Calculus proofs that baffle me utterly

1John5vs7

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I'm taking a summer intensive cram course of Calculus 2. It's been 12 years since I took Calc 1. I have a headache from doing 8 hours of math homework per day for the last 4 days. Please, can you have pity on me and help?

I'm working on finding a formula from the back of Stewart's "Calculus: Concepts and Contexts, 4th edition" that lets me integrate:

Integral[ cotx / Root (1 + 2sinx)]dx

Apparently there is one. I've tried looking at the trig identities in the front and back of the book as well as all of the formulas and I don't see anything that applies to a cotangent in the numerator with a radical denominator. Moreover, I tried checking to see if the sin-squaredx + cos-squaredx = 1 identity would fit somehow...doesn't look like double angle or half angle apply here...I'm pretty lost.

I'm also stuck trying to prove the following:

a.) Integral[ (Root (a^2 + u^2)/u]du = (Root(a^2 + u^2) + a*ln |(Root(a^2 + u^2) -a)/u| + C

I get that the numerator on the left hand side is part of a trig substitution where it can be replaced by a*secant(u), but I'm sort of mystified beyond that.


b.) Integral[ (Root(a^2 - u^2))/u^2)]du = -(1/u)*(Root(a^2 - u^2)) - arcsin(u/a) + C


This is part of a HUGE homework set consisting of dozens of problems and I have been sitting here for 4 or 5 hours working on it without making any real headway. Even if you can just provide a hint, that would be a HUGE boon to me.

Many thanks.


P.S. The reason I'm being a moron and taking a cram class after being away from it for 12 years is because I was recently accepted to a medical physics program, but they told me quasi-last-minute that it required me to take several undergrad classes to catch me up to speed on physics, one of which was Calc II. They *generously* offered to charge me $333/credit at 3 credits for the class, but I work for a university and get a free class per semester, so I'm getting Calc II at $0/credit, but to start my medical physics program this summer, I've had to take the summer Calc II class which is just 32 flavors of brutal and then some. Thank you for understanding...I'm not trying to be stupid or bullheaded by insisting on taking the shortest possible class...if I had another option, I'd be taking a semester-length class, but you can't beat free!
 
Last edited:
I'm taking a summer intensive cram course of Calculus 2. It's been 12 years since I took Calc 1. I have a headache from doing 8 hours of math homework per day for the last 4 days. Please, can you have pity on me and help?

I'm working on finding a formula from the back of Stewart's "Calculus: Concepts and Contexts, 4th edition" that lets me integrate:

Integral[ cotx / Root (1 + 2sinx)]dx

Apparently there is one. I've tried looking at the trig identities in the front and back of the book as well as all of the formulas and I don't see anything that applies to a cotangent in the numerator with a radical denominator. Moreover, I tried checking to see if the sin-squaredx + cos-squaredx = 1 identity would fit somehow...doesn't look like double angle or half angle apply here...I'm pretty lost.

I'm also stuck trying to prove the following:

a.) Integral[ (Root (a^2 + u^2)/u]du = (Root(a^2 + u^2) + a*ln |(Root(a^2 + u^2) -a)/u| + C

I get that the numerator on the left hand side is part of a trig substitution where it can be replaced by a*secant(u), but I'm sort of mystified beyond that.


b.) Integral[ (Root(a^2 - u^2))/u^2)]du = -(1/u)*(Root(a^2 - u^2)) - arcsin(u/a) + C


This is part of a HUGE homework set consisting of dozens of problems and I have been sitting here for 4 or 5 hours working on it without making any real headway. Even if you can just provide a hint, that would be a HUGE boon to me.

Many thanks.


P.S. The reason I'm being a moron and taking a cram class after being away from it for 12 years is because I was recently accepted to a medical physics program, but they told me quasi-last-minute that it required me to take several undergrad classes to catch me up to speed on physics, one of which was Calc II. They *generously* offered to charge me $333/credit at 3 credits for the class, but I work for a university and get a free class per semester, so I'm getting Calc II at $0/credit, but to start my medical physics program this summer, I've had to take the summer Calc II class which is just 32 flavors of brutal and then some. Thank you for understanding...I'm not trying to be stupid or bullheaded by insisting on taking the shortest possible class...if I had another option, I'd be taking a semester-length class, but you can't beat free!
I am not seeing the 1st one!
For a) use tan v =u/a and for b) use sin v = u/a

EDIT: Oh my, I thought that the denominator was 1=sin2x, not 1+2sinx. Oh well. Now it is probably easy if you make the substitution u=1+2sinx or u=sinx
 
Last edited:
I'm taking a summer intensive cram course of Calculus 2. It's been 12 years since I took Calc 1. I have a headache from doing 8 hours of math homework per day for the last 4 days. Please, can you have pity on me and help?

I'm working on finding a formula from the back of Stewart's "Calculus: Concepts and Contexts, 4th edition" that lets me integrate:

Integral[ cotx / Root (1 + 2sinx)]dx

Apparently there is one. I've tried looking at the trig identities in the front and back of the book as well as all of the formulas and I don't see anything that applies to a cotangent in the numerator with a radical denominator. Moreover, I tried checking to see if the sin-squaredx + cos-squaredx = 1 identity would fit somehow...doesn't look like double angle or half angle apply here...I'm pretty lost.

