How do you evaluate the integral $int 1/xcos(lnx)dx$ ?

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Answer 1 · 2017-02-14T23:42:30

$int 1/x cos(lnx)dx = sin(lnx)+C$

Substitute:

$t=lnx$
$dt =dx/x$

so that:

$int 1/x cos(lnx)dx = int cost dt =sint +C$

Undo the substitution and we have:

$int 1/x cos(lnx)dx = sin(lnx)+C$

Answer 2 · 2017-02-14T23:43:27

The integral equals $sin(lnx) + C$

This is a substitution problem. Note that $d/dxlnx = 1/x$ , so if we let $u = lnx$ , then $du = 1/xdx$ and $dx = xdu$ .

$=>int 1/xcos(u) * xdu$

$=> int cosu du$

$=>sinu + C$

$=>sin(lnx) + C$

Hopefully this helps!

How do you evaluate the integral int 1/xcos(lnx)dxint 1/xcos(lnx)dx?