How do you integrate #(sec(x)^2-1)(sin(x)/cos(x))#?

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Answer 1 · 2017-01-13T06:19:51

#int(sec^2x-1)(sinx/cosx)dx=tan^2x/2+lncosx+c#

#int(sec^2x-1)(sinx/cosx)dx#

#=int(sec^2x xxsinx/cosx)dx-intsinx/cosxdx#

#=intsec^2xtanxdx-intsinx/cosxdx#

Let us first solve #intsec^2xtanxdx#

let #u=tanx#, then #du=sec^2xdx#

Hence #intsec^2xtanxdx=intudu#

#=u^2/2=tan^2x/2#

For #intsinx/cosxdx#, let #v=cosx# than #dv=-sinxdx# and

#intsinx/cosxdx=int-dv/v=-lnv=-lncosx#

Hence #int(sec^2x-1)(sinx/cosx)dx=tan^2x/2+lncosx+c#