`intcos(logx)dx`

Options

A. ` ( 1 ) / ( 2 ) [ sin ( log x ) + cos (log x ) ] + C `
B. ` ( x ) /( 2 ) [ sin ( log x ) + cos ( log x ) ] + C `
C. ` ( x )/( 2 ) [sin ( log x ) - cos ( log x ) ] + C `
D. ` ( 1 ) / ( 2 ) [sin ( log x ) - cos ( log x ) ] + C `

Answer

B

Explanation

Let ` I= int cos (log x )* 1 dx " " ` … (i)
Use integral by parts,
` I = cos ( log x ) * x - int [ - sin ( log x)]* ( 1 ) /( x ) * x dx `
`=x * cos (log x ) + int sin ( log x ) * 1 dx `
` = x * cos (log x ) + [ sin ( log x ) * x - int cos (logx ) * (1 )/(x ) * xdx] + C `
` = x* cos ( log x ) + [ x sin (logx)- int cos (log x ) dx] + C `
` = x { sin (log x ) + cos (log x) } - I+ C `[from Eq. (i) ]
`rArr I = (x )/( 2 ) { sin (log x ) + cos ( log x ) } + C `