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Question Number 185935 by cortano1 last updated on 30/Jan/23

$$\int \frac{1-\cos x}{(1+\cos x)\cos x}dx=?$$

Answered by Ar Brandon last updated on 30/Jan/23

$$I=\int \frac{1-\cos x}{(1+\cos x)\cos x}dx=\int \frac{{(1-\cos x)}^2}{(1-{\cos }^{2}x)\cos x}dx$$$$=\int \frac{1-2\cos x+{\cos }^{2}x}{{\sin }^{2}x}\cdot \frac{1}{\cos x}dx=\int \frac{{2\cos }^{2}x+{\sin }^{2}x-2\cos x}{{\sin }^{2}x}\cdot \frac{dx}{\cos x}$$$$=\int (\frac{2\cos x}{{\sin }^{2}x}+\frac{1}{\cos x}-\frac{2}{{\sin }^{2}x})dx=2\cot x+\ln \mid \sec x+\tan x\mid -\frac{2}{\sin x}+C$$

Answered by ajfour last updated on 30/Jan/23

$$\int (\frac{a}{1+\cos x}+\frac{b}{\cos x})dx$$$$a+b=-1$$$$b=1\Rightarrow a=-2$$$$I=-\int \left(\sec {}^2\frac{x}{2}\right)dx+\int \sec xdx$$$$=-2\tan \left(\frac{x}{2}\right)+\ln \mid \tan (\frac{\pi }{4}+\frac{x}{2})\mid +c$$

Commented by ARUNG_Brandon_MBU last updated on 30/Jan/23

Nice

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