even more trig questions!

**maggiec** · June 26th 2010, 11:28 AM

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?

**Ackbeet** · June 26th 2010, 12:17 PM

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.

**maggiec** · June 26th 2010, 01:18 PM

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)?

**Ackbeet** · June 26th 2010, 01:50 PM

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?

**maggiec** · June 26th 2010, 02:02 PM

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 $\ln(\sec(\theta))$ ?

**Ackbeet** · June 26th 2010, 02:41 PM

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?

**maggiec** · June 26th 2010, 02:47 PM

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)..

**Ackbeet** · June 26th 2010, 02:49 PM

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

**maggiec** · June 26th 2010, 02:56 PM

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?

**Ackbeet** · June 26th 2010, 04:01 PM

That sounds like a plan. Try it!

**maggiec** · June 27th 2010, 01:39 AM

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) ?

**maggiec** · June 27th 2010, 02:51 AM

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!

even more trig questions!-screen-shot-2010-06-27-11.46.22.jpg

**Ackbeet** · June 28th 2010, 01:41 AM

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

**maggiec** · June 28th 2010, 01:55 AM

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)

**maggiec** · June 28th 2010, 02:01 AM

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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