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I would proceed by thus , let $y = [\sec (x)]^2 $


$$dy = 2 \cdot \sec(x) \cdot \sec(x) \cdot \tan(x) \cdot dx = 2 \cdot ( \sec (x))^2 \cdot \tan(x) \cdot dx $$

so, $$ 2 \tan^2(x) \sec^2 (x) dx = \sec(x) \cdot \tan(x) \cdot dy = y(y-1)^\frac{1}{2} \cdot dy $$

since $$\sec(x) = y^{\frac{1}{2}}$$ and by considering positive square roots only $\tan y = ( \sec^2(x) - 1)^{1/2} = (y - 1)^{1/2}$. Thus the substitution $y = \sec^2 x$ yields $$ 2 \int \tan^2 (x) \sec^3(x) dx = \int (y(y - 1) )^{1/2} dy $$ and this later form can be reduced to the standard form $\int(z^2 - a^2)^{1/2} dz$ since $$ y(y-1)=(y-(1/2))^2 - (1/2)^2 . $$ What are the other ways to integrate this expression, except for the substitution

$\tan^2(x)^2=\sec (x)^2 -1$ which gives $$ \sec^5(x) - \sec^3 (x) $$ in the integrand which I quite don't like.

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I feel like I'm missing something obvious here, but all I can think of so far is some integration by parts ugliness... – Mike Sep 28 '12 at 8:56
Perhaps ugliness cannot be avoided! Use reduction formula for Sec^n(x) and keep working. It can be done, and the result does not have to be cute. – Maesumi Sep 28 '12 at 9:15
Now I think I know what's throwing me. The sec exponent is odd and the tan exponent is even. Any other combination would be 100 times simpler. – Mike Sep 28 '12 at 9:56
up vote 7 down vote accepted

Let you want to solve $\int R\big(\sin(x),\cos(x)\big)dx$ and you know that $$R\big(\sin(x),-\cos(x)\big)\equiv -R\big(\sin(x),\cos(x)\big)$$ then you can take $\sin(x)=t$ for a good substitution. We have here $$\int \tan^2(x)\sec^3(x)dx=\int \frac{\sin^2(x)dx}{\cos^5(x)}$$ and we can see the above statement is true for the last integrand. By taking $\sin(x)=t$, we have $$\int\frac{t^2}{(1-t^2)^3}dt$$ which can be solve by fractions method.

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@ Babak Sorouh Hm, I like that approach. – Souvik Dey Sep 28 '12 at 9:18
@Souvik: If you have $\int R\big(\sin(x),\cos(x)\big)dx$ where in $$R\big(-\sin(x),\cos(x)\big)\equiv -R\big(\sin(x),\cos(x)\big)$$ then you can take $\cos(x)=t$ instead. – Babak S. Sep 28 '12 at 9:24

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