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Thread: Need help with integration

  1. #1 Need help with integration 
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    Hi everyone, this is my 1st post! I am returning back to school after 5 years off and reviewing my calc and physics to be ready. I'm having problems with this integration problem though. I found the right answer solving it one way but when trying another way to solve it I can't.



    integral {tan(z)/cos^2(z)}dz

    intergal {tan(z)}{sec^2(z)}dz

    so u=tan(z) and du=sec^2(z)dz

    intergal udu

    {(u^2)/2} + C
    (1/2){tan^2(z)} + C

    ok so that answer is correct, now I'm going to try it the other way I've been trying and can someone point out what I'm doing wrong


    integral {tan(z)/cos^2(z)}dz

    intergal {sin(z)/cos^3(z)}dz

    intergal {sin(z)}{cos^-3(z)}dz

    so u=cos(z) and du=-sin(z)dz

    - intergal {u^-3}du

    - {-1/2}{u^-2}
    {1/2}{1/cos^2}

    I just can't figure out what i'm doing wrong. Any help would be great thanks.

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  3. #2  
    Moderator Moderator AlexP's Avatar
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    Could you post the answer to it that you're given? I'm sure some here can help you quite easily, but there are some (aka me) who would like to help but actually have to work at it a little.


    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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  4. #3  
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    O sorry the answer given by the book is

    (1/2){tan^2(z)} + C
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  5. #4  
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    Your second answer is also correct (except that you need an aribitrary constant).

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  6. #5  
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    Quote Originally Posted by JaneBennet
    Your second answer is also correct (except that you need an aribitrary constant).

    Thank you, I just noticed what I did wrong.

    When I took the derative of

    {1/2}{1/cos^2(z)}

    to check my answer, i changed 2cos(z)sin(z)=sin2(z) making it harder for me to see the cancellation of cos(z0

    {(1/2)}{[2cos(z)sin(z)]/cos^4(z)} = {1/2}{sin2(z)/cos^4(z)}

    {(1/2)}{[2cos(z)sin(z)]/cos^4(z)} = {sin(z)/cos^3(z)} = {tan(z)/cos^2(z)}

    Just to make sure, you equaled both answers to prove they equal because

    (1/2){tan^2(z)} = {1/2}{1/cos^2(z)}

    (1/2){tan^2(z)} - {1/2}{1/cos^2(z)} = 0


    Once again thank you very much
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  7. #6  
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    Thanks for your help janebennet, but I have another question if that ok

    I'm trying to solve

    intergal 1/{x^2 +4}

    i started to work it like this

    intergal {1/4}/{[(x^2)/4] +1}

    1/2arctan(x/2) + C

    which is wrong the correct answer is

    1/12 arctan(x/2) +C

    Can you show me what I'm doing wrong, thanks again
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  8. #7  
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    Quote Originally Posted by 11rdc11
    the correct answer is

    1/12 arctan(x/2) +C
    Are you sure? Try differentiating that and see what you get. :?
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  9. #8  
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    Yep, you are right. The answer provided by the book was wrong. I just wished that i would have taken the derivative of the answer provided by the book and noticed it was wrong. It would have saved me the headache of attempting the problem over and over trying to match the answer in the book. Once again thank you
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  10. #9  
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    You’re welcome. Just remember to make a good habit, when integrating, of differentiating your answers to check if they’re correct. :wink:
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