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Thread: integration

  1. July 5th 2006, 12:12 PM #1
    macca101
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    integration

    I'm having problems integrating

    <br /> \int\frac {sin(x)}{cos^2(x)}dx<br />

    the answer in the book is

    <br /> \frac{1}{cos(x)}+c<br />

    I can't see how this answer is obtained.
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  2. July 5th 2006, 12:16 PM #2
    Jameson
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    Let u=cos(x) and du=-sin(x)

    This becomes \int \frac{1}{u^2}(-1du)

    When you integrate this, the negatives cancel out and you get \frac{1}{u}+C or \frac{1}{\cos(x)}+C
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  3. July 5th 2006, 12:24 PM #3
    macca101
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    Tremendous

    Thanks Jameson
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  4. July 5th 2006, 02:34 PM #4
    Soroban
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    Hello, macca101!

    Another method . . .

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

    The answer in the book is: . \frac{1}{\cos(x)}+C<br />
    I'm surprised that they left the answer like that . . .

    We have: . \int\frac{\sin x}{\cos^2x}\,dx\;=\;\int\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x}\,dx\;= \int\sec x\tan x\,dx \;= \;\sec x + C
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