`intsin(logx)dx=`

A

(a) `1/2x[cos(logx)-sin(logx)]+c`

B

(b) `cos(logx)-x+c`

C

(c) `1/2x[sin(logx)-cos(logx)]+c`

D

(d) `-cos(logx)+c`

Verified by Experts

The correct Answer is:

C

Similar Questions

Explore conceptually related problems

intcos(logx)dx=F(x)+c , where c is an arbitrary constant. Here F(x)=

underset(0)overset(1)int e^(2logx)dx=

intsin^(-1)xdx

int[sin(logx)+cos(logx)]dx

intlogx(logx+2)dx=

int(logx)/(x^3)dx=

For x gt 0 , if int(logx)^(5)dx=x[A(logx)^(5)+B(logx)^(4)+C(logx)^(3)+D(logx)^(2)+E(logx)+F] +constant, then A+B+C+D+E+F =