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Thread: Intregrals trig

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    Intregrals trig

    Intregrals trig-screenshot-2018-02-15-11.18.34.png Hello. Could someone explain what is happening here step by step? Thank you
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    Re: Intregrals trig

    they just rearranged the terms a bit and factored $\csc^4(x^2)$ into $\csc^3(x^2)\cdot \csc(x^2)$
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    Re: Intregrals trig

    Quote Originally Posted by RBB22 View Post
    Click image for larger version. 

Name:	Screenshot 2018-02-15 11.18.34.png 
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ID:	38524 Hello. Could someone explain what is happening here step by step? Thank you
    $\displaystyle \begin{align*} \int{x\csc^4{\left( x^2 \right) } \cot{ \left( x^2 \right) } \,\mathrm{d}x} &= \frac{1}{2} \int{ 2\,x\csc^4{\left( x^2 \right) } \cot{ \left( x^2 \right) } \,\mathrm{d}x} \end{align*}$

    Let $\displaystyle \begin{align*} u = x^2 \implies \mathrm{d}u = 2\,x\,\mathrm{d}x \end{align*}$ and the integral becomes

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

    Let $\displaystyle \begin{align*} v = \sin{ \left( u \right) } \implies \mathrm{d}v = \cos{ \left( u \right) } \,\mathrm{d}u \end{align*}$ and the integral becomes

    $\displaystyle \begin{align*} \frac{1}{2} \int{ \sin^{-5}{\left( u \right) } \cos{ \left( u \right) } \,\mathrm{d}u } &= \frac{1}{2} \int{ v^{-5}\,\mathrm{d}v } \\ &= \frac{1}{2} \left( \frac{v^{-4}}{-4} \right) + C \\ &= -\frac{1}{8\,v^4} + C \\ &= -\frac{1}{8 \sin^4{ \left( u \right) } } + C \\ &= -\frac{1}{8 \sin^4{\left( x^2 \right) }} + C \\ &= -\frac{\csc^4{ \left( x^2 \right) }}{8} + C \end{align*}$
    Thanks from RBB22
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    Re: Intregrals trig

    WOW!!!! This is perfect. Thank you so much, Prove It!
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