Question

integrate DX/sin x+ sin2x​

Answer 1

Step-by-step explanation:

We have ,

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

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

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

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

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

Putting $tan(\frac{x}{2})=t$

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

$= \int \frac{(1 + t ^{2} )dt}{ t(3 - t^{2} )}$

$= \int \frac{(1 + t ^{2} )dt}{ t( \sqrt{3} - t )( \sqrt{3} + t) }$

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

$= \frac{1}{3} \int \frac{dt}{t} + \frac{2}{3} \int \frac{dt}{( \sqrt{3} - t) } - \frac{2}{3} \int \frac{dt}{( \sqrt{3} + t)}$

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

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