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Math Help - Substitution method for antiderivatives.

  1. #1
    Member Marconis's Avatar
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    Substitution method for antiderivatives.

    Tried my best with Latex, took forever!

    1) \int12x(6x^2+1)^3dx
    Let u= 6x^2+1
    \frac {du}{dx}=12x
    du=12xdx

    \int(u)^312xdx= \int(u)^3du= \frac {u^4}{4} + C

    =  \frac {1}{4}(6x^2+1)^4 + C


    2) \int \frac {3x^2dx}{\sqrt[3]{x^3+1}}

    Let u= (x^3+1)
     du= 3x^2dx

    \int \frac {du}{\sqrt[3]{u}} = u^{-1/3}du
    = \frac {u^{2/3}}{2/3} + C

    = \frac {3}{2}(x^3+1)^{2/3} + C

    3)
    \int 2x\cos{(x^2)}dx

    Let u= (x^2) du= 2xdx
    <br />
\int \cos{x^2}dx2x= \cos{(u)}du<br /> <br />
=\sin{x^2} + C


    **Edit** One more

    4)  \int \sin{x}^5 \cos{x}dx

    Answer is (1/6) (sinx) ^6
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  2. #2
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    These look good. You can always find dy/dx of the answer to check them.
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  3. #3
    Member Marconis's Avatar
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    Awesome, thanks! I appreciate your help. I wish I could try more...I emailed my prof. and he said all of the ones in the book are too high a level and he hasn't taught it yet. Guess I'll just have to wait. Also, note that I added one more question.
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  4. #4
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    I assume 4. is \displaystyle \int{\sin^5{x}\cos{x}\,dx}

    Make the substitution \displaystyle u = \sin{x}.
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  5. #5
    Member Marconis's Avatar
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    Yeah, that's it. I did make that substitution. What did I do wrong? I thought sin^5x and sinx^5 were the same thing? Or am I seriously mistakin'?
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  6. #6
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    If you write \displaystyle \sin{x}^5 we can't tell if it's \displaystyle \sin{(x^5)} or \displaystyle (\sin{x})^5...


    So after you make the substitution \displaystyle u = \sin{x}, what's \displaystyle \frac{du}{dx}? What's \displaystyle du?
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  7. #7
    Member Marconis's Avatar
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    Quote Originally Posted by Prove It View Post
    If you write \displaystyle \sin{x}^5 we can't tell if it's \displaystyle \sin{(x^5)} or \displaystyle (\sin{x})^5...


    So after you make the substitution \displaystyle u = \sin{x}, what's \displaystyle \frac{du}{dx}? What's \displaystyle du?
    cos(x)dx
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  8. #8
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    So what is your new integral in terms of u?
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  9. #9
    Member Marconis's Avatar
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    Quote Originally Posted by Prove It View Post
    So what is your new integral in terms of u?
    (1/6) (sinx) ^6 + C.

    Is this not the answer? All the work I'd give would leave to that.
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  10. #10
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    \displaystyle y= \frac{1}{6}\sin ^6x+C \implies \frac{dy}{dx} = \sin^5 x \cos x

    Looks good.
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  11. #11
    Member Marconis's Avatar
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    Sweet, thanks! Can't wait for the more challenging ones.
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  12. #12
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    Quote Originally Posted by Marconis View Post
    Sweet, thanks! Can't wait for the more challenging ones.
    Try this one

    \displaystyle \int\frac{x+1}{x-1}~dx
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  13. #13
    Member Marconis's Avatar
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    Quote Originally Posted by pickslides View Post
    Try this one

    \displaystyle \int\frac{x+1}{x-1}~dx
    Seems like it's hard, lol. I'll get back to you tomorrow. Time to enjoy my Friday night.
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  14. #14
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    Hey, so I didn't get a chance do the question you provided, had a busy weekend. I will post the following, though:

    \int \cos^4{x} \sin{x} dx

    u= \cos{x}
    du=- \sin{x}dx

    -1 \int (u)^4 du
    =-1 \int \frac {u^{5}}{5} + C
    = \frac {-1}{5} (\cos{x})^5 + C

    What do you guys think?
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  15. #15
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    Quote Originally Posted by Marconis View Post
    Hey, so I didn't get a chance do the question you provided, had a busy weekend. I will post the following, though:

    \int \cos^4{x} \sin{x} dx

    u= \cos{x}
    du=- \sin{x}dx

    -1 \int (u)^4 du
    =-1 \int \frac {u^{5}}{5} + C Get rid of the integration sign here
    = \frac {-1}{5} (\cos{x})^5 + C

    What do you guys think?
    With this one small change everything is fine
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