Anyone know how to integrate (Tanx)^2(Secx) thanks

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To integrate (tan(x))^2(sec(x)), start by rewriting tan(x) and sec(x) in terms of sine and cosine, leading to the expression sin^2(x)/cos^3(x). By factoring out a cosine, the integral simplifies to sin^2(x)cos(x)/cos^4(x), which can be transformed using the substitution u = sin(x). Alternatively, using the substitution u = tan(x/2) allows for a different approach, yielding an integral that can be solved through partial fractions. The final result involves integrating sec^3(x) and sec(x), with the latter evaluated using logarithmic identities.
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(Tanx)^2(Secx)
 
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Not necessarily the simplest: tan(x)= sin(x)/cos(x) and sec(x)= 1/cos(x) so tan^2(x) sec(x)= sin^2(x)/cos^3(x). That involves cosine to an odd power so we can "factor out" one to use with a substitution. Specifically, multiply both numerator and denominator by cos(x) to get sin^2(x)cos(x)/cos^4(x). cos^4(x)= (cos^2(x))^2=(1- sin^2(x))^2 so let u= sin(x). Then du= cos(x) dx so the integral becomes
\int \frac{u^2}{(1-u^2)^2} du
That integral can be done by partial fractions.
 
Here's another substitution you could make, inspired by stereographic projection, and it always works when you have a rational function of trigonometric functions of x.

Let u = \tan(x/2). From that, using various trigonometric identities, obtain the following:
\sin x = \frac{2u}{1+u^2}; \cos x = \frac{1-u^2}{1+u^2}; dx = \frac{2 \,du}{1+u^2}
and in particular
\tan x = \frac{2u}{1-u^2}; \sec x = \frac{1+u^2}{1-u^2}.
Then your integral becomes
\int \tan^2 x \sec x \,dx = \int \frac{8u^2}{(1-u^2)^3} \,du
which can be done by partial fractions. Admittedly, you've ended up with a higher degree denominator than what HallsofIvy gets, but again it always reduces a rational function of trig functions to a rational function of u, which can be done by partial fractions (or if you're lucky, substitution).
 
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Expanding using Pythagorean Identities yields

\int \tan^2 x \sec x dx = \int \sec^3 x dx - \int \sec x dx

The second integral, upon multiplying both the numerator and denominator by (sec x + tan x), evaluates to log |sec x + tan x|+ C

The first requires integration by parts. You should be able to see the integral you want on both sides of the equation now, so isolate it and solve.
 
My calculus book gives an algorithm that can be used to solve all such kinds of trigonometric integrals. Doesn't yours have a similar thing?
 
You're probably thinking of the u = tan(x/2) thing I mentioned above. :-)
 
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