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Question Number 143032 by ZiYangLee last updated on 09/Jun/21

$$\int {\sin }^{-5}xdx=?$$

Answered by Olaf_Thorendsen last updated on 09/Jun/21

$$\mathrm{F}\left(x\right)=\int \frac{dx}{{\sin }^{5}x}$$$$\mathrm{Let}t=\tan \frac{x}{2}$$$$\mathrm{F}\left(t\right)=\int \frac{1}{{\left(\frac{2t}{1+t^2}\right)}^5}.\frac{2dt}{1+t^2}$$$$\mathrm{F}\left(t\right)=\frac{1}{16}\int \frac{{(1+t^2)}^4}{t^5}dt$$$$\mathrm{F}\left(t\right)=\frac{1}{16}\int \frac{1+4t^2+6t^4+4t^6+t^8}{t^5}dt$$$$\mathrm{F}\left(t\right)=\frac{1}{16}\int (\frac{1}{t^5}+\frac{4}{t^3}+\frac{6}{t}+4t+t^3)dt$$$$\mathrm{F}\left(t\right)=\frac{1}{16}(-\frac{1}{4t^4}-\frac{2}{t^2}+6\ln t+2t^2+\frac{t^4}{4})+\mathrm{C}$$$$\mathrm{F}\left(x\right)=-\frac{1}{{64\tan }^4\frac{x}{2}}-\frac{1}{{8\tan }^2\frac{x}{2}}+$$$$\frac{3}{8}\ln \left(\tan \frac{x}{2}\right)+\frac{1}{8}{\tan }^2\frac{x}{2}+\frac{{\tan }^4\frac{x}{2}}{64}+\mathrm{C}$$$$\mathrm{F}\left(x\right)=\frac{1}{8}\ln \left(\tan \frac{x}{2}\right)-(\frac{1}{{4\sin }^{4}x}+\frac{3}{{8\sin }^{2}x})\cos x+\mathrm{C}$$

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