Question

integrate cos(logx) dx

Answer 1

∫ cos ( log x ) dx 

integrate by parts 

u = cos (log x) 
du = - sin(log x) dx/x 

dv = dx 
v = x 


∫ cos ( log x ) dx = x cos(log x) + ∫ sin (log x) dx 

again integrate by parts 

u = sin(log x) 
du = cos(log x ) dx/x 

dv = dx 
v = x 

∫ cos ( log x ) dx = x cos(log x) + x sin (log x ) - ∫ cos (log x ) dx 

= 2∫ cos ( log x ) dx = x cos(log x) + x sin (log x ) 

= ∫ cos ( log x ) dx = (1/2)x [ cos(log x) + sin (log x ) ] + C

Answer 2

∫ cos ( log x ) dx
integrate by parts
u = cos (log x)
du = - sin(log x) dx/x
dv = dx
v = x
∫ cos ( log x ) dx = x cos(log x) + ∫ sin (log x) dx
again integrate by parts
u = sin(log x)
du = cos(log x ) dx/x
dv = dx
v = x
∫ cos ( log x ) dx = x cos(log x) + x sin (log x ) - ∫ cos (log x ) dx
= 2∫ cos ( log x ) dx = x cos(log x) + x sin (log x )
= ∫ cos ( log x ) dx = (1/2)x [ cos(log x) + sin (log x ) ] + C

Answer


$\frac{1}{2}x ( \cos( log(x) ) + \sin( log(x) ) + c$