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Math Help - Few more integral questions.

  1. #1
    Junior Member cinder's Avatar
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    Few more integral questions.

    First one, \int x^4(x^5+2)^7dx. Is that as straight forward as it seems? Expand that out x^{39}+14x^{34}+84x^{29}+... and then integrate as normal? Or should I substitute, setting u=x^5+2 and \frac{du}{5}=x^4dx?

    Second one, \int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2(\sq  rt{\theta})} d\theta. No idea where to start.
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  2. #2
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    Quote Originally Posted by cinder View Post
    First one, \int x^4(x^5+2)^7dx. Is that as straight forward as it seems? Expand that out x^{39}+14x^{34}+84x^{29}+... and then integrate as normal? Or should I substitute, setting u=x^5+2 and \frac{du}{5}=x^4dx?
    Yes substitute,
    u=x^5+2 because its derivative appears almost outside.
    Second one, \int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2(\sq  rt{\theta})} d\theta. No idea where to start.
    Hint: First substitute t=\sqrt{\theta}.

    Hint: There is a second substitute
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  3. #3
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    Hello, cinder!

    Looks like you're a bit shaky about substitution . . .


    \int x^4(x^5+2)^7\,dx
    Let u = x^5+ 2\quad\Rightarrow\quad du = 5x^4\,dx\quad\Rightarrow\quad dx = \frac{du}{5x^4}

    Substitute: . \int x^4\cdot u^7\left(\frac{du}{5x^4}\right) \;=\;\frac{1}{5}\int u^7\,du \;= \;\frac{1}{40}u^8 + C

    Back-substitute: . \frac{1}{40}(x^5 + 2)^8 + C



    \int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2\!(\  sqrt{\theta})} d\theta

    Let \sqrt{\theta} = u\quad\Rightarrow\quad \theta = u^2\quad\Rightarrow\quad d\theta = 2u\,du

    Substitute: . \int\frac{\cos u}{u\sin^2\!u}(2u\,du) \;=\;2\int\frac{\cos u}{\sin^2\!u}\,du \;=\;2\int\left(\frac{1}{\sin u}\!\cdot\!\frac{\cos u}{\sin u}\right)\,du

    . . = \;2\int\csc u\cot u\,du \;=\;-2\csc u + C

    Back-substitute: . -2\csc(\sqrt{\theta}) + C

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  4. #4
    Junior Member cinder's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, cinder!

    Looks like you're a bit shaky about substitution . . .
    Yes, I'm just not sure what I should be substituting.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by cinder View Post
    First one, \int x^4(x^5+2)^7dx. Is that as straight forward as it seems? Expand that out x^{39}+14x^{34}+84x^{29}+... and then integrate as normal? Or should I substitute, setting u=x^5+2 and \frac{du}{5}=x^4dx?

    Second one, \int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2(\sq  rt{\theta})} d\theta. No idea where to start.
    Observe that 5 x^4 is the derivative of x^5+2 so:

    \int x^4(x^5+2)^7dx=\frac{1}{5}\int \left[\frac{d}{dx}(x^5+2)\right](x^5+2)^7dx=\frac{1}{40}(x^5+2)^8

    For:

    \int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2(\sq  rt{\theta})} d\theta,

    ask yourself what is the derivative of:

    \frac{1}{\sin(\sqrt{\theta})}

    RonL
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  6. #6
    MHF Contributor red_dog's Avatar
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    For the second integral you can make the substitution \displaystyle \sin\sqrt{\theta}=u\Rightarrow \frac{\cos\sqrt{\theta}}{\sqrt{\theta}}d\theta=2du.
    Then \displaystyle \int\frac{2}{u^2}du=-\frac{2}{u}+C
    So \displaystyle \int\frac{\cos\sqrt{\theta}}{\sqrt{\theta}\sin ^2\sqrt{\theta}}d\theta=-\frac{2}{\sin\sqrt{\theta}}+C
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