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$\begingroup$ en.wikipedia.org/wiki/Tangent_half-angle_substitution $\endgroup$– lab bhattacharjeeJan 8, 2014 at 15:13
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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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$\begingroup$ I think the final terms are getting harder to integrate. ;-) $\endgroup$– MikasaJan 8, 2014 at 15:18
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1$\begingroup$ $\sin x/ (\cos x)^2$ can be integrated by a simple substitution, and $(\tan x)^2$ can be integrated by writing $(\tan x)^2 = [ (\tan x)^2 + 1] - 1$. I think this is a good way to do it if one doesn't know about the Weierstrass substitution $\endgroup$– UlrikJan 8, 2014 at 15:21
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$\begingroup$ $\sec x - \tan x = (1- \sin x)/(\cos x)$. This is a 0/0-expression when $x = \pi/2$, so you can evaluate it as a limit by using L'hôpital's rule for instance. $\endgroup$– UlrikJan 8, 2014 at 15:54
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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
$$\int dx - \int\frac{2}{u^2+2u+1}\,du $$
answered Jan 8, 2014 at 15:17