# help integrating

• May 18th 2008, 12:51 PM
cowboys111
help integrating
Hi im having trouble on where to start, im trying to itegrate sin(x)/cos^2(x). Should i make a u subsitution, if so for what?
• May 18th 2008, 12:52 PM
Moo
Hello,

Quote:

Originally Posted by cowboys111
Hi im having trouble on where to start, im trying to itegrate sin(x)/cos^2(x). Should i make a u subsitution, if so for what?

You can notice that $\displaystyle \sin x$ is almost the derivative of $\displaystyle \cos x$.

A direct substitution will be $\displaystyle t=\cos x$

--> $\displaystyle dt=-\sin x dx \implies dx=-\frac{1}{\sin x} dt$

(Wink)
• May 18th 2008, 01:06 PM
Chris L T521
Quote:

Originally Posted by Moo
Hello,

You can notice that $\displaystyle \sin x$ is almost the derivative of $\displaystyle \cos x$.

A direct substitution will be $\displaystyle t=\cos x$

--> $\displaystyle dt=-\sin x dx \implies dx=-\frac{1}{\sin x} dt$

(Wink)

You can also see that it can be written as : $\displaystyle \frac{sin(x)}{cos(x)}\cdot \frac{1}{cos(x)} = sec(x)tan(x)$

Then,

$\displaystyle \int sec(x)tan(x) \,dx = sec(x)+C$
• May 18th 2008, 01:09 PM
cowboys111
cool thank you!!
• May 18th 2008, 01:13 PM
Moo
Quote:

Originally Posted by Chris L T521
You can also see that it can be written as : $\displaystyle \frac{sin(x)}{cos(x)}\cdot \frac{1}{cos(x)} = sec(x)tan(x)$

Then,

$\displaystyle \int sec(x)tan(x) \,dx = sec(x)+C$

Actually, I've never dealt with csc, sec, etc...
What's the rule in here ? (Worried)
• May 18th 2008, 01:26 PM
galactus
I have had several tell me that in Europe they do not deal with sec,csc,cot at all.

Wonder why?.

$\displaystyle \frac{d}{dx}[csc(x)]=-cot(x)csc(x)$

$\displaystyle \frac{d}{dx}[cot(x)]=-csc^{2}(x)$

$\displaystyle \frac{d}{dx}[sec(x)]=tan(x)sec(x)$
• May 18th 2008, 01:28 PM
Moo
Quote:

Originally Posted by galactus
I have had several tell me that in Europe they do not deal with sec,csc,cot at all.

Wonder why?.

In France, we've never seen such formulae... Only dealing with cos, tan, sin, and that was enough ^^ (in my high school...can't tell about all the others). Why ? I don't know, maybe because cot, csc, sec are known with sin, cos, tan... ~
It's the same for the substitution. It seems that for you it's common to substitute, but here, we just recognize the derivative and apply formulae of antiderivatives we know :p

Actually, by "here" in my previous post, I meant the equality of Chris where an antiderivative of sec*tan is sec..

Edit : thanks a bunch !
• May 18th 2008, 01:32 PM
galactus
In my previous post, I went ahead and posted the derivatives of those functions.
• May 18th 2008, 01:38 PM
galactus
Here are their antiderivatives, Moo:

$\displaystyle \int{cot(x)}dx=ln(|sin(x)|)+C$

$\displaystyle \int{sec(x)}dx=ln(|sec(x)+tan(x)|)+C$

$\displaystyle \int{csc(x)}dx=ln(|csc(x)-cot(x)|)+C=ln(|tan(\frac{x}{2})|)+C$
• May 18th 2008, 01:41 PM
Moo
Really weird stuff :D

Thanks (Wink)
• May 18th 2008, 01:43 PM
galactus
Quote:

Why ? I don't know, maybe because cot, csc, sec are known with sin, cos, tan... ~
I reckon that's so. Same reason most calculators do not have them.
One thing I noticed about calculators that do have them (such as the Voyage 200, TI-92, TI-89), they have the functions but only display in terms of sine and cosine. For instance, if you wanted the antiderivative of sec(x), the calculator displays it as $\displaystyle ln(\frac{-cos(x)}{sin(x)-1})$ instead of
$\displaystyle ln(sec(x)+tan(x))$
• May 18th 2008, 04:55 PM
Chris L T521
Quote:

Originally Posted by Moo
Actually, I've never dealt with csc, sec, etc...
What's the rule in here ? (Worried)

Ah, I see that Galactus has explained it to you. They never taught that to you? At least you learned something new today! :D