Question

∫ cos(log x) dx=......+c ,Select correct option from the given options.
(a) x/2[cos(log x)+sin(log x)]
(b) x/4[cos(log x)+sin(log x)]
(c) x/2[cos(log x)-sin(log x)]
(d) x/2[sin(log x)-cos(log x)]

Answer 1

we have to find the value of $\int{cos(logx)}\,dx$

Let logx = P => e^P = x.......(1)
differentiate both sides ,
1/x = dP/dx
dx = xdP.......(2)
from equations (1) and (2),
now, $\int{cos(logx)}\,dx=\int{e^P.cosP}\,dP$

$I=\int{e^P.cosP}\,dP$.......(3)

$=\int{e^P\{cosP-sinP+sinP\}}\,dP$

$=\int{e^P\{cosP-sinP\}}\,dP+\int{e^PsinP}\,dP$

$=e^PcosP+\int{e^P\{sinP+cosP-cosP\}}\,dP$

[use $\int{e^{(ax+b)}\{f(x)+f'(x)\}}\,dx=\frac{1}{a}e^{(ax+b)}f(x)+C$ ]

$=xcos(logx)+\int{e^P\{sinP+cosP\}}\,dP-\int{e^PcosP}\,dP$

$=xcos(logx)+xsinx(logx)-I$

$I+I=x[cos(logx)+sin(logx)]$

$I=\frac{x}{2}[cos(logx)+sin(logx]$

hence, option (a) is correct.