What is the integral of #cos^2x * csc^3x#?

1 Answer
Oct 12, 2016

#intcos^2x*csc^3xdx=-1/2cscxcotx+1/2ln(cscx-cotx)+c#

Explanation:

#cos^2x*csc^3x=cos^2x*1/sin^3x#

= #cos^2x/sin^2x*cscx=cot^2x*cscx=(csc^2x-1)*cscx# ....(A)

Hence #intcos^2x*csc^3xdx=int(csc^2x-1)*cscxdx#

= #intcsc^2xcscxdx-intcscxdx#

Let us process the them separately

First Part - We use integration by parts for #intcsc^2xcscxdx#, considering #u=cscx# and #v=-cotx# and then #du=-cotxcscxdx# and #dv=csc^2xdx# and integrating by parts, as #intudv=uv-intvdu# we have

#intcscx*csc^2xdx=-cscxcotx-int(-cotx)(-cotxcscxdx)#

= #-cscxcotx-intcot^2xcscxdx# ....(B)

Second Part #intcscxdx=int(cscx(cscx-cotx))/((cscx-cotx))dx#

= #int(csc²x-cscxcotx)/(cscx-cotx)dx#

= #int(-cscxcotx+csc²x)/(cscx-cotx)dx#

As numerator is differential of denominator this is

= #ln(cscx-cotx)# ....(C)

Combining (B) and (C)

#intcos^2x*csc^3xdx=-cscxcotx-intcot^2xcscxdx+ln(cscx-cotx)#

Observe from (A) that #cos^2x*csc^3x=cot^2xcscx#.

Hence, this becomes

#2intcos^2x*csc^3xdx=-cscxcotx+ln(cscx-cotx)# and

#intcos^2x*csc^3xdx=-1/2cscxcotx+1/2ln(cscx-cotx)+c#