I have the solution but I don't understand.
[ csc(x - π/2) ]˛
= [ -csc( -(x - π/2) ) ] ˛
= [ -csc(π/2 - x ) ]˛
= [ -sec(x) ]˛ = sec˛(x)
∫ sec˛(x) dx = tan(x) + C
From the change of cosec to sec, I am blur!
Then you need to go back and review trig functions!
sine= "opposite side over hypotenuse"
cosine= "near side over hypotenuse"
sec= "hypotenuse over near side"
csc= "hypotenuse over opposite side"
switching from csc to sec or sin to cos, you are swapping "near side" and "opposite side" which is the same as changing form one acute angle to the other in the same right triangle. The two angles are complementary: $\displaystyle \theta+ \phi= \pi/2$ so $\displaystyle \theta= \pi/2- \phi$
$\displaystyle sec(\theta)= csc(\pi/2- \theta)$ and $\displaystyle csc(\theta)= sec(\pi/2- \theta)$.