I stack about following problem... $\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\cot^5x\,\csc^3xdx$

I tried to change $\cot^5x=\frac{\cos^5x}{\sin^5x}$ I got

$$\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\frac{\cos^5x}{\sin^5x}\csc^3xdx$$ but after that I couldn't get good idea to integral this function. Anyone has any idea for the problem

Thank you !

  • $\begingroup$ The two places show different limits of integration. Minor, but could be fixed! $\endgroup$ – user21436 Feb 12 '12 at 23:18
  • $\begingroup$ Write $\cos^5 x=\cos x(1-\sin^2 x)^2$, $\csc^3 x={1\over \sin^3 x}$ and let $u=\sin x$ $\endgroup$ – David Mitra Feb 12 '12 at 23:20

Here's one way to find the indefinite integral.

Write everything in terms of $\sin$ and $\cos$. If one of those functions is raised to an odd positive power, say $\cos$, factor one term of $\cos$ out and use the Pythagorean identity to write the other factor in terms of $\sin$. Then use a $u$-substitution with $u=\sin x$:

$$\eqalign{ \int\cot^5 x\,\csc^3 x\,dx&= \int {\cos^5 x\over \sin^5 x}{1\over \sin^3 x}\,dx\cr &=\int {\cos x \cdot(\cos^2 x)^2\over \sin^8 x} \,dx\cr &=\int {\cos x \cdot(1-\sin^2 x)^2\over \sin^8 x} \,dx\cr &\buildrel{u=\sin x}\over=\int { (1-u^2)^2\over u^8 } \,du\cr &=\int { 1-2u^2+u^4 \over u^8 } \,du\cr &=\int {( u^{-8}-2u^{-6}+u^{-4} )} \,du\cr &= {{ -u^{-7}\over7}+2{u^{-5}\over 5}-{u^{-3}\over3} } +C\cr &= {{ -\sin^{-7} x\over7}+2{\sin^{-5}x\over 5}-{\sin^{-3}x\over3} } +C.\cr } $$


Note that it’s not necessary to convert to sines and cosines:

$$\cot^5 x\,\csc^3 x=\cot^4x\,\csc^2 x(\cot x\csc x)= (\csc^2 x-1)^2\csc^2 x(\csc x\cot x)\;,$$

and $d(\csc x)=-\csc x\cot x dx$, so you can simply let $u=\csc x$. The indefinite integral then becomes

$$\begin{align*} \int(\csc^2 x-1)^2\csc^2 x(\csc x\cot x)dx&=-\int(u^2-1)^2u^2du\\ &=-\int(u^6-2u^4+u^2)du\\ &=-\frac{u^7}7+\frac{2u^5}5-\frac{u^3}3+C\\ &=-\frac17\csc^7 x+\frac25\csc^5 x-\frac13\csc^3 x+C\;. \end{align*}$$

  • $\begingroup$ Thank you for sharing idea !! so I can convert $cot^2x = csc^2x-1$ that seems more easier to find the answer :9 $\endgroup$ – Ryu Feb 12 '12 at 23:48

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