1. Intregrals trig

Hello. Could someone explain what is happening here step by step? Thank you

2. Re: Intregrals trig

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

3. Re: Intregrals trig

Originally Posted by RBB22
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*}

4. Re: Intregrals trig

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