MHB Integral Convergence and Divergence I

[ardentmed]

Hello,

I'm doubting a couple of my answers for these questions. Some of them seem relatively simple, but there are slight nuances that I'm not sure of.

This thread is only for question 4.

For 4a, I just used the (a^2) - (x^2) => x=asin(Ø) rule and substituted 3sin(Ø) for x. I ended up getting 9sin(Ø) - 6sin^3(Ø) + (9/5)sin^5(Ø) + C.
Likewise, for 4b, I used x=tan(Ø) and got (-1/4)cot(Ø) - (Ø) + C. I'm really not sure about this one. By the way, I promise I'll get a hang of this latex coding soon. I'm having trouble with it because using the $ sign seems to mess up the rest of the post (everything after it turns into latex).

 

[Jameson]

Hi there! Welcome to MHB. :)

As you've heard, we prefer to have 1 question per thread but since there are so many here I'm not going to split them up into 6 separate threads. So here's what I suggest - pick 2-3 of these problems to start with. Which one do you want to be in this thread? I'll edit your post to focus on the one you choose. Then you can make 2 other threads for the other 2 problems. Once those are completed you can move on to the other ones.

We are very happy to help and glad you are here, just want to go about this in an efficient way.

 

[Prove It]

Hello,

I'm doubting a couple of my answers for these questions. Some of them seem relatively simple, but there are slight nuances that I'm not sure of.

For 4a, I just used the (a^2) - (x^2) => x=asin(feta) rule and substituted 3sin(feta) for x. I ended up getting 9sin(feta) - 6sin^3(feta) + (9/5)sin^5(feta) + C.
Likewise, for 4b, I used x=tan(feta) and got (-1/4)cot(feta) - (feta) + C. I'm really not sure about this one.

As for number 5 part a, after tediously expanding the partial fraction expression, I ended up getting c=1, d=0, b=1, and c=1, ultimately resulting in: ln(x) - (1/x^2) + c. I really don't think this looks right.

As for 5b, I obtained b=-1, c=-1, a=2, and 2lnx - (1/2)ln(x^2 +3) - (1/3)arctan(x/√3)

6a and 6b seemed relatively simple, with 6c seeming to be the most troublesome. I got 1/4 arcsin (x^2 / 2 ) + c for 6a after substituting u for x^2.
As for 6b, I got (x^2)/2 -2ln(x^2 +4) + c after using algebraic division and taking u=x^2 + 4.

Finally, for 6c, I took u as √(x), followed by algebraic division since the nominator should always be smaller than the denominator in order for partial fractions to be usable. This gave me 2√(x) + 2ln(√x -2 ) - 2ln(√x + 2) + c.

Finally, for 8, I got divergent (to -infinity), convergent (to π/6), and divergent (to infinity, since the first part's sum is 1/3, but lim negative infinity gives infinity, thus the summation of the two integrals gives a divergent integral). I'm sure these are right, but I'd appreciate some help, especially for 8c.I really appreciate the help guys. Thanks in advance.

By the way, I promise I'll get a hang of this latex coding soon. I'm having trouble with it because using the $ sign seems to mess up the rest of the post (everything after it turns into latex).

First of all, "feta" is a type of cheese. This guy: $\displaystyle \begin{align*} \theta \end{align*}$ is the greek letter "theta".

Second, please edit your post so that no more than two questions are in each thread, as Mark instructed in your previous thread.

 

[ardentmed]

First of all, "feta" is a type of cheese. This guy: $\displaystyle \begin{align*} \theta \end{align*}$ is the greek letter "theta".

Second, please edit your post so that no more than two questions are in each thread, as Mark instructed in your previous thread.

Gotcha. Edited accordingly. Sorry about that.

 

[Prove It]

Hello,

I'm doubting a couple of my answers for these questions. Some of them seem relatively simple, but there are slight nuances that I'm not sure of.

This thread is only for question 4.

For 4a, I just used the (a^2) - (x^2) => x=asin(Ø) rule and substituted 3sin(Ø) for x. I ended up getting 9sin(Ø) - 6sin^3(Ø) + (9/5)sin^5(Ø) + C.
Likewise, for 4b, I used x=tan(Ø) and got (-1/4)cot(Ø) - (Ø) + C. I'm really not sure about this one. By the way, I promise I'll get a hang of this latex coding soon. I'm having trouble with it because using the $ sign seems to mess up the rest of the post (everything after it turns into latex).

Your substitution for 4. a) is good. So when you substitute $\displaystyle \begin{align*} x = 3\sin{ \left( \theta \right) } \implies \mathrm{d}x = 3\cos{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$ you get

$\displaystyle \begin{align*} \int{ x^3\,\sqrt{ 9 - x^2 } \, \mathrm{d}x } &= \int{ \left[ 3\sin{ \left( \theta \right) } \right] ^3 \, \sqrt{ 9 - \left[ 3\sin{ \left( \theta \right) } \right] } \, 3\cos{ \left( \theta \right) } \, \mathrm{d}\theta } \\ &= \int{ 27\sin^3{ \left( \theta \right) } \, 3\cos{ \left( \theta \right) } \, 3\cos{ \left( \theta \right) } \, \mathrm{d}\theta } \\ &= 243 \int{ \sin^3{ \left( \theta \right) } \cos^2{ \left( \theta \right) } \, \mathrm{d}\theta} \\ &= 243 \int{ \sin{ \left( \theta \right) } \left[ 1 - \cos^2{ \left( \theta \right) } \right] \cos^2{ \left( \theta \right) } \, \mathrm{d}\theta} \\ &= -243 \int{ -\sin{ \left( \theta \right) } \left[ \cos^2{ \left( \theta \right) } - \cos^4{ \left( \theta \right) } \right] \, \mathrm{d}\theta} \end{align*}$

