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Ellen's question at Yahoo! Answers ( Int 4 (tan ^3 x) dx )

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Ellen,

We have: $$\begin{aligned}\int 4\tan^3 x\;dx&=4\int (\tan^2x)(\tan x)\:dx\\&=4\int(\sec^2x-1)(\tan x)dx\\&=4\int(\sec^2x)(\tan x)\:dx-4\int\tan x\;dx\end{aligned}$$ If $t=\tan x$, then $dt=\sec^2x\;dx$ so $$\int(\sec^2x)(\tan x)\:dx=\int t\:dt=\frac{t^2}{2}=\frac{\tan^2x}{2}$$ On the other hand:
$$\int\tan x\;dx=\int \frac{\sin x}{\cos x}dx=-\ln |\cos x|$$ As a consequence: $$\boxed{\;\displaystyle\int 4\tan^3 x\;dx=2\tan^2x+4\ln|\cos x|+C\;}$$

P.S. Something must be wrong with Yahoo Answers, when I'm logged in, I can't access to the corresponding page.

Edit: Now, that is all right.
 
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