Question

please solve no spamming ​

Answer 1

Step-by-step explanation:

We have,

$i = \int \sin( log( x ) ) dx$

$i= \int \sin( log(x) ) .1dx$

$i = \sin( log(x) ) \int1.dx - \int( \frac{d}{dx} ( \sin( log(x) ) ) \int1.dx)dx$

$i = x \sin( log(x) ) - \int \frac{ \cos( log(x) ) }{x} .xdx$

$\implies \: i = x \sin( log(x) ) - \int \cos( log(x) ) dx$

$\implies \: i = x\sin( log(x) ) - \cos( log(x) ) \int1.dx + \int( \frac{d}{dx} ( \cos( log(x) )) \int1.dx)dx$

$\implies \: i = x \sin( log(x) ) - x \cos( log(x) ) - \int \frac{ \sin( log(x) ) }{x} .xdx$

$\implies \: i = x \sin( log( x ) ) - x \cos( log(x) ) - \int \sin( log(x) ) dx$

$\implies \: i = x \sin( log(x) ) - x \cos( log(x) ) - i$

$\implies2i = x \sin( log(x) ) - x \cos( log(x) )$

$\implies \: i = \frac{x}{2} ( \sin( log( x ) ) - \cos( log(x) )$