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Math Help - How would i integrate this.

  1. #1
    Junior Member
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    How would i integrate this.

    \frac{cosx}{sinx} * cos x dx

    Im so lost that second cos x is really thorwing me off
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  2. #2
    Super Member Deadstar's Avatar
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    My initial guess would be to simplify it to...
    <br />
\frac{\cos^2(x)}{\sin(x)} = \frac{1-\sin^2(x)}{\sin(x)} = \frac{1}{\sin(x)} - \sin(x)

    But integrating 1/sin(x) is not exactly trivial.

    Will have another think
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  3. #3
    Super Member Deadstar's Avatar
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    Nah that's still the best I can come up with.

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

    Let u = \cos(x), du = \sin(x)dx

    Hence integral becomes...

    \int \frac{1}{1-u^2}du = arctanh(u)
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  4. #4
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    Hello, x5pyd3rx!

    \int \frac{\cos x}{\sin x}\cdot\cos x\, dx

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

    . . . . . . . . =\;\int(\csc x - \sin x)\,dx \;\;=\;\;\ln|\csc x - \cot x| + \cos x + C

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  5. #5
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, x5pyd3rx!


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

    . . . . . . . . =\;\int(\csc x - \sin x)\,dx \;\;=\;\;\ln|\csc x - \cot x| + \cos x + C

    That.



    Is easier.
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