# Indefinite Integral - I

1. Apr 28, 2013

### Saitama

1. The problem statement, all variables and given/known data
$$\int \sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}} \frac{\sec x}{\sqrt{1+2\sec x}}dx$$

2. Relevant equations

3. The attempt at a solution
The integral can be simplified to:
$$\int \sqrt{\frac{1-\cos x}{1+\cos x}} \frac{1}{\sqrt{\cos x} \sqrt{\cos x+2}}dx$$
Using $\cos x=2\cos^2x/2-1=1-2\sin^2x/2$, I end up with
$$\int \frac{\tan(x/2)}{\sqrt{2\cos^2(x/2)-1}\sqrt{2\cos^2(x/2)+1}}dx=\int \frac{\tan(x/2)}{\sqrt{4\cos^4(x/2)-1}}dx$$
I am stuck here.

Any help is appreciated. Thanks!

2. Apr 28, 2013

### SteamKing

Staff Emeritus
Your algebra has a slight error at the beginning of item 3. above. sec x = 1 / cos x. When this is combined with 1 / SQRT (cos x + 2), you should obtain 1 / SQRT (cos^2 x * (cos x + 2)).

3. Apr 28, 2013

### haruspex

I don't see an error there.
Pranav-Arora, you can eliminate the trig, if it helps. tan(x)dx = -d(cos(x))/cos(x). (I think that after a few substitutions you can get it to integrating sech.)

4. Apr 28, 2013

### Saitama

Substituting x/2=t, dx=2dt
$$2\int \frac{\tan(t)}{\sqrt{4\cos^4(t)-1}}dt$$
$\because \tan t=-d(\cos x)/(\cos x)$

$$2\int \frac{\tan(t)}{\sqrt{4\cos^4(t)-1}}dt=\int \frac{-d(\cos x)}{\cos x\sqrt{\cos^4(t)-(1/2)^2}}$$

Is this equivalent to integrating $\displaystyle \int \frac{-dt}{t\sqrt{t^4-a^2}}$ where a is some constant.

I am asking this because I have never used this type of method to solve integrals.

5. Apr 28, 2013

### SteamKing

Staff Emeritus
sec x is still not equal to 1 / SQRT(cos x).

6. Apr 28, 2013

### Saitama

I never wrote that it is.
I should have not skipped the steps. I simplified it the following way:
$$\frac{\sec x}{\sqrt{1+2\sec x}}=\frac{1}{\cos x \sqrt{1+2/\cos x}}=\frac{1}{\sqrt{\cos x} \sqrt{2+\cos x}}$$

7. Apr 28, 2013

### haruspex

Good. Now, what can you do after rewriting that as $\int \frac{-tdt}{t^2\sqrt{t^4-a^2}}$?

8. Apr 29, 2013

### Saitama

Let t^2=z or 2tdt=dz. The integral can be written as
$$\frac{-1}{2} \int \frac{dz}{z\sqrt{z^2-a^2}}$$
My notes say that the above integral evaluates to -1/(2a)(arcsec(x/a)) but the answer is in terms of arcsin.

9. Apr 29, 2013

### haruspex

well, arcsec(x/a) = arccos(a/x), and there's a pretty easy step from there to arcsin form.

10. Apr 30, 2013

### scurty

Furthermore, if you draw the right angled triangle and label the x and a sides correctly, you can relate all six trig functions to each other in terms of x and a.

11. Apr 30, 2013

### SammyS

Staff Emeritus
I know that we're past this but there is a significant error in the middle of this post.

You wrote that $\ \tan t=-d(\cos x)/(\cos x)\ .$

That should have been $\ \tan(t)\,dt=-d(\cos t)/(\cos t)\,,\$ which makes the next line:

$\displaystyle 2\int \frac{\tan(t)}{\sqrt{4\cos^4(t)-1}}dt=\int \frac{-d(\cos t)}{\cos (t)\ \sqrt{\cos^4(t)-(1/2)^2}}\ .$​

.

12. May 1, 2013

### Saitama

Ah, I should have taken care of that. I did not notice that while I was writing that post. Thank you!

Substituting cos(t)=z

$$\int \frac{-d(\cos t)}{\cos (t)\ \sqrt{\cos^4(t)-(1/2)^2}}=\frac{-1}{2} \int \frac{dz}{z\sqrt{z^2-(1/2)^2}}=-\sec^{-1}(2z)+C=-\arccos\left(\frac{1}{2z}\right)+C$$
Adding and subtracting $\pi/2$
$$\frac{\pi}{2}-\arccos\left(\frac{1}{2z}\right)-\frac{\pi}{2}+C$$
$$\arcsin\left(\frac{1}{2z}\right)+K=\arcsin\left(\frac{1}{2}\sec^2\frac{x}{2}\right)+K$$

Thank you haruspex and SammyS!