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

[coverband]

(Tanx)^2(Secx)

 

[HallsofIvy]

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.

 

[adriank]

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).

 

[Gib Z]

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.

 

[Hurkyl]

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?

 

[adriank]

You're probably thinking of the u = tan(x/2) thing I mentioned above. :-)