# Thread: even more trig questions!

1. You couldn't use the power rule on $\displaystyle x^{-u}$, because $\displaystyle u$ wasn't constant. However, because the function in question is a composition of functions, you use the chain rule. You can always use the chain rule on compositions of functions, unless there are domain/behavior issues (not likely encountered in Calc classes; such pathological behaviors are usually reserved for upper level courses in analysis). I like your $\displaystyle du/d\theta$ and your integrand. How do you propose to integrate?

2. using integration by parts again?

3. No, I don't think that approach would be very helpful here. I would use the trig identities $\displaystyle \sin^{2}(\theta)+\cos^{2}(\theta)=1$ and $\displaystyle \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$ to rewrite the integrand as a simple sum.

4. sec(theta) - cos(theta) d(theta) ?

ive gone through this now and my result is incomplete again
this is what i get:

√3(ln2) - 2ln(2 + √3) + √3

5. Excellent, but be careful with your parentheses, just to be on the safe side. So, how to integrate those two guys?

6. So we have

ln(sec(theta) + tan(theta)) - sin(theta) between 0 and 60 degrees

7. There you go. Now, relating all this back to the problem at hand, the area you are after is equal to the area under the curve minus the area under the triangle. Once you exhibit that difference, you are done with part b, assuming your answer corresponds to the answer required. Does all this make sense?

8. Originally Posted by Ackbeet
There you go. Now, relating all this back to the problem at hand, the area you are after is equal to the area under the curve minus the area under the triangle. Once you exhibit that difference, you are done with part b, assuming your answer corresponds to the answer required. Does all this make sense?
Because it's been so long and i've been doing other questions I'd completely forgotten about that triangle!

I have gotten to the correct answer now, thank you so much for all your help!

9. You're very welcome. Have a good one!

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