= & gty = - ln| cos(x+C 1)| + C2

10.continue: 1+ 1/p2 = E(-2x-2c 1)

= & gtp = 1/[e^(-2x-2c 1]- 1 ]^( 1/2)

= & gty ' = e^(x+c 1)/[ 1-e^(2x+2c 1)]^( 1/2)

= & gty = arcsin[e^(x+c 1]]+C2

The results of both questions are consistent with those in the book. However, 10. If you choose the method you mentioned, the integration process will be much simpler.

But usually this kind of topic is chosen as y' = p(x).