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Thread: Few more integral questions.

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

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

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

    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 . . .


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

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

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



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

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

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

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

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

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

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

    $\displaystyle \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:

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

    ask yourself what is the derivative of:

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

    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 $\displaystyle \displaystyle \sin\sqrt{\theta}=u\Rightarrow \frac{\cos\sqrt{\theta}}{\sqrt{\theta}}d\theta=2du$.
    Then $\displaystyle \displaystyle \int\frac{2}{u^2}du=-\frac{2}{u}+C$
    So $\displaystyle \displaystyle \int\frac{\cos\sqrt{\theta}}{\sqrt{\theta}\sin ^2\sqrt{\theta}}d\theta=-\frac{2}{\sin\sqrt{\theta}}+C$
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