How do you find #int (cosx-sinx)^2(cscx/tanx) #?

Question

1514 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-22T16:46:36

#I=-[cscx+2cosx+2ln|cscx-cotx|+c#

Here,

#I=int(cosx-sinx)^2(cscx/tanx)dx#

#=int(cos^2x-2sinxcosx+sin^2x)((1/sinx)/(sinx/cosx))dx#

#=int(1-2sinxcosx)(cosx/sin^2x)dx#

#=int{cosx/sin^2x-(2sinxcos^2x)/sin^2x}dx#

#=int{1/sinx*cosx/sinx-(2cos^2x)/sinx}dx#

#=int{1/sinx*cosx/sinx-(2(1-sin^2x))/sinx}dx#

#=int{cscxcotx-2/sinx+(2sin^2x)/sinx}dx#

#=int{cscxcotx-2cscx+2sinx}dx#

#=-cscx-2ln|cscx-cotx|-2cosx+c#

#:.I=-[cscx+2cosx+2ln|cscx-cotx|+c#