int \frac{1\cos^2(x)}{\cos^3(x)}~d(\cos(x)) &,&u=\cos(x) \\ &= \int\left(\frac{1}{u^3}\frac{1}{u}\right)~du \end{split}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Somehow it seems wrong though.

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1@SithsAndGiggles ok.

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1I forgot a basic step, you were right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@wio seems right to me. \[\int\frac{du}{u^3}+\int\frac{du}{u}=\frac{1}{2u^2}+\lnu+C=\frac{\sec^2x}{2}+\ln\sec x+C\]
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