- #1
stargazer843
- 10
- 0
∫1+sinx/(cosx)² dx
I made u = cos x.
du = -sinxdx
-du = sinxdx
so:
∫1+sinxdx/(cosx)² = ∫1-du/u² = ∫ (1/u²) * (1-du)
This is where I got stuck. the 1-du is throwing me off. distributing would get me nowhere and I don't know how to get rid of the 1.
Please help!
I made u = cos x.
du = -sinxdx
-du = sinxdx
so:
∫1+sinxdx/(cosx)² = ∫1-du/u² = ∫ (1/u²) * (1-du)
This is where I got stuck. the 1-du is throwing me off. distributing would get me nowhere and I don't know how to get rid of the 1.
Please help!