Question

Intigration of sinx/(1+2sinx)

Answer 1

Step-by-step explanation:

We have,

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

$= \frac{1}{2} \int \frac{2 \sin(x) + 1 - 1 }{1 + 2 \sin( x) } dx$

$= \frac{1}{2} \int \frac{2 \sin(x) + 1 }{1 + 2 \sin( x) } dx - \frac{1}{2} \int \frac{dx}{1 + 2 \sin(x) }$

$= \frac{1}{2} \int dx - \frac{1}{2} \int \frac{dx}{1 + \frac{4 \tan( \frac{x}{2} ) }{1 + \tan^{2} ( \frac{x}{2} ) } }$

$= \frac{x}{2} - \frac{1}{2} \int \frac{(1 + \tan ^{2} ( \frac{x}{2} )) dx}{ \tan^{2} ( \frac{x}{2} ) + 4 \tan( \frac{x}{2} ) + 1 }$

$= \frac{x}{2} - \frac{1}{2} \int \frac{ \sec^{2} ( \frac{x}{2} ) dx}{ \tan^{2} ( \frac{x}{2} ) + 4 \tan( \frac{x}{2} ) + 4 - 3 }$

$= \frac{x}{2} - \frac{1}{2} \int \frac{ \sec^{2} ( \frac{x}{2} ) dx}{( \tan ( \frac{x}{2} ) + 2) ^{2} - 3 }$

$Let \tan(\frac{x}{2}) + 2= t\\ \implies \frac{1}{2}\sec^{2}(\frac{x}{2}) dx= dt$

$= \frac{x}{2} - \int \frac{dt}{( t) ^{2} - { (\sqrt{3} )}^{2} }$

$= \frac{x}{2} - \frac{1}{2 \sqrt{3} } ln | \frac{t + \sqrt{3} }{t - \sqrt{3} } | + c$

$= \frac{x}{2} - \frac{1}{2 \sqrt{3} } ln | \frac{ \tan( \frac{x}{2} ) + \sqrt{3} }{ \tan( \frac{x}{2} ) - \sqrt{3} } | + c$