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Math Help - Trigonometric Subsitution

  1. #1
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    Trigonometric Subsitution

    Hello peeps,

    What I need help on is clarifying if the answer for this integral is correct because I have did it in two ways and I'm not sure if both of them are correct.

    \int x^3\sqrt{9-x^2}\,dx

    x = 3sin\Theta
    dx = 9sin^2\Theta d\Theta\

    for the square root I have:
    \sqrt{9 - x^2}\ = \sqrt{9- 9 sin^2\Theta}\ = \sqrt{9(1-sin^2\Theta)}\ = \sqrt{9cos^2\Theta}\ = 3cos\Theta

    <br />
\int 27sin^3\Theta 3cos\Theta 9sin^2\Theta d\Theta\<br />
=<br />
\int 729sin^3\Theta cos\Theta sin^2\Theta d\Theta\<br />

    From here I did the equation 2 ways, please let me know if one of the ways is incorrect.

    1st way:
    <br />
\int 27sin^3\Theta 3cos\Theta 9sin^2\Theta d\Theta\<br />

    <br />
u = sin\Theta<br />
du = cos\Theta d\Theta<br />


    <br />
729\int u^3 u^2 du\<br />
=<br />
729\int u^5 du\<br />
= 729 \frac{u^6}6 +C<br />
= 729 \frac{sin^6}6 +C<br />
= 729 \frac{\frac{x}{3}^6}{6} + C<br />


    2nd way

    <br />
\int 729sin^3\Theta 3cos\Theta sin^2\Theta d\Theta\<br />
=<br />
\int 729sin^4\Theta cos\Theta sin\Theta d\Theta\<br />

    <br />
u = cos\Theta<br />
du = -sin\Theta d\Theta<br />

    <br />
\int 729 (1-cos^2\Theta)^2 cos\Theta sin\Theta d\Theta\<br />
=<br />
-\int 729 u(1-u^2)^2 du<br />

    <br />
= -(1 - 2u^2 + u^4)u<br />
= (1 + 2u^2 - u^4)u<br />
= u + 2u^3 - u^5 <br />

    <br />
= \frac{u^2}{2} + 2\frac{u^4}{4} - \frac{u^6}{6} <br />
= \frac{cos\Theta^2}{2} + 2\frac{cos\Theta^4}{4} - \frac{cos\Theta^6}{6} + C<br />

    <br />
=729 \frac{\frac{\sqrt{9-x^2}^2}{3}}{2} + \frac{\frac{\sqrt{9-x^2}^4}{3}}{4} +<br />
\frac{\frac{\sqrt{9-x^2}^6}{3}}{6} + C<br />
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ff4930 View Post
    Hello peeps,

    What I need help on is clarifying if the answer for this integral is correct because I have did it in two ways and I'm not sure if both of them are correct.

    \int x^3\sqrt{9-x^2}\,dx

    x = 3sin\Theta
    \color {red}dx = 9sin^2\Theta d\Theta\ ...
    you messed up in this line, so everything afterwards is incorrect

    what is the derivative of 3 \sin \theta ?
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  3. #3
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    .... 3 cos \Theta
    I feel like an idiot.
    I was thinking of x^2 for some reason, thanks anyways.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ff4930 View Post
    .... 3 cos \Theta
    I feel like an idiot.
    we all make mistakes. try again
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  5. #5
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    Ok, on the second try...

    I get

    <br />
\int 3^3 sin^3\Theta 3cos\Theta 3cos\Theta d\Theta\<br />
=<br />
\int 3^5 sin^2\Theta sin\Theta cos^2\Theta d\Theta\<br />
    <br />
=<br />
3^5 \int (1-cos^2\Theta) sin\Theta cos^2\Theta d\Theta\<br />
let u = cos\Theta<br />
du = -sin\Theta d\Theta<br />
    <br />
-3^5 \int (1-u^2)u^2du\<br />
=<br />
3^5 \int u^2+u^4du\<br />

    <br />
=  3^5 \int \frac{u^3}{3}+\frac{u^5}{5}du\<br />
= 3^5 \frac{cos\Theta^3}{3}+\frac{cos\Theta^5}{5}du<br />
    <br />
= 3^5 \frac{\frac{\sqrt{9-x^2}^3}{3}}{3} +<br />
\frac{\frac{\sqrt{9-x^2}^5}{3}}{5}<br />

    Is this correct? is an even problem on my textbook so I cant look it up, thanks for your patience.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ff4930 View Post
    Ok, on the second try...

