Trig Integration Help: What & How to Use Trig Substitutions

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Trigonometric substitution involves replacing a variable in an integral with a trigonometric function, such as x = sin(u) or x = tan(u), to simplify the integral using trigonometric identities. After performing the substitution and integrating, the original variable is substituted back to obtain the final answer. A sample problem illustrates this process, demonstrating how to transform the integral ∫ dx/√(a² - x²) using x = a sin(θ). Participants discuss specific integrals and the simplification steps involved, emphasizing the importance of trigonometric identities in the process. The conversation highlights the challenges and solutions encountered while working through calculus problems.
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help wat is mean by trigonometric substituations? how do i use it/? thx
 
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hah, another part of calc 2 that I've sort of forgotten.., let me get out my calc book and ill get back :)
 
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
 
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
 
i do remember that
sin^2 + cos^2 = 1

so there for, you can rearrange...

... wait, I am thinking of trig identities.

hrm, maybe i shouldn't be doing this calc stuff... i have too many brain farts.
 
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
 
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
 
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? wat happen to my latex graphic?
 
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
 
  • #11
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 let's 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.
 
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  • #13
That keeps going to the index of the General Math forum, not a specific thread. Which specific thread in General Math?

cookiemonster
 
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