What is the integral of #( cos x +sec x )^2#?

Question

7321 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-01T17:41:52

#1/4sin(2x)+5/2x+tanx+C#

Note that #(cosx+secx)^2=cos^2x+2secxcosx+sec^2x=cos^2x+2+sec^2x#.

Thus, we have:

#intcos^2xdx+2intdx+intsec^2dx#

The last two are common integrals:

#=intcos^2xdx+2x+tanx#

For the remaining integral, use the following identity:

#cos(2x)=2cos^2x-1" "=>" "cos^2x=(cos(2x)+1)/2#

Thus the integral equals:

#=1/2intcos(2x)dx+1/2intdx+2x+tanx#

#=1/2intcos(2x)dx+1/2x+2x+tanx#

#=1/2intcos(2x)dx+5/2x+tanx#

For the first integral, let #u=2x#, so #du=2dx#.

#=1/4int2cos(2x)dx+5/2x+tanx#

#=1/4intcos(u)du+5/2x+tanx#

#=1/4sin(u)+5/2x+tanx+C#

#=1/4sin(2x)+5/2x+tanx+C#