$\int \frac{\sin x}{1+2\sin x}dx$

Question

calculate: $$\int \frac{\sin x}{1+2\sin x}dx$$ I tried using $\sin x=\dfrac{2u}{u^2+1}$, $u=\tan \dfrac{x}2$ and after Simplification: $$\int \frac{2u}{u^2+4u+1}×\frac{2}{u^2+1}du$$ and I am not able to calulate that.

Answer 1

Before doing that substitution I might say

$\int \frac{\sin x}{1+2\sin x} \ dx\\ \int \frac 12 \frac{2\sin x + 1 - 1}{1+2\sin x} \ dx\\ \int \frac 12 - \frac{1}{1+2\sin x} \ dx\\ \int \frac 12 \ dx - \frac 12 \int \frac{1}{1+2\sin x} \ dx$

Now when we do the substitution it isn't as messy.

$\frac x2 - \int \frac{1}{u^2 + 4u + 1} \ du$

Answer 2

Hint:

Use the partial fractions method (I think that is the name of it):

$$\frac{2u}{u^2+4u+1}×\frac{2}{u^2+1} =\frac{au+b}{u^2+4u+1}+\frac{cu+d}{u^2+1} $$

where you have to find $a,b,c,d$.

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How do we do that? First get rid of the fractions:

$$4u = (au+b)(u^2+1)+(cu+d)(u^2+4u+1)$$

and now get rid of the brackets and then compare the coefficents...

Answer 3

Since $u=\tan\frac{x}{2}$ gives$$\int\frac{dx}{1+2\sin x}=\int\frac{2 du}{(u+2)^2-3}=\frac{-2}{\sqrt{3}}\operatorname{artanh}\frac{u+2}{\sqrt{3}}+C=\frac{-1}{\sqrt{3}}\ln\frac{\sqrt{3}+2+\tan\frac{x}{2}}{\sqrt{3}-2-\tan\frac{x}{2}}+C,$$we have$$\int\frac{\sin xdx}{1+2\sin x}=\frac12x+\frac{1}{2\sqrt{3}}\ln\frac{\sqrt{3}+2+\tan\frac{x}{2}}{\sqrt{3}-2-\tan\frac{x}{2}}+C.$$

Answer 4

Simplify first before substituting,

$$\int \frac{\sin x}{1+2\sin x}dx =\frac12\int dx -\frac12 \int \frac{1}{1+2\sin x}dx$$ $$=\frac12 x -\frac12 \int \frac{1+\tan^2\frac x2}{\tan^2\frac x2 +4\tan\frac x2 +1}dx$$ $$=\frac12 x -\int \frac{d(\tan \frac x2)}{\left(\tan^2\frac x2 +2\right)^2 -3}$$ $$=\frac12 x +\frac{1}{\sqrt3}\tanh^{-1}\left( \frac{ \tan\frac x2 +2}{\sqrt3}\right)+C$$