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Math Help - integration by subsitiution

  1. #1
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    integration by subsitiution

    secx^3(tan x) dx

    I'm not sure if my work is correct, but I know the answer is right because I checked it in my book. Can someone check the work in blod type, please?

    u = secx
    du/dx = secxtanx
    du = secxtanx dx
    du/secx = tanx dx

    u^3*du/secx

    secx^3/secx du

    secx^2 du

    secx^3/3

    1/3(secx^3) + C
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by zachb View Post
    secx^3(tan x) dx

    I'm not sure if my work is correct, but I know the answer is right because I checked it in my book. Can someone check the work in blod type, please?

    u = secx
    du/dx = secxtanx
    du = secxtanx dx
    du/secx = tanx dx

    u^3*du/secx

    secx^3/secx du

    secx^2 du

    secx^3/3

    1/3(secx^3) + C
    How can you integrate sec(x^2) with respect to du??

    I presume this is supposed to be:
    Int[sec^3(x) * tan(x) dx]

    = Int[sin(x)/cos^4(x) dx]

    Let u = cos(x) ==> du = -sin(x) dx

    Int[sec^3(x) * tan(x) dx] = Int[sin(x)/cos^4(x) dx] = Int[1/u^4 * (-du)]

    = -Int[1/u^4 du] = -(-1/(3u^3)) + C

    = 1/(3*cos^3(x)) + C = (1/3)*sec^3(x) + C

    You apparently have the correct answer, but it looks like you had the wrong way to get it.

    -Dan
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  3. #3
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    Int[sin(x)/cos^4(x) dx]
    I actually thought of writing it that way while I was away from my computer.
    -Int[1/u^4 du] = -(-1/(3u^3)) + C
    I would have done it like this:

    -Int[1/u^4 du] = -Int[u^-4 du] = -(-1/3u^-3)) +C

    But I'm assuming you left it the other way so that when you put cosx back into the function it would become secx.

    Thanks.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by zachb View Post
    I would have done it like this:

    -Int[1/u^4 du] = -Int[u^-4 du] = -(-1/3u^-3)) +C

    But I'm assuming you left it the other way so that when you put cosx back into the function it would become secx.

    Thanks.
    Actually no. How you do the integral is a matter of taste. It's just the way I think of how to do the integral is all.

    -Dan
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