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

Question

1619 views around the world

You can reuse this answer
Creative Commons License

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#