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Thread: Integration by Substitution.

  1. #1
    Junior Member pearlyc's Avatar
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    Integration by Substitution.

    Hey there, the given question is ..

    $\displaystyle \int^2_1 \sqrt{x^2 + 1} \frac{dx}{x}$

    I am confused with the $\displaystyle \frac{dx}{x}$ part.

    Is the above equivalent to ..

    $\displaystyle \int^2_1 \sqrt{x^2 + 1} \frac{1}{x} dx$ ?

    If so, the approach for this question would be ..

    Letting $\displaystyle x = sinh^2 (x)$?

    -----

    What is this question trying to say? Why are the boundaries in x variable while the function is in t?

    $\displaystyle \frac{d}{dx} \int^{e^x}_{sin x} arcsin(t) log(t) dt$

    Thank you!
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  2. #2
    Eater of Worlds
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    For the last one, it is the second fundamental rule of calc.

    Here is a way to look at it:

    $\displaystyle \frac{d}{dx}\int_{h(x)}^{g(x)}f(t)dt=f(g(x))g'(x)-f(h(x))h'(x)$

    We get: $\displaystyle sin^{-1}(e^{x})log(e^{x})e^{x}-sin^{-1}(sin(x))log(sin(x))cos(x)$

    Now, simplify.

    For the first one, that should be $\displaystyle \int\frac{\sqrt{x^{2}+1}}{x}dx$.

    let $\displaystyle x=tan(t), \;\ dx=sec^{2}(t)dt$
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  3. #3
    Junior Member pearlyc's Avatar
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    Oh, so it is true that the $\displaystyle \frac{dx}{x}$ is equivalent to $\displaystyle \frac{1}{x} dx$?

    Thanks a lot for your help!!
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  4. #4
    Junior Member pearlyc's Avatar
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    one more question,

    is differentiating $\displaystyle sin[arctan(x)]$
    = $\displaystyle \frac{cos[arctan(x)]}{1 - x^2}$?

    is there anyway to further simplify this?
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  5. #5
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    Hello, pearlyc!

    Is the derivative of $\displaystyle \sin[\arctan(x)] \:=\:\frac{\cos[\arctan(x)]}{1 - x^2}$ ? . . . . no

    Is there anyway to further simplify this? . . . . yes
    The derivative is: .$\displaystyle \frac{\cos[\arctan(x)]}{1\: {\color{red}+}\: x^2} $


    $\displaystyle \text{We have: }\;\cos\underbrace{[\arctan(x)]}_{\text{angle }\theta}$

    Then: .$\displaystyle \theta \:=\:\arctan(x) \quad\Rightarrow\quad\tan\theta \:=\:\frac{x}{1} \:=\:\frac{opp}{adj}$

    So $\displaystyle \theta$ is in this right triangle:
    Code:
                            *
               ____      *  *
              √1+x   *     *
                   *        * x
                *           *
             * θ            *
          *  *  *  *  *  *  *
                    1

    We have: .$\displaystyle opp = x,\;adj = 1$
    From Pythagorus, we have: .$\displaystyle hyp = \sqrt{1+x^2}$
    Hence: .$\displaystyle \cos\theta \:=\:\frac{1}{\sqrt{1+x^2}}$

    And your answer becomes: .$\displaystyle \frac{\frac{1}{\sqrt{1+x^2}}}{1+x^2} \;=\;\frac{1}{(1+x^2)^{\frac{3}{2}}} $

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  6. #6
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    Hello, pearlyc!

    Oh, so it is true that the $\displaystyle \frac{dx}{x}$ is equivalent to $\displaystyle \frac{1}{x} dx$?

    Sorry, but I have to snicker . . .

    You're in Calculus and you're not sure if: .$\displaystyle \frac{y}{2} \:=\:\frac{1}{2}y$ . . . . ?

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  7. #7
    Junior Member pearlyc's Avatar
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    For the first one, that should be .

    let

    For that one, I got $\displaystyle \frac {sec^2 t}{sin t}$
    Am I on the right track ?

    To Soroban, it's okay to snicker It's through this I'll learn right! Tee hee. Thanks for your help too. I was out of touch for a long time, so I just wanna clear my doubts and reassure myself Thank you!
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