# Finding ∫sec²x dx with basic techniques.

1. Jun 6, 2012

### Bohrok

Not really a homework problem, but this is just a problem that's been bugging me for a while: how would you find ∫sec2x dx without "cheating" and using the simple fact that d/dx tanx = sec2x? I know it's possible using integration by parts, but I'm looking for a way using calc I level techniques, like u-substitution. So far no success. Any ideas?

2. Jun 6, 2012

### LCKurtz

But the most basic of all Calc I methods is to know the derivatives of the "standard" functions so you can recognize those antiderivatives on sight. So knowing the antiderivative of $\sec^2x$ on sight isn't "cheating". It is the standard Calc I method.

Last edited: Jun 6, 2012
3. Jun 6, 2012

### SammyS

Staff Emeritus
I don't see how arriving at an anti-derivative by recognizing that the integrand is the derivative of a standard function is "cheating" , but try the following.

This gets pretty ugly, but here goes ...

$\displaystyle \sec^2(x) = 1 + \tan^2(x)$
$\displaystyle = 1 + \frac{\sin^2(x)}{\cos^2(x)}$

$\displaystyle = 1 + \frac{1-\cos(2x)}{1+\cos(2x)}$​
Integrate that.

Unless I missed something, WolframAlpha does it with only substitutions (many of them). See the link: http://www.wolframalpha.com/input/?i=integrate+%281-cos%282+x%29%29%2F%281%2Bcos%282+x%29%29.

4. Jun 6, 2012

### vela

Staff Emeritus
The substitution u=sec x works.

5. Jun 6, 2012

### BloodyFrozen

Slightly deviating from SammyS's method and using substitutions,

Remember $sec^{2}(x) = 1 + tan^{2}(x)$

$$\int sec^{2}(x) dx = \int 1+ tan^2(x)dx = \int 1 + \frac{sin^{2}(x)}{cos^{2}(x)}dx = \int dx + \int sin(x)\frac{sin(x)}{cos^{2}(x)} dx$$

Integrating by Parts (second integral)
u = sin(x)
u'= cos(x)
v' = sin(x)/cos2(x)
v = 1/cos(x)

Last edited: Jun 6, 2012
6. Jun 7, 2012

### Bohrok

I got some great answers, thanks.
vela, u=secx was just what I was looking for

SammyS, that method looks promising (I was close to working it out like that before), but I kept running into integrating cot2x or csc2x, which are about the same level as integrating sec2x.

BloodyFrozen, I know parts will work; I was just seeing if it was possible using more basic integration techniques in calc I.

I know in a way this problem may seem a little silly by not just using u = tanx since du = sec2x dx in the first place It's kind of like trying to work out a limit problem algebraically without resorting to l'Hôpital's rule; those are the kinds of challenging problems that I like to try working out on my own.