trig intergration

[expscv]

help wat is mean by trigonometric substituations? how do i use it/? thx

[KSCphysics]

hah, another part of calc 2 that i've sorta forgotten.., let me get out my calc book and ill get back :)

[cookiemonster]

You make a substitution of the form x = some trig function (like sin(u) or tan(u)) and then use trig properties to simplify the integral to a manageable form. Then, after integrating, you make the inverse substitution into the solution to get your final answer.

Look at the table at the bottom of Mathworld's entry and just make an up example.

Trigonometric Substitution:
http://mathworld.wolfram.com/TrigonometricSubstitution.html

Better yet, here's a sample problem for you. Try it out.

\int \frac{dx}{\sqrt{a^2 - x^2}}

cookiemonster

[cookiemonster]

Figured I should work out my own sample problem.

\int \frac{dx}{\sqrt{a^2 - x^2}}
x = a\sin{\theta}
dx = a\cos{\theta}d\theta
\int\frac{a\cos{\theta}d\theta}{\sqrt{a^2(1-\sin^2{\theta})}}
\int\frac{a\cos{\theta}d\theta}{a\cos{\theta}}
\int d \theta
\theta = \arcsin{\frac{x}{a}}

cookiemonster

[KSCphysics]

i do remember that
sin^2 + cos^2 = 1

so there for, you can rearrange...

....... wait, im thinking of trig identities.

hrm, maybe i shouldnt be doin this calc stuff.... i have too many brain farts.

[cookiemonster]

Trig identities are good. Keep going.

You'll notice that that particular trig identity is used to get from the 4th step in my example to the 5th step.

cookiemonster

[expscv]

oh cool thx
this was my problem in solving it

\int \frac{1}{x^2\sqrt{1+x^2}} dx --->

x=tan{\theta}
my step after ur help

\int \frac {1}{tan^2{\theta}sec{\theta}}dx

x=tan{\theta} then dx=sec^2{\theta} d{\theta}

\int \frac {sec^2{\theta}}{Tan^2{\theta}sec{\theta}}d{\theta}

\int \frac {sec{\theta}}{tan^2{\theta}} d{\theta}

\frac{1}{2}\int \frac {2sec{\theta}}{1-sec^2{\theta}} d{\theta}


is this correct ? thx

[expscv]

??? wat happen to my latex graphic?

[cookiemonster]

You're working too hard.

Try to simplify

\int \frac {sec{\theta}}{tan^2{\theta}} d{\theta}

a bit. Once you do, there's an obvious substitution that completes the integral.

cookiemonster

[expscv]

~~i cant simplfied any more the last step i got is Ln(something)

btw i need some help in here
http://www.physicsforums.com/forumdisplay.php?f=73

thx

[cookiemonster]

What do you mean you can't simplify anymore?

\int \frac{\sec{\theta}}{\tan^2{\theta}} \, d\theta = \int \frac{1}{\cos{\theta}} \cdot \frac{\cos^2{\theta}}{\sin^2{\theta}} \, d\theta = \int \frac{\cos{\theta}}{\sin^2{\theta}} \, d\theta

There's a substitution that lets you evaluate this integral...

And which thread are you referring to?

cookiemonster

Edit: Gaah! Why won't LaTeX work?

Edit^2: Guess it does work.

[expscv]

oh thx subituation twice ~ wa sa

btw it is the link

http://www.physicsforums.com/forumdisplay.php?f=73

[cookiemonster]

That keeps going to the index of the General Math forum, not a specific thread. Which specific thread in General Math?

cookiemonster

[expscv]

oh its in hwk help grade k12 under

help needed maths thx
expscv

http://www.physicsforums.com/showthread.php?t=17453