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Question Number 136440 by liberty last updated on 22/Mar/21

$$\int \frac{dx}{\sin {}^{6}x}?$$

Answered by rs4089 last updated on 22/Mar/21

$$\int {cosec}^{6}x.dx$$$$\int {(1+{cot}^{2}x)}^2{cosec}^{2}x.dx$$$$letcotx=t\Rightarrow -{cosec}^{2}x.dx=dt$$$$\int {(1+t^2)}^2(-dt)$$$$-\int (1+t^4+2t^2)dt$$$$-t-\frac{t^5}{5}-2\frac{t^3}{3}+C$$$$-cotx-\frac{{cot}^{5}x}{5}-2\frac{{cot}^{3}x}{3}+C$$

Answered by EDWIN88 last updated on 22/Mar/21

$$\mathrm{E}=\int \frac{\mathrm{dx}}{\sin {}^6\mathrm{x}}=\int \frac{{(\sin {}^2\mathrm{x}+{\cos }^2\mathrm{x})}^2}{\sin {}^6\mathrm{x}}\mathrm{dx}$$$$\mathrm{E}=\int \frac{\sin {}^4\mathrm{x}+2\sin {}^2\mathrm{x}\cos {}^2\mathrm{x}+\cos {}^4\mathrm{x}}{\sin {}^6\mathrm{x}}\mathrm{dx}$$$$\mathrm{E}=\int \mathrm{cosec}{}^2\mathrm{x}\mathrm{dx}+\int \cot {}^4\mathrm{x}\mathrm{cosec}{}^2\mathrm{x}\mathrm{dx}+2\int \cot {}^2\mathrm{x}\mathrm{cosec}{}^2\mathrm{x}\mathrm{dx}$$$$\mathrm{E}=-\cot \mathrm{x}-\int \cot {}^4\mathrm{x}\mathrm{d}\left(\cot \mathrm{x}\right)-2\int \cot {}^2\mathrm{x}\mathrm{d}\left(\cot \mathrm{x}\right)$$$$\mathrm{E}=-\cot \mathrm{x}-\frac{\cot {}^5\mathrm{x}}{5}-\frac{2\cot {}^3\mathrm{x}}{3}+\mathrm{c}$$

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