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

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To evaluate the integral ∫ 4(tan^3 x) dx, the process begins by rewriting it as 4∫(tan^2 x)(tan x) dx. This is further transformed using the identity tan^2 x = sec^2 x - 1, leading to the separation of the integral into two parts: 4∫(sec^2 x)(tan x) dx and -4∫tan x dx. By substituting t = tan x, the first integral simplifies to 2tan^2 x, while the second integral evaluates to -4ln|cos x|. The final result is ∫ 4(tan^3 x) dx = 2tan^2 x + 4ln|cos x| + C.
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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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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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