How do you evaluate the integral from 0 to #pi/4# of #(1 + cos^2 x) / (cos^2 x) dx#?

1 Answer
Apr 10, 2015

Just do simple math : #int_0^(pi/4)(1+cos^2(x))/cos^2(x) dx #

#=int_0^(pi/4)1/cos^2(x)+1 dx#

#=(tan(x)+x)~|_0^(pi/4)#

#=1+pi/4#

EDIT : Here the "Funny" answer !

#int_0^(pi/4)(1+cos^2(x))/cos^2(x)dx = int_0^(pi/4)(2+cos^2(x)-1)/cos^2(x)dx#

Factorize : #int_0^(pi/4)(-(1-cos^2(x))+2)/cos^2(x)dx = int_0^(pi/4)(-sin^2(x)+2)/cos^2(x)dx#

#=-int_0^(pi/4)sin^2(x)/cos^2(x) dx+2int_0^(pi/4)1/cos^2(x) dx#

#=-int_0^(pi/4)1+tan^2(x)-1 dx + 2int_0^(pi/4)1/cos^2(x)dx#

#=-(tan(x)-x)~|_0^(pi/4)+2(tan(x))~|_0^(pi/4)#

#=-1+pi/4+2 = 1+pi/4#