1. ## Integrating Sec(x)

Here is the original problem

(Sin(x))^2/ cos(x)

I used an identity and converted the numerator to 1 - (cos(x))^2

I broke apart the fraction and got

1 / cos(x) - cos(x)

I know how to integrate cos(x) but I haven't a clue as to how to integrate sec(x). Any help or hints is appreciated!

2. It is mostly just memorized. $\int{sec(x)}dx=ln(|sec(x)+tan(x))$

But, there is a cool way to go about it I am sure.

Let's integrate it then this way.

$\int{sec(x)}dx$

Rewrite as $\int{sec(x)\left(\frac{sec(x)+tan(x)}{sec(x)+tan(x )}\right)}dx$

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

Now, let $u=sec(x)+tan(x), \;\ du=(sec^{2}(x)+sec(x)tan(x))dx$

Now, make the subs and it is rather easy.

$\int\frac{1}{u}du=ln(u)$

Resub:

$\boxed{ln|sec(x)+tan(x)|+C}$

3. Do you know how to integrate tanx? And can you do integration by parts?

Because another way of reading the question is that

sin^2(x)/cosx = sinx.tanx

But without knowledge of integration by parts, this may not help.

4. Thanks alot!!!

5. Originally Posted by JonathanEyoon
I know how to integrate cos(x) but I haven't a clue as to how to integrate sec(x). Any help or hints is appreciated!
$\int secx .dx$
multiply the top and the bottom by $secx + tanx$ so you'll have
$\int secx . \frac{secx+tanx}{secx+tanx} .dx = \int \frac{sec^2x + secxtanx}{secx+tanx} .dx$

Notice that $sec^2x+secxtanx$ is the derivative of $secx+tanx$ which means that when you integrate the function, you'll get the log of that funcion...

which leads us to this solution:

$\int secx .dx = ln[secx+tanx] + C$

from that, we can get these solutions, which sometimes are more helpful:

$ln[tan\frac{x}{2} + \frac{\pi}{4}] + C$ or $ln[\frac{1+tan\frac{x}{2}}{1-tan{x}{2}}] + C$

6. Ugh I was late

7. Originally Posted by Gaal Dornick
Do you know how to integrate tanx? And can you do integration by parts?

Because another way of reading the question is that

sin^2(x)/cosx = sinx.tanx

But without knowledge of integration by parts, this may not help.
You wouldn't be able to use parts for this one. Not that I can see anyway...

8. Well, you could, but I realise now you still have to integrate secx

Put u=tanx and dv/dx=sinx

then du/dx=sec^2x and v= -cosx

so the integral of sinx.tanx= -tanx.cosx + (integral of)secx

So yup, not much use if you don't know secx, hence the post I guess.

Ah damn, my first post as well!