# Thread: even more trig questions!

1. Originally Posted by Ackbeet
So, number 4 works like this. You've got your integral,

$\int_{0}^{\sqrt{3}}\ln\left(\frac{2}{\sqrt{4-x^{2}}}\right)dx.$

Now then, suppose you draw your triangle with theta as the lower left angle. The squared terms have opposite signs, so you let 2 be the hypotenuse, and x the opposite side to theta. That means the adjacent side has length $\sqrt{4-x^{2}}$. You need to transform the limits, the integrand, and the differential so that there are no more x's, only theta's. Here are some relevant equations that let you do that:

$\frac{x}{2}=\sin(\theta)$

$\frac{2}{\sqrt{4-x^{2}}}=\sec(\theta)$.

So, hopefully, getting the integrand to have only theta's should be fairly straight-forward. How do you propose to get the limits and the differential to have only theta's?

Can I replace $\frac{2}{\sqrt{4-x^{2}}}$ found in the integral with sec(theta) such that I am integrating ln(sec(theta))?

In terms of the limits, they have no xs in them so can they be left as they are?

2. Yes, you can do the replacement with $\sec(\theta)$. No, you cannot leave the limits as they are. They are values of x, because of the differential. You have to change them to values of $\theta$. Or, if you want, you can convert the antiderivative back into x's before you evaluate at the limits. I generally find the first alternative less work, but you can do as you wish.

3. Originally Posted by Ackbeet
Yes, you can do the replacement with $\sec(\theta)$. No, you cannot leave the limits as they are. They are values of x, because of the differential. You have to change them to values of $\theta$. Or, if you want, you can convert the antiderivative back into x's before you evaluate at the limits. I generally find the first alternative less work, but you can do as you wish.
Thus the limits change:

when x=√3, sin $\theta$ = (√3)/2 hence $\theta$ = 60 degrees
when x = 0, $\theta$ = 0

so the new limits are 0 and 60?

any hints on how to integrate ln(sectheta)?

4. Your limits look correct to me. But $\ln(\sec(\theta))$ isn't all that's in the integrand, if I'm not mistaken. What does the new differential look like?

5. Originally Posted by Ackbeet
Your limits look correct to me. But $\ln(\sec(\theta))$ isn't all that's in the integrand, if I'm not mistaken. What does the new differential look like?
I thought it was y= $\ln\left(\frac{2}{\sqrt{4-x^{2}}}\right)$

but then sec(theta)= $\frac{2}{\sqrt{4-x^{2}}}\right)$

thus the differential becomes $\ln(\sec(\theta))$?

6. No, that's the integrand. You've successfully put the integrand into thetas. What you have to do now is find the differential $dx$. This is the same thing you have to do with any substitution (u substitution or whatever). In order to find the differential, I would recommend finding the simplest relation possible between x and theta from the triangle you drew. What do you get?

7. Originally Posted by Ackbeet
No, that's the integrand. You've successfully put the integrand into thetas. What you have to do now is find the differential $dx$. This is the same thing you have to do with any substitution (u substitution or whatever). In order to find the differential, I would recommend finding the simplest relation possible between x and theta from the triangle you drew. What do you get?

$\sin(\theta)=\frac{x}{2}$

Oh I have to find dx/d(theta)..

8. Excellent. Now take the differentials of both sides to get the needed relation.

9. 2sin(theta) = x
hence dx/d(theta) = 2 cos(theta)

hence the integral becomes

integral of: 2ln(sec(theta)).cos(theta).d(theta) between 0 and 60 degrees

Now integration by parts?

10. That sounds like a plan. Try it!

11. Originally Posted by Ackbeet
That sounds like a plan. Try it!
Can I just check, differentiating ln(sec(theta)) you get [sec(theta).tan(theta)]/(e^u) ?

12. I've sort of gotten there but half way... theres a final term I lost somewhere but Im not sure where!
Here's what I've done, hope it's clear to see!

13. I don't agree with your differentiation of $\ln(\sec(\theta))$. Use the Chain Rule. What do you get?

14. Originally Posted by Ackbeet
I don't agree with your differentiation of $\ln(\sec(\theta))$. Use the Chain Rule. What do you get?
I used that exact same method earlier on in the question and got the correct answer for the gradient?

I thought I couldn't use the chain rule because u isn't a constant..

but using the chain rule I get: du/d(theta) = tan(theta)

15. I still get sin(theta)tan(theta) having to be integrated thus giving me the same result, so both differentials lead me to the same overall integral, not sure why..

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