For 0≤x<(π/2), an antiderivative of 2tanx is:
Choices:
a. ln(sec2x)
b. 2(secx)^2
c. ln((secx)^2)
d. 2ln(cosx)
e. ln(2secx)
What I did was, i used u substitution and made u=tanx...
du=(secx)^2 dx
I then put 1/(secx)^2 on the outside of the integral, substituted 2tanx with 2u, found the anti derivative of 2u which is (2u^2)/2...
then I put tanx back in, and got (tanx)^2...which I divided by (secx)^2 since 1/(secx)^2 was on the outside...
((tanx)^2)/((secx)^2)=(sinx)^2 but that is not one of the choices... so I probably did something wrong... not sure how to do this problem correctly... help...
thanks!!!
2007-04-28
10:40:00
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5 answers
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asked by
Empty Spaces
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Mathematics