# Trig Substitution

• January 25th 2010, 09:14 AM
Mike9182
Trig Substitution
What did I do wrong?

$\int{\frac{dx}{x\sqrt{x^2+3}}}$

$\sin{\Theta}=\frac{\sqrt{3}}{\sqrt{x^2+3}}$

$\cos{\Theta}=\frac{x}{\sqrt{x^2+3}}$

$\tan{\Theta}=\frac{\sqrt{3}}{x}$

$=\int{\frac{1}{\frac{\sqrt{3}}{\tan{\Theta}}*\frac {x}{\cos{\Theta}}}*\frac{x^2*\sec^2{\Theta}}{-\sqrt{3}}d\Theta}$

$=\int{\frac{\tan{\Theta}}{\sqrt{3}}*\frac{\cos{\Th eta}}{x}*\frac{x^2*\sec^2{\Theta}}{-\sqrt{3}}d\Theta}$

$=\int{\frac{\tan{\Theta}*\cos{\Theta}*x*\sec^2{\Th eta}}{-3}d\Theta}$

$=\frac{-1}{3}\int{\tan{\Theta}*{\sec{\Theta}{d\Theta}}}$

The correct answer is $\frac{1}{\sqrt{3}}\int{\csc{\Theta}d\Theta}$
• January 25th 2010, 09:23 AM
Jhevon
Quote:

Originally Posted by Mike9182
What did I do wrong?

$\int{\frac{dx}{x\sqrt{x^2+3}}}$

$\sin{\Theta}=\frac{\sqrt{3}}{\sqrt{x^2+3}}$

$\cos{\Theta}=\frac{x}{\sqrt{x^2+3}}$

$\tan{\Theta}=\frac{\sqrt{3}}{x}$

$=\int{\frac{1}{\frac{\sqrt{3}}{\tan{\Theta}}*\frac {x}{\cos{\Theta}}}*\frac{x^2*\sec^2{\Theta}}{-\sqrt{3}}d\Theta}$

$=\int{\frac{\tan{\Theta}}{\sqrt{3}}*\frac{\cos{\Th eta}}{x}*\frac{x^2*\sec^2{\Theta}}{-\sqrt{3}}d\Theta}$

$=\int{\frac{\tan{\Theta}*\cos{\Theta}*x*\sec^2{\Th eta}}{-3}d\Theta}$

$=\frac{-1}{3}\int{\tan{\Theta}*{\sec{\Theta}{d\Theta}}}$

The correct answer is $\frac{1}{\sqrt{3}}\int{\csc{\Theta}d\Theta}$

i don't know what you're doing. whatever it is, it is not trig sub.

the correct way is as follows:

$x = \sqrt 3 \tan \theta$

$\Rightarrow dx = \sqrt 3 \sec^2 \theta ~d \theta$

$\int \frac {\sqrt 3 \sec^2 \theta ~d \theta}{\sqrt 3 \tan \theta \cdot \sqrt 3 \sec \theta}$

$= \frac 1{\sqrt 3} \int \csc \theta ~d \theta$
• January 25th 2010, 09:32 AM
Mike9182
I set up a triangle and put the three main trig functions I found in my post, I just didn't place the triangle itself in my post because I wasn't sure how to do so using the latex editor.
• January 25th 2010, 09:32 AM
Krizalid
Quote:

Originally Posted by Jhevon
$= \frac 1{\sqrt 3} \int \csc \theta ~d \theta$

That's why I don't prefer the trig. substitution much, since people doesn't spot a quickly way in order to compute this integral, it's tricky.

So I stay away when they post and say "trig. sub.," since there're other ways to tackle a certain integral.
• January 25th 2010, 09:48 AM
Jhevon
Quote:

Originally Posted by Mike9182
I set up a triangle and put the three main trig functions I found in my post, I just didn't place the triangle itself in my post because I wasn't sure how to do so using the latex editor.

you set up the triangle after you see what equation you should use. you should have seen that $x = \sqrt 3 \tan \theta$ and set up the triangle based on that. then you wouldn't have this $\tan \theta = \frac {\sqrt 3}x$ business. be sure to review trig sub so you know what equation to start with and hence how to set up the right triangle (no pun intended, well, maybe a little :D)

Quote:

Originally Posted by Krizalid
That's why I don't prefer the trig. substitution much, since people doesn't spot a quickly way in order to compute this integral, it's tricky.

So I stay away when they post and say "trig. sub.," since there're other ways to tackle a certain integral.

Haha, i know. i hate trig sub as well. This is what i thought of when i first saw the integral (then i sighed, grit my teeth and did what the OP asked for)

$\int \frac {dx}{x \sqrt {x^2 + 3}} = \int \frac {x~dx}{x^2 \sqrt {x^2 + 3}}$

Let $u^2 = x^2 + 3$ and the integral becomes

$\int \frac {du}{u(u^2 - 3)} = \frac 13 \int \left( \frac u{u^2 - 3} - \frac 1u \right)~du$

is there an easier way?
• January 25th 2010, 10:00 AM
Mike9182
Quote:

Originally Posted by Mike9182

$\int{\frac{dx}{x\sqrt{x^2+3}}}$

I set up the triangle so that the hypotenuse was $\sqrt{x^2+3}$, the adjacent angle to theta was $x$, and the opposite angle to theta was $\sqrt{3}$. I'm guessing that when the triangle is set up the constant should be placed adjacent to theta when possible in order to make the process of working the problem out easier. Is this correct?
• January 25th 2010, 10:09 AM
Jhevon
Quote:

Originally Posted by Mike9182
I set up the triangle so that the hypotenuse was $\sqrt{x^2+3}$, the adjacent angle to theta was $x$, and the opposite angle to theta was $\sqrt{3}$. I'm guessing that when the triangle is set up the constant should be placed adjacent to theta when possible in order to make the process of working the problem out easier. Is this correct?

no, you don't decide where the things go. your equation does.

you had the expression $x^2 + 3 = x^2 + (\sqrt 3)^2$

this is of the form $x^2 + a^2$. the proper trig sub is $x = a \tan \theta$

So that is $x = \sqrt 3 \tan \theta$ here.

This means that $\tan \theta = \frac x{\sqrt 3} = \frac {\text{opposite}}{\text{adjacent}}$ which means that:

Your acute angle is $\theta$
The opposite side is $x$
The adjacent side is $\sqrt 3$ and
The hypotenuse is $\sqrt {x^2 + 3}$

The radical expression is not always the hypotenuse, our expression $x^2 + a^2$ dictated where we put stuff