# Thread: $$\int\sec{x}dx$$

1. ## [solved]

My professor gave me this one.

MAT 1740 – Assignment #1 Name:

Due: 9/22 at the beginning of class. Remember to include all details as listed in the “How To” worksheet for application problems for full credit.

1. Find the exact volume of the solid formed by revolving the region bounded by x = 0, , and y = 0 about the x-axis using
a) a vertical representative element (analytically).

Ok. Everything here is straightforward except for one little thing:

How in the hEck do I go about integrating sec(x)?

I know that I must have missed something here because we haven't done anything like this before.

I'm stuck. We haven't done trig sub. yet, so I'm at a loss. No u-substitution that I can think of is helping. If I could just get a little nudge from somebody, that'd be grrrrrrrrrreeeat! Thanks.

2. It's a standard integral isn't it? List of integrals of trigonometric functions - Wikipedia, the free encyclopedia

$\int \sec{ax} \, dx = \frac{1}{a}\ln{\left| \sec{ax} + \tan{ax}\right|}+C$

3. Originally Posted by e^(i*pi)
It's a standard integral isn't it? List of integrals of trigonometric functions - Wikipedia, the free encyclopedia

$\int \sec{ax} \, dx = \frac{1}{a}\ln{\left| \sec{ax} + \tan{ax}\right|}+C$
Yeah it is. But that word analytically is freaking me out. I know that we haven't learned a technique yet that would enable me to do this analytically. I'm gonna e-mail my professor and see what's up.

4. multiply by $\sec x+\tan x$ top and bottom, and then you'll see what to do next.

5. The trick is $\int \sec x\,dx = \int\sec x\cdot\frac{\sec x+\tan x}{\sec x+\tan x}\,dx = \int\frac{\sec^2x+\sec x\tan x}{\sec x+\tan x}\,dx$

Now let $u=\sec x+\tan x$. Therefore $du=\sec x\tan x+\sec^2x\,dx$. Now the integral is:

$\int\frac{1}{u}\,du = \ln|u|=\boxed{\ln|\sec x+\tan x|}$

6. Originally Posted by redsoxfan325
The trick is $\int \sec x\,dx = \int\sec x\cdot\frac{\sec x+\tan x}{\sec x+\tan x}\,dx = \int\frac{\sec^2x+\sec x\tan x}{\sec x+\tan x}\,dx$

Now let $u=\sec x+\tan x$. Therefore $du=\sec x\tan x+\sec^2x\,dx$. Now the integral is:

$\int\frac{1}{u}\,du = \ln|u|=\boxed{\ln|\sec x+\tan x|}$
Tricks are good but you have to know them and there's a slightly different trick for each one ....

The non-trick (well, actually there is a trick but it's the same trick all the time) standard way of integrating rational functions of sin x and cos x is to use the Weierstrass substitution:

Weierstrass Substitution Formulas

You will find examples at MHF if you search the threads.

7. other way to solve this is to note that

$\sec x=\frac{\cos x}{1-\sin ^{2}x}=\frac{1}{2}\left( \frac{1+\sin x+1-\sin x}{1-\sin ^{2}x} \right)\cos x=\frac{1}{2}\left( \frac{\cos x}{1-\sin x}+\frac{\cos x}{1+\sin x} \right).$

Thus,

$\int{\sec x\,dx}=\frac{1}{2}\ln \left| \frac{1+\sin x}{1-\sin x} \right|+k.$