    I get

    <br />
\int 3^3 sin^3\Theta 3cos\Theta 3cos\Theta d\Theta\<br />
=<br />
\int 3^5 sin^2\Theta sin\Theta cos^2\Theta d\Theta\<br />
    <br />
=<br />
3^5 \int (1-cos^2\Theta) sin\Theta cos^2\Theta d\Theta\<br />
let u = cos\Theta<br />
du = -sin\Theta d\Theta<br />
    <br />
-3^5 \int (1-u^2)u^2du\<br />
=<br />
3^5 \int u^2+u^4du\<br />

    <br />
=  3^5 \int \frac{u^3}{3}+\frac{u^5}{5}du\<br />
= 3^5 \frac{cos\Theta^3}{3}+\frac{cos\Theta^5}{5}du<br />
    <br />
= 3^5 \frac{\frac{\sqrt{9-x^2}^3}{3}}{3} +<br />
\frac{\frac{\sqrt{9-x^2}^5}{3}}{5}<br />

    Is this correct? is an even problem on my textbook so I cant look it up, thanks for your patience.
    you can actually check your answer here. just type in your function, you would enter "(x^3)*Sqrt[9 - x^2]" --without the quotes of course. i will look over your work in a little while
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  7. #7
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    Quote Originally Posted by ff4930 View Post
    \int x^3\sqrt{9-x^2}\,dx
    Are you sure you need a trig. substitution?

    They're nice, but in this case, I would do the following

    According to the change of variables defined by u=9-x^2\implies du=-2x\,dx, the integral becomes to

    \int x^3\sqrt{9-x^2}\,dx=-\frac12\int(9-u)\sqrt u\,du

    Does that make sense?

    P.S.: by the way ff4930, you can use \theta, looks better.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Are you sure you need a trig. substitution?

    They're nice, but in this case, I would do the following

    According to the change of variables defined by u=9-x^2\implies du=-2x\,dx, the integral becomes to

    \int x^3\sqrt{9-x^2}\,dx=-\frac12\int(9-u)\sqrt u\,du

    Does that make sense?

    P.S.: by the way ff4930, you can use \theta, looks better.
    that's the way i would have done it. i don't like trig. substitution that much. but ff4930 wants it. i think he's required to do it that way.

    Quote Originally Posted by ff4930 View Post
    Ok, on the second try...

    I get

    <br />
\int 3^3 sin^3\Theta 3cos\Theta 3cos\Theta d\Theta\<br />
=<br />
\int 3^5 sin^2\Theta sin\Theta cos^2\Theta d\Theta\<br />
    <br />
=<br />
3^5 \int (1-cos^2\Theta) sin\Theta cos^2\Theta d\Theta\<br />
let u = cos\Theta<br />
du = -sin\Theta d\Theta<br />
    <br />
-3^5 \int (1-u^2)u^2du\<br />
=<br />
\color {red} 3^5 \int u^2+u^4du\<br />
...
    you messed up the signs in the last line shown here
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  9. #9
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    How you get (9-u) from x^3?

    Also, the problems are from a textbook and is on the section of trig. substitution.

    First time using the math tags, took me half hour to do the first post =[

    rats, didn't see the sign error, other than that the steps were fine?
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  10. #10
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    Quote Originally Posted by ff4930 View Post
    \int x^3\sqrt{9-x^2}\,dx
    Well... the integrator gives -\frac15(x^2+6)(9-x^2)^{3/2}+k

    So, you should get that

    Sometimes, trigs. substitutions are really cool, but it depends of the integral, of course.
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ff4930 View Post
    How you get (9-u) from x^3?

    Also, the problems are from a textbook and is on the section of trig. substitution.

    First time using the math tags, took me half hour to do the first post =[

    rats, didn't see the sign error, other than that the steps were fine?
    he said u = 9 - x^2 \implies x^2 = 9 - u. since du = -2x ~dx, the du took away one of the x's from the x^3 and we were left with x^2. now we just replaced the x^2 with the function of u i mentioned above
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Well... the integrator gives -\frac15(x^2+6)(9-x^2)^{3/2}+k

    So, you should get that
    not necessarily. sometimes the result of the same integral can have hugely different forms depending on what integration technique you use. often you can simplify one form to get the other, but sometimes you can't, the difference is absorbed in the arbitrary constant.

    we just need to take it step by step here and make sure ff4930 made no errors. which i am pointing out has he goes along...

    Sometimes, trigs. substitutions are really cool, but it depends of the integral, of course.
    i agree
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  13. #13
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    Talking

    Thanks guys, I really appreciate your help.

    I've had this math professor since calculus 1 and he has always let us student leave the answer in such a mess and does not require us to simplify the answer. So now I'm so used to just to leaving a really long answer and not bother simplifying.
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  14. #14
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    Quote Originally Posted by Jhevon View Post
    not necessarily. sometimes the result of the same integral can have hugely different forms depending on what integration technique you use. often you can simplify one form to get the other, but sometimes you can't, the difference is absorbed in the arbitrary constant.
    Of course dear Jhevon, I knew that

    It's true, sometimes the integrator gives awful answers, and we get beauty ones
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