How do you use substitution to integrate #sec(v + pi/2) * tan(v + pi/2)#?

Question

2601 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-22T16:45:15

See the explanation.

#t=1/sec(v+pi/2)#

#dt=((sec(v+pi/2))^-1)'dv=#

#dt=-1/sec^2(v+pi/2)sec(v+pi/2)tan(v+pi/2)dv#

#dt=-(tan(v+pi/2)dv)/sec(v+pi/2)=-t tan(v+pi/2)dv#

#tan(v+pi/2)dv=-dt/t#

#int sec(v+pi/2) tan(v+pi/2) dv=int 1/t(-dt/t)=#

#=-int t^-2dt=1/t+C=sec(v+pi/2)+C#