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Question Number 114102 by bemath last updated on 17/Sep/20

$$\int \frac{dx}{\tan x-\sin x}$$

Answered by bobhans last updated on 17/Sep/20

$$\int \frac{dx}{\tan x-\sin x}=-\frac{1}{2}\mathrm{cosec}{}^{2}x-\frac{1}{2}\cot x\mathrm{cosec}x$$$$+\frac{1}{2}\ln \mid \cot x+\mathrm{cosec}x\mid +c$$

Answered by 1549442205PVT last updated on 17/Sep/20

$$\mathrm{Put}\tan \frac{\mathrm{x}}{2}=\mathrm{t}\Rightarrow \mathrm{dt}=\frac{1}{2}(1+{\mathrm{t}}^2)\mathrm{dx}$$$$\int \frac{dx}{\tan x-\sin x}=\int \frac{2\mathrm{dt}}{(\frac{2\mathrm{t}}{1-{\mathrm{t}}^2}-\frac{2\mathrm{t}}{1+{\mathrm{t}}^2})(1+{\mathrm{t}}^2)}$$$$=\int \frac{(1-{\mathrm{t}}^2)}{{2\mathrm{t}}^3}\mathrm{dt}=\int (\frac{1}{{2\mathrm{t}}^3}-\frac{1}{2\mathrm{t}})\mathrm{dt}=-\frac{1}{{4\mathrm{t}}^2}-\frac{1}{2}\ln \mid \mathrm{t}\mid$$$$=-\frac{1}{4{\mathbf{tan}}^2\frac{\mathbf{x}}{2}}-\frac{1}{2}\mathbf{ln}\mid \mathbf{tan}\frac{\mathbf{x}}{2}\mid +\mathbf{C}$$

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