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Math Help - Is this integration correct?

  1. #1
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    Is this integration correct?

    Hi,

    I think I solved this integration but wolframalpha seems to be saying I'm wrong. This integral is: \int cot^5(x) sin^4(x) dx

    My solution is:

    \int (csc^2(x) -1)^2sin^4(x)cot(x)

    expanding and cancelling trig functions:

    \int cot(x) -2sin(x)cos(x) + sin^3(x)cos(x)

     ANS = \ln{\mid sin(x) \mid} - sin^2(x) +\frac{1}{4}sin^4(x)

    Does that work or have I made a mistake?

    Thanks!
    Last edited by KevinShaughnessy; April 17th 2014 at 08:58 PM. Reason: Made a typo, wrote + where I meant -
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  2. #2
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    Re: Is this integration correct?

    If your integral is $\displaystyle \begin{align*} \int{\cot^5{(x)}\sin^4{(x)}\,dx} \end{align*}$, I would do...

    $\displaystyle \begin{align*} \int{\cot^5{(x)}\sin^4{(x)}\,dx} &= \int{ \frac{\cos^5{(x)}}{\sin^5{(x)}} \cdot \sin^4{(x)} \,dx } \\ &= \int{ \frac{\cos^5{(x)}}{\sin{(x)}} \,dx } \\ &= \int{ \frac{\cos{(x)} \left[ \cos^2{(x)} \right] ^2 }{ \sin{(x)} } \, dx } \\ &= \int{ \frac{\left[ 1 - \sin^2{(x)} \right] ^2 }{ \sin{(x)} } \cdot \cos{(x)} \, dx } \end{align*}$

    Now with the substitution $\displaystyle \begin{align*} u = \sin{(x)} \implies du = \cos{(x)} \, dx \end{align*}$ we have

    $\displaystyle \begin{align*} \int{ \frac{\left[ 1 - \sin^2{(x)} \right] ^2}{\sin{(x)}} \cdot \cos{(x)}\,dx } &= \int{ \frac{\left( 1 - u^2 \right) ^2}{u} \, du } \\ &= \int{ \frac{1 - 2u^2 + u^4}{u} \, du } \\ &= \int{\frac{1}{u} - 2u + u^3 \, du} \\ &= \ln{|u|} - u^2 + \frac{1}{4}u^4 + C \\ &= \ln{ \left| \sin{(x)} \right| } - \sin^2{(x)} + \frac{1}{4}\sin^4{(x)} + C \end{align*}$
    Thanks from prasum and KevinShaughnessy
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  3. #3
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    Re: Is this integration correct?

    Quote Originally Posted by Prove It View Post
    If your integral is $\displaystyle \begin{align*} \int{\cot^5{(x)}\sin^4{(x)}\,dx} \end{align*}$, I would do...

    $\displaystyle \begin{align*} \int{\cot^5{(x)}\sin^4{(x)}\,dx} &= \int{ \frac{\cos^5{(x)}}{\sin^5{(x)}} \cdot \sin^4{(x)} \,dx } \\ &= \int{ \frac{\cos^5{(x)}}{\sin{(x)}} \,dx } \\ &= \int{ \frac{\cos{(x)} \left[ \cos^2{(x)} \right] ^2 }{ \sin{(x)} } \, dx } \\ &= \int{ \frac{\left[ 1 - \sin^2{(x)} \right] ^2 }{ \sin{(x)} } \cdot \cos{(x)} \, dx } \end{align*}$

    Now with the substitution $\displaystyle \begin{align*} u = \sin{(x)} \implies du = \cos{(x)} \, dx \end{align*}$ we have

    $\displaystyle \begin{align*} \int{ \frac{\left[ 1 - \sin^2{(x)} \right] ^2}{\sin{(x)}} \cdot \cos{(x)}\,dx } &= \int{ \frac{\left( 1 - u^2 \right) ^2}{u} \, du } \\ &= \int{ \frac{1 - 2u^2 + u^4}{u} \, du } \\ &= \int{\frac{1}{u} - 2u + u^3 \, du} \\ &= \ln{|u|} - u^2 + \frac{1}{4}u^4 + C \\ &= \ln{ \left| \sin{(x)} \right| } - \sin^2{(x)} + \frac{1}{4}\sin^4{(x)} + C \end{align*}$
    Thanks!
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