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# little help watch

1. how do u intergrate secx again ????

cant remember

thanks
2. ∫ secx dx

= ∫ [secx(secx+tanx)]/(secx+tanx) dx

= ∫ (sec²x +secxtanx)/(tanx+secx) dx

Top is derivative of the bottom, so

∫ secx dx = ln(secx+tanx) dx
3. Or: I = ∫1/cosxdx = ∫cosx/cos²x dx = ∫du/(1-u²) = ∫du/(1-u).(1+u)
4. (Original post by BCHL85)
Or: I = ∫1/cosxdx = ∫cosx/cos²x dx = ∫du/(1-u²) = ∫du/(1-u).(1+u)
Or you can use the t = tan(x/2) substitution
5. Or if you're doing edexcel, just look in the formula book.

http://www.edexcel.org.uk/VirtualCon...cification.pdf

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