Calc 2 question?

Question

I was doing my homework, and on problem 45 in Early Trancend. Calc 5th edition. I got stuck.

I'm supposed to integrate (1-(tanx)^2)/(secx)^2

I tried out some different things(changing it to cos and sin, multipling top and bottom with secs and tans) but none of them worked. The answer in the back of the book says (1/2)sin(2x) + C. I don't understand how to start. Any help?

ok i think i figured it out

to i multiply top and bottom by (secx)^2, then change the bottom (secx)^2 to 1+(tanx)^2?

Well i tried that out, set u=tanx and du=(secx)^2. But I'm stuck at integrating: (u^2-1)/(1+2u^2+u^4). Maybe its simple, but i dont see it

Answer 1

Integrate (1- tan²x) / sec²x.

First simplify the expression then integrate.

(1- tan²x) / sec²x = (1 - sin²x/cos²x) / (1/cos²x)
= [(cos²x - sin²x)/cos²x] / (1/cos²x)
= cos²x - sin²x = cos 2x

Now we can integrate.

∫[(1- tan²x) / sec²x] dx = ∫(cos 2x) dx = ½sin 2x + C

Answer 2

Easier:
∫ (1 - sin²θ/cos²θ)/(1/cos²θ) dθ
= ∫ (cos²θ - sin²θ) dθ .... use the double angle formula for cosine
= ∫ cos(2θ) dθ
then the final answer comes out....

d: