How do you integrate $int [(x-1) / (cos^2 x)]dx$ ?

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Answer 1 · 2018-06-02T10:09:15

$I=(x-1)tanx-ln|secx|+c$

Here,

$I=int[(x-1)/cos^2x]dx$

$int(x-1)sec^2xdx$

$"Using "color(blue)"Integration by Parts :"$

$color(blue)(intu*vdx=uintvdx-int(u'intvdx)dx$

Let,

$u=x-1 and v=sec^2x=>u'=1 and intvdx=tanx$

So,

$I=(x-1)*tanx-int1*tanxdx$

$I=(x-1)tanx-ln|secx|+c$

How do you integrate int [(x-1) / (cos^2 x)]dxint [(x-1) / (cos^2 x)]dx?