Indefinite integral of secant cubed

I need to calculate the following indefinite integral:

$$I=\int \frac{1}{\cos^3(x)}dx$$

I know what the result is (from Mathematica):

$$I=\tanh^{-1}(\tan(x/2))+(1/2)\sec(x)\tan(x)$$

but I don't know how to integrate it myself. I have been trying some substitutions to no avail.

Equivalently, I need to know how to compute:

$$I=\int \sqrt{1+z^2}dz$$

which follows after making the change of variables $z=\tan x$.

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We have an odd power of cosine. So there is a mechanical procedure for doing the integration. Multiply top and bottom by $\cos x$. The bottom is now $\cos^4 x$, which is $(1-\sin^2 x)^2$. So we want to find $$\int \frac{\cos x\,dx}{(1-\sin^2 x)^2}.$$ After the natural substitution $t=\sin x$, we arrive at $$\int \frac{dt}{(1-t^2)^2}.$$ So we want the integral of a rational function. Use the partial fractions machinery to find numbers $A$, $B$, $C$, $D$ such that $$\frac{1}{(1-t^2)^2}=\frac{A}{1-t}+\frac{B}{(1-t)^2}+ \frac{C}{1+t}+\frac{D}{(1+t)^2}$$ and integrate.

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 cool.. a wikipedia page just for one integral – becko Jun 7 '12 at 2:22

Hint: rewrite the integral as

$$\int \sec ^3 (x) \, dx$$

Recall the identity $\sec^2(x)=\tan^2(x)+1$.

So, substituting, you get

$$\int\sec(x)(\tan^2(x)+1) \, dx=\int\tan(x)\tan(x)\sec(x) \, dx+\int\sec(x) \, dx.$$

The first integral can be solved by $u$-substitution and integration by parts, while the second, is an identity.

$$\int\tan(x) \, d\sec(x) = \tan(x)\sec(x)-\int\sec(x) \, d\tan(x)$$

But $\int\sec(x) \, d\tan(x)$ is the original integral. So write an equation and solve for $\int \sec^3(x)dx$

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