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Question Number 103512 by bemath last updated on 15/Jul/20

$$\int \frac{xdx}{{(\cot x+\tan x)}^2}=$$$$\left(a\right)\frac{x}{16}-\frac{x\sin 4x}{32}-\frac{\cos 4x}{128}+c$$$$\left(b\right)\frac{x}{16}+\frac{x\sin 4x}{32}-\frac{\cos 4x}{128}+c$$$$\left(c\right)\frac{x}{16}+\frac{x\sin 4x}{64}+\frac{\cos 4x}{128}+c$$$$\left(d\right)\frac{x}{16}+\frac{x\cos 4x}{32}+\frac{\sin 4x}{128}+c$$

Commented by bobhans last updated on 15/Jul/20

$$nothinganswertoo$$

Commented by bemath last updated on 15/Jul/20

$$yes.ithinkthequestionwrong$$

Answered by Dwaipayan Shikari last updated on 15/Jul/20

$$\int \frac{x}{{(\frac{cosx}{sinx}+\frac{sinx}{cosx})}^2}dx=\frac{1}{4}\int {xsin}^{2}2x=\int \frac{x}{8}(1-cos4x)$$$$\int \frac{x}{8}-\frac{1}{8}\int xcos4x$$$$\frac{x^2}{16}-\frac{1}{32}xsin4x+\frac{1}{32}\int sin4x$$$$\frac{x^2}{16}-\frac{1}{32}xsin4x-\frac{1}{128}cos4x+\mathrm{Constant}$$

Answered by bramlex last updated on 15/Jul/20

$$\cot \mathrm{x}+\tan \mathrm{x}=\frac{\cos \mathrm{x}}{\sin \mathrm{x}}+\frac{\sin \mathrm{x}}{\cos \mathrm{x}}$$$$=\frac{2}{\sin 2\mathrm{x}};\mathrm{I}=\int \frac{\mathrm{x}\mathrm{dx}}{\left(\frac{4}{\sin {}^{2}2\mathrm{x}}\right)}$$$$=\frac{1}{4}\int \mathrm{x}(\frac{1}{2}-\frac{1}{2}\cos 4\mathrm{x})\mathrm{dx}$$$$=\frac{1}{8}\int (\mathrm{x}-\mathrm{xcos}4\mathrm{x})\mathrm{dx}$$$$=\frac{1}{8}(\frac{{\mathrm{x}}^2}{2}-\frac{\mathrm{xsin}4\mathrm{x}}{4}-\frac{\cos 4\mathrm{x}}{16})+\mathrm{c}$$$$=\frac{{\mathrm{x}}^2}{16}-\frac{\mathrm{xsin}4\mathrm{x}}{32}-\frac{\cos 4\mathrm{x}}{128}+\mathrm{c}◼$$

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