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Math Help - Substitution Integrals Help!

  1. #1
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    Substitution Integrals Help!

    Hi, I am struggling in Chapter 5, Section 5.
    I've been absent due to illness, so I'm trying to teach myself.
    I've been doing pretty well on most of the problems, but one in particular has been difficult.
    Would someone please help me?

    integral sinx/ cos^2x dx

    Thank you!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by brittneyymariee View Post
    Hi, I am struggling in Chapter 5, Section 5.
    I've been absent due to illness, so I'm trying to teach myself.
    I've been doing pretty well on most of the problems, but one in particular has been difficult.
    Would someone please help me?

    integral sinx/ cos^2x dx

    Thank you!
    begin with a substitution of u = \cos x

    do you see why this works?

    can you take it from there?
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  3. #3
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    well i just tried doing it a different way...and my answer is

    1/2 tanxsecx ^2

    is that acceptable or has something gone horribly wrong?


    scratch that...new answer at secx.
    yes? no?
    Last edited by mr fantastic; February 18th 2009 at 06:47 PM. Reason: Merged posts
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by brittneyymariee View Post
    scratch that...new answer at secx.
    yes? no?
    correct.

    you could have also noticed that the original integrand was \sec x \tan x, which is the derivative of \sec x

    so yes, the answer is \sec x + C.

    good job
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  5. #5
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    hint ...

    \frac{\sin{x}}{\cos^2{x}} = \frac{1}{\cos{x}} \cdot \frac{\sin{x}}{\cos{x}} = \, ?
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  6. #6
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    thanks! when i realized that...it was pretty simple.



    i got it!
    thanks!
    Last edited by mr fantastic; February 18th 2009 at 06:48 PM. Reason: Merged posts
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  7. #7
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    Quote Originally Posted by brittneyymariee View Post
    Hi, I am struggling in Chapter 5, Section 5.
    I've been absent due to illness, so I'm trying to teach myself.
    I've been doing pretty well on most of the problems, but one in particular has been difficult.
    Would someone please help me?

    integral sinx/ cos^2x dx

    Thank you!
    \int{\frac{\sin{x}}{\cos^2{x}}\,dx}.

    Let u = \cos{x}

    Therefore \frac{du}{dx} = -\sin{x}.


    \int{\frac{\sin{x}}{\cos^2{x}}\,dx} = -\int{\frac{1}{\cos^2{x}}(-\sin{x})\,dx}

     = -\int{\frac{1}{u^2}\frac{du}{dx}\,dx}

     = -\int{u^{-2}\,du}

     = u^{-1} + C

     = \frac{1}{\cos{x}} + C.
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