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Math Help - Two substitution integral evaluation questions

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    Two substitution integral evaluation questions

    I've attached a picture, there's a few types of problems I'm having trouble with. I can easily solve simple integral problems involving substitution but these two are stumping me. For the first one I don't even know the deriv. of arc tan or know of a derivative that will be part of something else in the problem.

    As for the second one I don't even know what to make u or du because again no part of the problem is the other part's derivative.
    Attached Thumbnails Attached Thumbnails Two substitution integral evaluation questions-picture-1.png  
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    1. You should be able to find the derivative of arctan(x) in your calc book or Google it. I would suggest learning to derive it. Either way, it's not a secret to find if you have internet access. Once you look this up, the substitution will be obvious I hope.

    2. u=x-3
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    For the first, you should know that \frac d{dx}[\arctan u] = \frac{u'}{1+u^2}

    Do you see this form in your integrand?

    For the second one, let u = x - 3\text. Then du = dx and the limits become -3 to 3:

    \int_0^6(x-3)^{23}\,dx = \int_{-3}^3u^{23}\,du
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    Quote Originally Posted by Jameson View Post
    1. You should be able to find the derivative of arctan(x) in your calc book or Google it. I would suggest learning to derive it. Either way, it's not a secret to find if you have internet access. Once you look this up, the substitution will be obvious I hope.

    2. u=x-3


    1. Ok it's 5/25x^2+5 but that still doesn't divide evenly into 25x^2+1

    2.Wow that was easy, stupid oversight.
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    Quote Originally Posted by Reckoner View Post
    For the first, you should know that \frac d{dx}[\arctan u] = \frac{u'}{1+u^2}

    Do you see this form in your integrand?

    For the second one, let u = x - 3\text. Then du = dx and the limits become -3 to 3:

    \int_0^6(x-3)^{23}\,dx = \int_{-3}^3u^{23}\,du
    Yes but even knowing that it still doesn't divide evenly into another term in the problem
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    Quote Originally Posted by fattydq View Post
    Yes but even knowing that it still doesn't divide evenly into another term in the problem
    Let u = \arctan5x\Rightarrow du=\frac{5\,dx}{1+(5x)^2}=\frac{5\,dx}{1+25x^2}\te  xt.

    Then we have

    \int\frac{\arctan5x}{1+25x^2}dx

    = \frac15\int u\,du

    Simple, right?

    If you wanted, you could still do u=5x, but then you would need to do a second substitution ( v=\arctan u).
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    Quote Originally Posted by Reckoner View Post
    Let u = \arctan5x\Rightarrow du=\frac{5\,dx}{1+(5x)^2}=\frac{5\,dx}{1+25x^2}\te  xt.

    Then we have

    \int\frac{\arctan5x}{1+25x^2}dx

    = \frac15\int u\,du

    Simple, right?

    If you wanted, you could still do u=5x, but then you would need to do a second substitution ( v=\arctan u).

    It seems simple to you but I still don't understand. how does 25x^2+1 divide into 25x^2+5 and turn out to be five. I mean 2+4 divided by 2+ 2 isn't 2 just because the 2's can cancel out, if that makes sense.
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    Quote Originally Posted by fattydq View Post
    It seems simple to you but I still don't understand. how does 25x^2+1 divide into 25x^2+5 and turn out to be five. I mean 2+4 divided by 2+ 2 isn't 2 just because the 2's can cancel out, if that makes sense.
    Sorry, let me make it more clear.

    We are using u=\arctan5x, which gives du=\frac{5\,dx}{1+25x^2}\Rightarrow dx=\frac15(1+25x^2)\,du\text. Directly after the substitution, we have

    \int\frac u{1+25x^2}\left[\frac15(1+25x^2)\,du\right]

    =\int u\left(\frac15\right)\left(\frac{1+25x^2}{1+25x^2}  \right)du\text.

    Now, we can simplify and pull the constant out of the integrand to get

    \frac15\int u\,du\text.

    Does that help? If not, I'm afraid you'll have to wait a few hours before I can respond again, as I'll be leaving in a couple minutes.
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    I still don't understand where the 1/5 comes from. What do you have to multiply by 1/5? Normally I treat these problems as you want your u's derivative (du) to contain part of the function so that you can divide it out, and 25x^2+1 just doesn't divide nto 25x^5 evenly how does this become 1/5.
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    Quote Originally Posted by fattydq View Post
    I still don't understand where the 1/5 comes from. What do you have to multiply by 1/5? Normally I treat these problems as you want your u's derivative (du) to contain part of the function so that you can divide it out, and 25x^2+1 just doesn't divide nto 25x^5 evenly how does this become 1/5.
    Okay, real quick before I go: the \frac15 comes from the substitution for dx\text. The reason it is there is because of the use of the chain rule in differentiating \arctan5x\text. The 5 was a result of the chain rule, and when solving for dx it becomes \frac15, which is then inserted into the integral when we substitute.
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