Evaluate the definite integral and it exists

[afcwestwarrior]

a=0, b= pie/3 sin theta/ cos^2 theta

i let u= cos theta
du/dx= -sin theta
-du = sin theta

sin theta/u^2 theta

then i anti differentiated it

-cos theta/ (1/3)(cos)^3 theta
this is where i got stuck

 

[afcwestwarrior]

i know how to do this problem but need help with the antidifferentiation

 

[Hurkyl]

a=0, b= pie/3 sin theta/ cos^2 theta

Did you mean
\int_{0}^{\pi / 3} \frac{ \sin \theta }{\cos^2 \theta}
? If so, that makes no sense: there is a very important part of this notation that has been omitted.

It looks like you were doing a substitution -- is there any particular reason why you didn't convert the expression entirely into a function of u?

 

[afcwestwarrior]

yes that is what i meant, the reason is because sin and cos confuse me

 

[2ltben]

I think I had this same problem not too long ago. sin(u)/cos(u) is a trig property that equals tan(u), and 1/cos(u) is the definition of sec(u). Therefore, the problem condenses to tan(u)sec(u)du, which you should recognize if you're on up to integration in your studies.