Partial fractions to determine antiderivative of sec x

[hpnhpluv]

Homework Statement

Derive a formula for the antiderivative of sec x using the identity that sec x= cos x/ (1-sin^2x). Use a substitution for sin x and then partial fractions. Then multiply the solution by (1+sin x)/ (1+sin x) to obtain the more familiar formula for the antiderivative.

Homework Equations

Known antiderivative of sec x: ln (tanx+secx)+c

The Attempt at a Solution

U= sinx, du =cosxdx
intergral of du/1-u^2
integral of A/(1-u) + integral of B/(1+u) = integral of 1/(1-u^2)
A+B=1
Au-Bu=0
A= 1/2, B=1/2
plugging back in and integrating I got 1/2(ln(1-sinx)+ln(1+sinx))... can this be simplified more?

I am confused about how multiplying my solution by (1+sinx)/(1+sinx) is going to give me the formula I want... perhaps I made a math error? Please help!

 

[Dick]

Yep, math error. The integral of 1/(1-u) isn't log(1-u). There's a sign error. Do you see it?

 

[hpnhpluv]

Yep, math error. The integral of 1/(1-u) isn't log(1-u). There's a sign error. Do you see it?

I still don't see it...

 

[gabbagabbahey]

As Dick said, you have a sign error...once you've corrected it you can combine your two log terms using the following property of logs :

$$\ln (a) - \ln (b) = \ln \left( \frac{a}{b} \right)$$

 

[gabbagabbahey]

I still don't see it...

$$\frac{d}{du} \ln (1-u) = \frac{-1}{1-u}$$

 

[hpnhpluv]

Ok...so the integral in question should have been -ln(1-u)?

 

[Dick]

I still don't see it...

The integral of 1/(1-u) is -log(1-u). Check it by differentiating. Use the chain rule.

 

[hpnhpluv]

Oh, didn't see your last reply. Thanks. :)

 

[Dick]

Ok...so the integral in question should have been -ln(1-u)?

Sure.

 

[hpnhpluv]

When I fixed my solution, I got 1/2*ln((1+sinx)/(1-sinx)).

Does that look right?

 

[Dick]

Yes. Can you finish it? Use the hint.

 

[gabbagabbahey]

When I fixed my solution, I got 1/2*ln((1+sinx)/(1-sinx)).

Does that look right?

Looks fine, now what do you get when you multiply the argument of you logarithm by (1+sinx)/(1+sinx)?

 

[Dick]

Looks fine, now what do you get when you multiply the argument of you logarithm by (1+sinx)/(1+sinx)?

Slow night out there, gabbagabba?

 

[gabbagabbahey]

Slow night out there, gabbagabba?

All too often

*quietly butts out of thread*

 

[hpnhpluv]

Looks fine, now what do you get when you multiply the argument of you logarithm by (1+sinx)/(1+sinx)?

I know 1/2ln ((1+sinx)/(1-sinx)) is equivalent to saying ln ((1+sinx)/(1-sinx))^1/2...

 

[Dick]

Sure, can you get rid of those 1/2's by making the argument of the log a perfect square?

 

[Dick]

All too often

*quietly butts out of thread*

No problem, but is was kind of echoy. We are saying exactly the same thing. Know anything about Banach spaces? There's a Carl140 problem way down the list I haven't had much time to think about.

 

[hpnhpluv]

Sure, can you get rid of those 1/2's by making the argument of the log a perfect square?

By completing the square? I'm not quite sure how to do that in this case...

 

[Dick]

I told you, use the hint. Like gabbagabba said, multiply (1+sin(x))/(1-sin(x)) by (1+sin(x))/(1+sin(x)). Don't expand the numerator. Do expand the denominator.

 

[hpnhpluv]

I told you, use the hint. Like gabbagabba said, multiply (1+sin(x))/(1-sin(x)) by (1+sin(x))/(1+sin(x)). Don't expand the numerator. Do expand the denominator.

Ok, so I got the denominator to be cos^2(x), which I know is correct. My numerator is (1+sinx)*(1+sinx)^(1/2). I know this has to equal sin(x). I just don't see how...

Thank you for being so patient with me.

 

[Dick]

No problem. The numerator is (1+sin(x))^2. The denominator is cos(x)^2. So you have (1/2)*log((1+sin(x))^2/cos(x)^2). I think you might be tired.

 

[hpnhpluv]

No problem. The numerator is (1+sin(x))^2. The denominator is cos(x)^2. So you have (1/2)*log((1+sin(x))^2/cos(x)^2). I think you might be tired.

Ahh, I see now! I got it! Thank you so incredibly much for your help!

And you are right, I am tired. It's been a long week...