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

then

$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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    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.2012-09-28

2 Answers 2

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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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    @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.2012-09-28
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We can do this by intgration by parts

$ I=\int tan^2 x \cdot sec^3x \space dx$ $=\int (sec^2 x-1)\cdot sec^3 x \space dx$ $=\int sec^5 x \space dx-\int sec^3 x \space dx$ $=sec^3 x \cdot tan x-\int 3sec^2 x \cdot sec x tan x \cdot tan x \space dx-\int sec^3 x \space dx $ $= sec^3x \cdot tanx-3I-I_1$ $ or, 4I= sec^3x \cdot tanx-I_1$

We can now find out the second integral

$I_1=\int sec^3 x ~dx=\int sec x ~sec^2x~dx=sec x~tan x-\int sec x~(sec^2 x-1)dx=sec x~tanx-I_1+\int sec x~dx$

$or, 2I_1=secx~tan x+ln|sec x+tan x|$

Now back to original problem:

$I={1 \over 4}sec^3x\cdot tanx-{1 \over 8}(sec x~tan x+ln|sec x+tan x|)+c$