- #1

- 81

- 0

using that i get to .... int[1+cos2u]= u + sin2u/2 = u +sin u cos u... but subing back in with trig doesnt give me the right answer... any help with this method?

- Thread starter badtwistoffate
- Start date

- #1

- 81

- 0

using that i get to .... int[1+cos2u]= u + sin2u/2 = u +sin u cos u... but subing back in with trig doesnt give me the right answer... any help with this method?

- #2

- 584

- 0

Instead of setting x = 3tan(u) try...

[tex]

x = \sqrt 3 \tan \theta \Rightarrow dx = \sqrt 3 \sec ^2 \theta d\theta

[/tex]

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