How do you integrate $int cos6x dx$ ?

Question

23070 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-18T10:46:39

$intcos(6x)dx=1/6sin(6x)+C$

Using integration by substitution along with the integral $intcos(x)dx = sin(x)+C$

let $u = 6x => du = 6dx$

then

$intcos(6x)dx = 1/6intcos(6x)6dx$

$=1/6intcos(u)du$

$=1/6sin(u)+C$

$=1/6sin(6x)+C$

How do you integrate int cos6x dxint cos6x dx?