I'm also stuck trying to prove the following:

a.) Integral[ (Root (a^2 + u^2)/u]du = (Root(a^2 + u^2) + a*ln |(Root(a^2 + u^2) -a)/u| + C

I get that the numerator on the left hand side is part of a trig substitution where it can be replaced by a*secant(u), but I'm sort of mystified beyond that.


b.) Integral[ (Root(a^2 - u^2))/u^2)]du = -(1/u)*(Root(a^2 - u^2)) - arcsin(u/a) + C


This is part of a HUGE homework set consisting of dozens of problems and I have been sitting here for 4 or 5 hours working on it without making any real headway. Even if you can just provide a hint, that would be a HUGE boon to me.

Many thanks.


P.S. The reason I'm being a moron and taking a cram class after being away from it for 12 years is because I was recently accepted to a medical physics program, but they told me quasi-last-minute that it required me to take several undergrad classes to catch me up to speed on physics, one of which was Calc II. They *generously* offered to charge me $333/credit at 3 credits for the class, but I work for a university and get a free class per semester, so I'm getting Calc II at $0/credit, but to start my medical physics program this summer, I've had to take the summer Calc II class which is just 32 flavors of brutal and then some. Thank you for understanding...I'm not trying to be stupid or bullheaded by insisting on taking the shortest possible class...if I had another option, I'd be taking a semester-length class, but you can't beat free!

For \(\displaystyle \int\, \frac{cot(x)}{\sqrt{1\, +\, 2\, sin(x)}}dx\)
try a substitution of u = sin(x) and see if it helps.

For (a) and (b), if what you need to do is actually prove it (as opposed to derive it), just take the derivative of the right hand side and see if that works out.

You might ignore the following as it really has nothing to do with the problems. As an interesting aside (at least to me), in later studies you may run across complex numbers and and at that time you will be able to note the correspondence between the ln function and the sin-1. If we let c = i a where i2 = -1, note that you can write (c) as
\(\displaystyle \int\, \frac{\sqrt{i^2 c^2 - u^2}}{u} du\, =\, \int\, \frac{\sqrt{i^2 c^2 + i^2 u^2}}{u} du\, =\, \int\, \frac{\sqrt{i^2 (c^2 + u^2)}}{u} du\, =\, i\, \int\, \frac{\sqrt{c^2 + u^2}}{u} du\)
 
I am not seeing the 1st one!
For a) use tan v =u/a and for b) use sin v = u/a
I will try your advice. Thank you. I don't see how I can justify those substitutions based on what I know about calculus thus far, but I'll go look at trig again...maybe that's just an identity I missed? I mean, I recognize trig is all referencing sides of triangles and the related angles such that sin = the side opposite the angle divided by the hypotenuse etc. etc. But I don't see how these are justifiable. Still, I'll plug them in and see what happens.
 
For \(\displaystyle \int\, \frac{cot(x)}{\sqrt{1\, +\, 2\, sin(x)}}dx\)
try a substitution of u = sin(x) and see if it helps.

For (a) and (b), if what you need to do is actually prove it (as opposed to derive it), just take the derivative of the right hand side and see if that works out.

You might ignore the following as it really has nothing to do with the problems. As an interesting aside (at least to me), in later studies you may run across complex numbers and and at that time you will be able to note the correspondence between the ln function and the sin-1. If we let c = i a where i2 = -1, note that you can write (c) as
\(\displaystyle \int\, \frac{\sqrt{i^2 c^2 - u^2}}{u} du\, =\, \int\, \frac{\sqrt{i^2 c^2 + i^2 u^2}}{u} du\, =\, \int\, \frac{\sqrt{i^2 (c^2 + u^2)}}{u} du\, =\, i\, \int\, \frac{\sqrt{c^2 + u^2}}{u} du\)
Thanks, I'll try this in the morning. Exhausted now and have an exam tomorrow, so this will be good prep.
 
1John5vs7;375216 Integral[ cotx / Root (1 + 2sinx) said:

substitute

\(\displaystyle \displaystyle{2*sin(x) = tan^2(\theta)}\)

\(\displaystyle \displaystyle{2*cos(x)\ \ dx \ = \ 2*tan(\theta) * sec^2(\theta) \ \ d(\theta)}\)

\(\displaystyle \displaystyle{\frac{cot(x)}{\sqrt{1+2sin(x)}}}\)

\(\displaystyle \displaystyle{\frac{cos(x)}{sin(x)*\sqrt{1+2sin(x)}}}\)

\(\displaystyle \displaystyle{\frac{cos(x)}{sin(x)*\sqrt{1+2sin(x)}}}\)

\(\displaystyle \displaystyle{\int \frac{cot(x)}{\sqrt{1+2sin(x)}}dx}\)

\(\displaystyle = \ \displaystyle{\int \frac{tan(\theta) * sec^2(\theta) \ \ d(\theta)}{\frac{tan^2(\theta)}{2}*sec(\theta)}}\)

\(\displaystyle = \ \displaystyle{\int \frac{2 * sec(\theta) \ \ d(\theta)}{tan(\theta)}}\)

\(\displaystyle = \ \displaystyle{2\int \frac{d(\theta)}{sin(\theta)}}\)

That integral should be familiar and substitute back.......
 
It is a shame that my detailed excellent post that would have helped the student (OP) immensely got removed. Oh well, I guess the student is on their own.
 
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