help integrating

**cowboys111** · May 18th 2008, 12:51 PM

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?

**Moo** · May 18th 2008, 12:52 PM

Hello,

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 $\sin x$ is almost the derivative of $\cos x$ .

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

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

**Chris L T521** · May 18th 2008, 01:06 PM

Originally Posted by Moo

Hello,

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

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

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

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

Then,

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

**cowboys111** · May 18th 2008, 01:09 PM

cool thank you!!

**Moo** · May 18th 2008, 01:13 PM

Originally Posted by Chris L T521

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

Then,

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

Actually, I've never dealt with csc, sec, etc...
What's the rule in here ?

**galactus** · May 18th 2008, 01:26 PM

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

Wonder why?.

For your benefit. Their derivatives:

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

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

$\frac{d}{dx}[sec(x)]=tan(x)sec(x)$

**Moo** · May 18th 2008, 01:28 PM

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

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 !

**galactus** · May 18th 2008, 01:32 PM

In my previous post, I went ahead and posted the derivatives of those functions.

**galactus** · May 18th 2008, 01:38 PM

Here are their antiderivatives, Moo:

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

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

$\int{csc(x)}dx=ln(|csc(x)-cot(x)|)+C=ln(|tan(\frac{x}{2})|)+C$

**Moo** · May 18th 2008, 01:41 PM

Really weird stuff

Thanks

**galactus** · May 18th 2008, 01:43 PM

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 $ln(\frac{-cos(x)}{sin(x)-1})$ instead of
$ln(sec(x)+tan(x))$

**Chris L T521** · May 18th 2008, 04:55 PM

Originally Posted by Moo

Actually, I've never dealt with csc, sec, etc...
What's the rule in here ?

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

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