What is #int cotxcosx dx#?

Question

14910 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-13T12:54:08

The answer is #=cosx-ln(|cscx+cotx|)+C#

#"Reminder"#

#cotx=cosx/sinx#

#cos^2x+sin^2x=1#

#intcscxdx=-ln(cscx+cotx)#

Therefore,

The integral is

#I=intcotxcosxdx=intcosx/sinx*cosxdx#

#=int(cos^2xdx)/sinx#

#=int(1-sin^2x)/sinxdx#

#=intdx/sinx-intsinxdx#

#=intcscxdx+cosx#

#=-ln(|cscx+cotx|)+cosx+C#