# Integral of $\int \tan(x) \sec^3 {x}\, dx$

I am trying to find the integral of $$\int \tan x \sec^3 x dx$$

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

This gets me nowhere since I get a $\sec^2 x$ derivative with tan subsitution so I try something else.

$$\int \frac {\sin x}{\cos x} \frac{1}{\cos^3x} dx$$

$$\int \frac {\sin x}{\cos^4 x} dx$$

$u = \cos x$ $du = -\sin x$

$$\int \frac {-1}{u^4} du$$

$$-1\int {u^{-4}} du$$

$$-1 \frac{u^{-3}}{3}$$ $$\frac{-1}{3\cos^3 x}$$

This for some reason is wrong.

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Your calculation is correct except for the last two lines. The integration of $-u^{-4}$ is $u^{-3}/3$, not $-u^{-3}/3$. – Sangchul Lee Jun 3 '12 at 17:24
You just missed a minus at the end. – Artem Jun 3 '12 at 17:24
(and for all of those constant-fans - add a +C) – H. M. Šiljak Jun 3 '12 at 17:26
Another method would be to note that $\tan x \, \sec^3 x = (\tan x \, \sec x)(\sec^2 x)$ and that the differential of $\sec x$ is $\tan x \, \sec x$. A method called spotting the presence of the derivative. – user26069 Jun 3 '12 at 17:30

You missed a $-$ sign in the denominator. $$\int x^{n} \ dx = \frac{x^{n+1}}{n+1} + C$$ when $n \neq -1$.
So when $n=-4$ you get $$\int x^{-4} \ dx = -\frac{1}{3} \cdot x^{-3} +C$$
There is still an easier method of doing this: $$\int \tan{x}\cdot \sec^{3}(x) \ dx = \int \tan{x} \cdot \sec{x} \cdot \sec^{2}(x) \ dx = \int t^{2} \ dt$$ by putting $t=\sec{x}$.
@jordan: Great that you were able to come up with a solution on your own. Here is an exercise : $\int\frac{1}{\sqrt{\sin{x} \cdot \cos{x}}} \ dx$ – user9413 Jun 3 '12 at 17:45