Now make the substitution $\displaystyle \begin{align*} u = \cos{ \left( \theta \right) } \implies \mathrm{d}u = -\sin{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} -243\int{ -\sin{ \left( \theta \right) } \left[ \cos^2{ \left( \theta \right) } - \cos^4{ \left( \theta \right) } \right] \, \mathrm{d}\theta} &= -243 \int{ u^2 - u^4\, \mathrm{d}u } \\ &= -243 \left( \frac{u^3}{3} - \frac{u^5}{5} \right) + C \\ &= -243 \left[ \frac{ \cos^3{ \left( \theta \right) } }{3} - \frac{\cos^5{ \left( \theta \right) }}{5} \right] + C \\ &= -243 \cos{ \left( \theta \right) } \left\{ \frac{\cos^2{ \left( \theta \right) }}{3} - \frac{\left[ \cos^2{ \left( \theta \right) } \right] ^2 }{5} \right\} + C \\ &= -243 \, \sqrt{ 1 - \sin^2{ \left( \theta \right) } } \left\{ \frac{1 - \sin^2{\left( \theta \right) } }{3} - \frac{\left[ 1 - \sin^2{ \left( \theta \right) } \right] ^2 }{5} \right\} + C \\ &= -243 \, \sqrt{ 1 - \left( \frac{x}{3} \right) ^2 } \left\{ \frac{1 - \left( \frac{x}{3} \right) ^2 }{3} - \frac{\left[ 1 - \left( \frac{x}{3} \right) ^2 \right] ^2 }{5} \right\} \end{align*}$

which you can now try to simplify if you want...

 

[ardentmed]

Your substitution for 4. a) is good. So when you substitute $\displaystyle \begin{align*} x = 3\sin{ \left( \theta \right) } \implies \mathrm{d}x = 3\cos{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$ you get

$\displaystyle \begin{align*} \int{ x^3\,\sqrt{ 9 - x^2 } \, \mathrm{d}x } &= \int{ \left[ 3\sin{ \left( \theta \right) } \right] ^3 \, \sqrt{ 9 - \left[ 3\sin{ \left( \theta \right) } \right] } \, 3\cos{ \left( \theta \right) } \, \mathrm{d}\theta } \\ &= \int{ 27\sin^3{ \left( \theta \right) } \, 3\cos{ \left( \theta \right) } \, 3\cos{ \left( \theta \right) } \, \mathrm{d}\theta } \\ &= 243 \int{ \sin^3{ \left( \theta \right) } \cos^2{ \left( \theta \right) } \, \mathrm{d}\theta} \\ &= 243 \int{ \sin{ \left( \theta \right) } \left[ 1 - \cos^2{ \left( \theta \right) } \right] \cos^2{ \left( \theta \right) } \, \mathrm{d}\theta} \\ &= -243 \int{ -\sin{ \left( \theta \right) } \left[ \cos^2{ \left( \theta \right) } - \cos^4{ \left( \theta \right) } \right] \, \mathrm{d}\theta} \end{align*}$

Now make the substitution $\displaystyle \begin{align*} u = \cos{ \left( \theta \right) } \implies \mathrm{d}u = -\sin{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} -243\int{ -\sin{ \left( \theta \right) } \left[ \cos^2{ \left( \theta \right) } - \cos^4{ \left( \theta \right) } \right] \, \mathrm{d}\theta} &= -243 \int{ u^2 - u^4\, \mathrm{d}u } \\ &= -243 \left( \frac{u^3}{3} - \frac{u^5}{5} \right) + C \\ &= -243 \left[ \frac{ \cos^3{ \left( \theta \right) } }{3} - \frac{\cos^5{ \left( \theta \right) }}{5} \right] + C \\ &= -243 \cos{ \left( \theta \right) } \left\{ \frac{\cos^2{ \left( \theta \right) }}{3} - \frac{\left[ \cos^2{ \left( \theta \right) } \right] ^2 }{5} \right\} + C \\ &= -243 \, \sqrt{ 1 - \sin^2{ \left( \theta \right) } } \left\{ \frac{1 - \sin^2{\left( \theta \right) } }{3} - \frac{\left[ 1 - \sin^2{ \left( \theta \right) } \right] ^2 }{5} \right\} + C \\ &= -243 \, \sqrt{ 1 - \left( \frac{x}{3} \right) ^2 } \left\{ \frac{1 - \left( \frac{x}{3} \right) ^2 }{3} - \frac{\left[ 1 - \left( \frac{x}{3} \right) ^2 \right] ^2 }{5} \right\} \end{align*}$

which you can now try to simplify if you want...

Edit: Got it. Thanks.

Thanks in advance.