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Math Help - quick help with 3 integrals

  1. #1
    Junior Member
    Joined
    Jan 2009
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    quick help with 3 integrals

    NOTE: I use ( as the beggining intergral sign... so ( x^2 dx is the integral of x^2.


    1.I have two integrals. (tan x sec^2 x dx and (sin x / cos^3 x dx
    -after looking at them I realize they are the same thing...
    -then why do i get different answers?
    1) u =tan x du=sec^2 x dx ((u)du = u^2 /2. Answer is (tan^2 x/2) + C
    2) u =cosx du=-sinx dx -du=sinx dx (-du/(u)^3= (u)^-2/2. Answer is (sec^2 x)/2 + C
    What am I doing wrong?

    2. (tan(3x) dx
    - I got the answer -1/3 ln cos(3x) +C. I was wondering if I should an absolute value around COS 3x or not. Should I?

    3.( x(2+x)^.5 dx
    -how do i do this one?
    - this is how far i got. u= 2+x du=1 ( du(x) (u)^.5 I am lost now.

    -Thanks for the help in advance
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  2. #2
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    Indiana
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    Quote Originally Posted by dandaman View Post
    NOTE: I use ( as the beggining intergral sign... so ( x^2 dx is the integral of x^2.


    1.I have two integrals. (tan x sec^2 x dx and (sin x / cos^3 x dx
    -after looking at them I realize they are the same thing...
    -then why do i get different answers?
    1) u =tan x du=sec^2 x dx ((u)du = u^2 /2. Answer is (tan^2 x/2) + C
    2) u =cosx du=-sinx dx -du=sinx dx (-du/(u)^3= (u)^-2/2. Answer is (sec^2 x)/2 + C
    What am I doing wrong?

    2. (tan(3x) dx
    - I got the answer -1/3 ln cos(3x) +C. I was wondering if I should an absolute value around COS 3x or not. Should I?

    3.( x(2+x)^.5 dx
    -how do i do this one?
    - this is how far i got. u= 2+x du=1 ( du(x) (u)^.5 I am lost now.

    -Thanks for the help in advance
    To answer your first question,

    \int \frac{\sin x}{\cos^3x}dx=\int -u^{-3}du=\frac{1}{2}u^{-2}=\frac{1}{2 \cos^2 x}=\frac{\sec^2x}{2}+C =\frac{\tan^2x+1}{2}+C=\frac{\tan^2x}{2}+\left(\fr  ac{1}{2}+C\right) <--It is just a different C.
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  3. #3
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    k thanks... any idea for the others?
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  4. #4
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    Dec 2008
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    Indiana
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    Quote Originally Posted by dandaman View Post
    NOTE: I use ( as the beggining intergral sign... so ( x^2 dx is the integral of x^2.


    1.I have two integrals. (tan x sec^2 x dx and (sin x / cos^3 x dx
    -after looking at them I realize they are the same thing...
    -then why do i get different answers?
    1) u =tan x du=sec^2 x dx ((u)du = u^2 /2. Answer is (tan^2 x/2) + C
    2) u =cosx du=-sinx dx -du=sinx dx (-du/(u)^3= (u)^-2/2. Answer is (sec^2 x)/2 + C
    What am I doing wrong?

    2. (tan(3x) dx
    - I got the answer -1/3 ln cos(3x) +C. I was wondering if I should an absolute value around COS 3x or not. Should I?

    3.( x(2+x)^.5 dx
    -how do i do this one?
    - this is how far i got. u= 2+x du=1 ( du(x) (u)^.5 I am lost now.

    -Thanks for the help in advance
    Quote Originally Posted by dandaman View Post
    k thanks... any idea for the others?
    Sorry I was taking off yesterday and did not have time to come back until now.

    For #2, I believe you should have absolute value since you are talking about \ln.
    For #3, with your choice of u,
    since u=2+x, du=dx. Thus x=u-2
    So \int x(2+x)^{\frac{1}{2}}dx=\int (u-2)u^{\frac{1}{2}}du =\int u^{\frac{3}{2}}-2u^{\frac{1}{2}}du=....
    I believe you can take it from here.
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