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

1 Answer
maganbhai P.
Jul 22, 2018

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

Explanation:

Here,

I=int(cosx-sinx)^2(cscx/tanx)dxI=∫(cosx−sinx)2(cscxtanx)dx

=int(cos^2x-2sinxcosx+sin^2x)((1/sinx)/(sinx/cosx))dx=∫(cos2x−2sinxcosx+sin2x)(1sinxsinxcosx)dx

=int(1-2sinxcosx)(cosx/sin^2x)dx=∫(1−2sinxcosx)(cosxsin2x)dx

=int{cosx/sin^2x-(2sinxcos^2x)/sin^2x}dx=∫{cosxsin2x−2sinxcos2xsin2x}dx

=int{1/sinx*cosx/sinx-(2cos^2x)/sinx}dx=∫{1sinx⋅cosxsinx−2cos2xsinx}dx

=int{1/sinx*cosx/sinx-(2(1-sin^2x))/sinx}dx=∫{1sinx⋅cosxsinx−2(1−sin2x)sinx}dx

=int{cscxcotx-2/sinx+(2sin^2x)/sinx}dx=∫{cscxcotx−2sinx+2sin2xsinx}dx

=int{cscxcotx-2cscx+2sinx}dx=∫{cscxcotx−2cscx+2sinx}dx

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

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

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