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Math Help - More integrals

  1. #1
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    More integrals

    In the attached picture for #5 what would I use as u, since no term from the problem can be divided out of the original function...

    And for #6 I have no idea what I would use as u either. Is there a method I can use to determine this easily in problems like this, because it seems like when people give me the first and second step I can do the rest easily.
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  2. #2
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    Hello, fattydq!

    Exactly where is your difficulty with #5?


    5)\;\;\int\cot(15x)\,dx

    There is a standard integration formula: . \int \cot u\,du \;=\;\ln|\sin u| + C

    And obviously, we let: u = 15x

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  3. #3
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    Quote Originally Posted by fattydq View Post
    In the attached picture for #5 what would I use as u, since no term from the problem can be divided out of the original function...

    And for #6 I have no idea what I would use as u either. Is there a method I can use to determine this easily in problems like this, because it seems like when people give me the first and second step I can do the rest easily.
    for Q5 just use u = 15x, then

    \int\cot(15x)\,dx = \frac{1}{15}\int\cot(u)\,du

    for Q6 I would write the integrand as

    \begin{array}{rcl}<br />
\dfrac{x}{(x+16)^{1/4}} &=& \dfrac{x+16-16}{(x+16)^{1/4}} \\<br />
&=& (x+16)^{3/4} - \dfrac{16}{(x+16)^{1/4}}\end{array}

    Hope this helps.
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  4. #4
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    Quote Originally Posted by Rincewind View Post
    for Q5 just use u = 15x, then

    \int\cot(15x)\,dx = \frac{1}{15}\int\cot(u)\,du

    for Q6 I would write the integrand as

    \begin{array}{rcl}<br />
\dfrac{x}{(x+16)^{1/4}} &=& \dfrac{x+16-16}{(x+16)^{1/4}} \\<br />
&=& (x+16)^{3/4} - \dfrac{16}{(x+16)^{1/4}}\end{array}

    Hope this helps.
    OK but how would I figure out the antiderivative of cot?
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  5. #5
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    Quote Originally Posted by fattydq View Post
    OK but how would I figure out the antiderivative of cot?
    It is is many tables of standard integrals. To do it from first principles.

    I = \int\cot(u)du = \int\frac{\cos(u)}{\sin(u)}du

    let t = \sin(u), therefore dt/du = \cos(u) so

    I = \int\frac{dt}{t} = \ln|t| + C = \ln|\sin u| + C

    Hope this helps.
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  6. #6
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    Quote Originally Posted by Rincewind View Post
    for Q5 just use u = 15x, then

    \int\cot(15x)\,dx = \frac{1}{15}\int\cot(u)\,du

    for Q6 I would write the integrand as

    \begin{array}{rcl}<br />
\dfrac{x}{(x+16)^{1/4}} &=& \dfrac{x+16-16}{(x+16)^{1/4}} \\<br />
&=& (x+16)^{3/4} - \dfrac{16}{(x+16)^{1/4}}\end{array}

    Hope this helps.
    OK I understand that I'd use (x+16)^(3/4) as u and thus du would be 3/4(x+16)^1/4 and I'd multiply it by 4/3 to even it out so I'm left with 4/3 the integral of -16, 4 times 16 is 64 so my final answer is -64/3 but it's incorrect apparently...where am I going wrong?
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  7. #7
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    Quote Originally Posted by fattydq View Post
    OK I understand that I'd use (x+16)^(3/4) as u and thus du would be 3/4(x+16)^1/4 and I'd multiply it by 4/3 to even it out so I'm left with 4/3 the integral of -16, 4 times 16 is 64 so my final answer is -64/3 but it's incorrect apparently...where am I going wrong?
    I'd just use u = (x+16)

    \begin{array}{rcl}\int\dfrac{x\,dx}{\sqrt[4]{x+16}} &=& \int u^{3/4} - 16u^{-1/4}\,du\\<br />
&=& \dfrac{4}{7}u^{7/4} - \dfrac{64}{3}u^{3/4} + C.\end{array}
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