Integrate $\sin x/(1+\sin x)$, how to?

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How can I integrate $\displaystyle\int{\dfrac{\sin x}{1+\sin x}}dx$? I have already tried by applying different trigonometric results.

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2 Answers

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If we write the expression in the following way: $$\frac{\sin x}{1+ \sin x} = \frac{\sin x(1 - \sin x)}{(1+\sin x)(1 - \sin x)} = \frac{\sin x - (\sin x)^2}{1 - (\sin x)^2} = \frac{\sin x - (\sin x)^2}{(\cos x)^2} \\= \frac{\sin x}{(\cos x)^2} - (\tan x)^2$$ Do you see how we can integrate these two terms?

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Hint:

Write $\frac{\sin{x}}{1+\sin{x}}$ as $$\frac{\sin{x}}{1+\sin{x}}= \frac{1+\sin{x}-1}{1+\sin{x}} =1 -\frac{1}{1+\sin{x}} $$

Now subsitute $u=\tan{\frac{x}{2}}$

With this substitution the integral will simplify to

$$\int dx - \int\frac{2}{u^2+2u+1}\,du $$

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