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Question Number 148137 by Willson last updated on 25/Jul/21

$$\int \frac{3{cos}^2\left(x\right)+1}{{sin}^5\left(x\right)}dx=???$$

Answered by puissant last updated on 25/Jul/21

$$=\int \frac{{4\cos }^2\left(\mathrm{x}\right)+{\sin }^2\left(\mathrm{x}\right)}{{\sin }^5\left(\mathrm{x}\right)}\mathrm{dx}=\int \frac{4+{\tan }^2\left(\mathrm{x}\right)}{{\tan }^2\left(\mathrm{x}\right){\sin }^3\left(\mathrm{x}\right)}\mathrm{dx}$$$$=\int \frac{\frac{4}{{\tan }^2\left(\mathrm{x}\right)}+1}{{\sin }^3\left(\mathrm{x}\right)}\mathrm{dx}=\int \frac{{4\cos }^2\left(\mathrm{x}\right)-3}{{\sin }^3\left(\mathrm{x}\right)}\mathrm{dx}$$$$=\int \frac{{4\mathrm{cotan}}^2\left(\mathrm{x}\right)-\frac{3}{{\sin }^2\left(\mathrm{x}\right)}}{\sin \left(\mathrm{x}\right)}\mathrm{dx}=\int \frac{\frac{1}{{\sin }^2\left(\mathrm{x}\right)}}{\sin \left(\mathrm{x}\right)}\mathrm{dx}=\int \frac{1}{{\sin }^3\left(\mathrm{x}\right)}\mathrm{dx}$$$$=\int \frac{1}{\sin \left(\mathrm{x}\right)(1-{\cos }^2\left(\mathrm{x}\right))}\mathrm{dx}$$$$\mathrm{t}=\cos \left(\mathrm{x}\right)\Rightarrow \mathrm{dt}=-\sin \left(\mathrm{x}\right)\mathrm{dx}\Rightarrow \mathrm{dx}=-\frac{\mathrm{dt}}{\sin \left(\mathrm{x}\right)}$$$$\Rightarrow \mathrm{I}=-\int \frac{\mathrm{dt}}{{(1-{\mathrm{t}}^2)}^2}$$$$=-\int (\frac{\mathrm{a}}{1-\mathrm{t}}+\frac{\mathrm{b}}{{(1-\mathrm{t})}^2}+\frac{\mathrm{c}}{1+\mathrm{t}}+\frac{\mathrm{b}}{{(1+\mathrm{t})}^2})\mathrm{dt}$$$$\mathrm{a}=\mathrm{b}=\mathrm{c}=\mathrm{d}=\frac{1}{4}$$$$\Rightarrow \mathrm{I}=\frac{1}{4}\ln \mid 1-\mathrm{t}\mid -\frac{1}{4(1-\mathrm{t})}-\frac{1}{4}\ln \mid 1+\mathrm{t}\mid +\frac{1}{4(1+\mathrm{t})}+\mathrm{C}$$$$\mathrm{I}=\frac{1}{4}\ln \mid \frac{1-\cos \left(\mathrm{x}\right)}{1+\cos \left(\mathrm{x}\right)}\mid -\frac{1}{{2\sin }^2\left(\mathrm{x}\right)}+\mathrm{C}..$$

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