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Math Help - trig sub and i need the integral of cot(x)

  1. #1
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    trig sub and i need the integral of cot(x)

    Hello I don't believe this was asked before but I was wondering the integral of cosx/sinx (or cotx)I think it is

    let u=cosx du=-sinx dx -1/sinxdu=dx and then i would sinply take the integral of u? or would a make u= cosx and then end up with 1/u and the intergral sould be ln(sinx)

    The hole problem I had was
    (integral) x+3/x^4+9x^2 dx some algebra
    (integral)x+3/9x^2(9x^2[{x^2/9}+1]})dx
    so tan^2(theta)=x^2/9 tan(theta)=x/3 3tan(theta)=x dx=3sec^2(theta) remember the idenity 1+tan^2(x)=sec^2(x) so substituting gives
    (integral) [3tan(theta)+3/81tan^2(theta)sec^2(theta)] 3sec^2(theta) d(theta)
    some more algebra (integral)18/81tan(theta) d(theta)
    Pulled the 18/81 out of the int and get (integral) 1/tan(theta) or cot(theta) or cos(theta)/sine(theta)
    I believe sin over cos is ln of sin(theta) and the going back into terms of x would give the final answer of

    ln[(sin{x/x^4+9x^2})/9]+c
    thanks in advance for any help
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  2. #2
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    Re: trig sub and i need the integral of cot(x)

    Quote Originally Posted by capita7 View Post
    let u=cosx du=-sinx dx -1/sinxdu=dx and then i would sinply take the integral of u? or would a make u= cosx and then end up with 1/u and the intergral sould be ln(sinx)
    Correct  \displaystyle \int \frac{\cos x }{\sin x } ~dx = \ln (\sin x) +C
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  3. #3
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    Re: trig sub and i need the integral of cot(x)

    Quote Originally Posted by pickslides View Post
    Correct  \displaystyle \int \frac{\cos x }{\sin x } ~dx = \ln |\sin{x}| +C
    ...
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