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Math Help - Trig Substitution

  1. #1
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    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}
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mike9182 View Post
    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

    So your integral becomes:

    \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
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  3. #3
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    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.
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    = \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.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mike9182 View Post
    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 )

    Quote Originally Posted by Krizalid View Post
    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?
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  6. #6
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    Quote Originally Posted by Mike9182 View Post

    \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?
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mike9182 View Post
    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
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