Integrating Trigonometric Functions

[Jetru]

How do i integrate trigonometric functions? Say 'cosec x'?Is there some general process to follow?

[Alvaro]

[syntax="cpp"]#include <cmath>

double cosec(double x){
return 1/std::sin(x);
}

[/syntax]

[Jetru]

That's not integration. I'm not talking about C++. I'm talking about normal mathematical integration.

[Alvaro]

That's not integration. I'm not talking about C++. I'm talking about normal mathematical integration.

Ooops! I'm sorry. Software engineers often say "integrate" meaning "add something to an existing program". Yeah, this question makes much more sense. Unfortunately, I never liked finding primitives and I don't even remember the few tricks that I managed to learn.

[Big Dog]

http://tutorial.math.lamar.edu/AllBrows ... thTrig.asp

Hope that helps for some. If i remember correclty, which I may not, if you have the cosec x, you change it to 1/sin(x) and then intrigrate from there.

http://people.hofstra.edu/faculty/Stefa ... trig4.html

here is another site that has more rules for you and even shows the intregral of cosec x

[Zen]

http://integrals.wolfram.com/ -> check if you have integrated correct

[BlackDeath]

Numerical method: Simpson's rule

[Jetru]

Oh, cool thanks...

Numerical method: Simpson's rule

No offense, but doing it numerically is such a pain

[GeekDog]

It's much easier to implement computationally than analytic integration

[Alvaro]

It's much easier to implement computationally than analytic integration

Especially when an analytic solution doesn't exists (like the integrals of "exp(-x^2)" and "x^x").

[Jetru]

Never knew x^x had no analytic soln. I was actually trying that a week ago...

[Jetru]

How do I integrate (sin x)^n with respect to x? Don't want the answer, just the hints. Tried Parts... and it didn't look very pretty.

[GeekDog]

Try substituting y = sin(x), and then integrate by parts. Not sure if it'll work, but it's an idea!

[Jetru]

Um...Turns out the integral of (sin x)^n is recursive. Oh well.

[MXP]

That can be a good thing. If you know that integrating one side eventually will give you a term that looks like the original integral then you can use that like this:

int(f(x)) = a0(x) + a1(x) + a2(x) + ... + an(x) - int(f(x))
2int(f(x)) = a0(x) + a1(x) + a2(x) + ... + an(x)
int(f(x)) = 1/2 * (a0(x) + a1(x) + a2(x) + ... + an(x))

The functions ai(x) are functions that you found in the process of integration (for instance, while using integration by parts).