# Simple Integral

1. Apr 10, 2006

### cscott

$$\int \tan x \cdot dx + \int \tan 2y \cdot dy = 0$$

I get

$$1 = C\sqrt{\cos 2y} \cos x$$

My init conditions were y(0) = pi/2 so I end up getting C as $\sqrt{-1}$ but the equation should be

$$-1 = \cos 2y \cos^2 x$$

Where's my error?

2. Apr 11, 2006

### HallsofIvy

Staff Emeritus
Did you forget that $\int \frac{du}{u}= ln |u|+ C$?

That is, did you forget the absolute value?

3. Apr 11, 2006

### cscott

Wouldn't I then have to add absolute signs around both cosines? Then C would equal 1 instead of -1.

I'm not sure where 1/u comes up... can't I just integrate tan x to get -ln |cos x| and tan 2y to get -1/2 ln |cos 2y|?

4. Apr 11, 2006

### HallsofIvy

Staff Emeritus
Yes, but the "1/u" is where ln|cos x| comes from: tanx= sin x/cos x. If you let u= cos x then du= -sin x dx so $\int tan x dx= -\int (1/u)du= --ln|u|+ C= -ln|cos x|+ C$. Of course, tan 2y= sin 2y/cos 2y. If you let u= cos 2y, then du= 2 sin 2y dy so $$\int tan 2y dy= -\frac{1}{2}\int (1/u)du= -\frac{1}{2}ln u+ C= -\frac{1}{2} ln|cos 2y|$$.

$$\int tan x dx+ \int tan 2y dy= -ln|cos x|- \frac{1}{2}ln|cos 2y|+ C$$

Of course, you didn't tell us what the actual problem is so I have no idea how you would decide what C is!

5. Apr 11, 2006

### cscott

Oops, I should have seen that.

Oh, but I did state the initial conditions in my OP! Here they are again: $y(0) = \pi/2$

I worked from where you left off (where we both agree) and I got
$$C^2 = |\cos 2y||\cos^2 x|$$

So with the initial conditions

$$C^2 = |-1||1| \Leftrightarrow C = \sqrt{1}$$

I can get -1 out of that, but why choose it over 1? Also, the answer stated in my OP (as given in my book) has no absolute value signs, why's that?

Thanks for your help.

Last edited: Apr 11, 2006
6. Apr 12, 2006

### benorin

Quick recap,

$$\int \tan x dx+ \int \tan 2y dy = -\ln |\cos x|- \frac{1}{2}\ln |\cos 2y|+ C=0$$​

notice that since

$$-1\leq \cos x\leq 1 \Rightarrow 0\leq |\cos x| \leq 1\Rightarrow -\infty \leq \ln |\cos x| \leq 0 \Rightarrow 0 \leq -\ln |\cos x| \leq \infty ,$$​

and likewise we have $$0 \leq - \frac{1}{2}\ln |\cos 2y| \leq \infty$$ so that

$$0 \leq -\ln |\cos x| - \frac{1}{2}\ln |\cos 2y| \leq \infty$$​

but

$$-\ln |\cos x| - \frac{1}{2}\ln |\cos 2y| +C=0$$​

so it must be that $$C\leq 0,$$ so choose the value $$C=-1$$.

7. Apr 12, 2006

### cscott

Thanks for your help.