Integration Problem (pun intended) : Integral( x^3 * (x^2 + 4)^(1/2) dx )

[eighsse]

Note: I'm sorry that I am using longhand text for the mathematical expressions, but my computer and this site don't seem to get along (i.e., when I type, only 75% of the letters actually show up), so I am writing this in Notepad.


Hi everyone, I am having a lot of trouble with one particular problem in my Calculus book (_Calculus and Its Applications_, Marvin L. Bittinger). It asks you to evaluate:


Integral( x^3 * (x^2 + 4)^(1/2) dx )


using any method. Section 5.1 taught basic integration (polynomial, b*e^ax, etc.), Section 5.5 taught substitution, Section 5.6 taught integration by parts, and Section 5.7 gave a table of 26 formulae. The back of the book says that the solution is:


1/15 * (3x^2 - 8) * (x^2 + 4)^(3/2) + C


and that Sections 5.6 and 5.7 (by parts and tables) should be used.


I have had very little trouble with the whole chapter, but cannot figure this out, even knowing the solution AND the means! I also tried substitution, in case the answer key gave an incorrect reference, or perhaps there were an alternate way to find the answer.

The entire function itself did not fit any formula on the given table, so apparently by-parts must be the first thing to do.
The way I see it, there are a few ways to separate it into parts, the most sensible being:


x^3 and (x^2 + 4)^(1/2) dx


It seems to make sense that you would want to differentiate x^3, and integrate the other side, because (x^2 + a^2)^(1/2) is one of the forms found on the given integration table.

However, the result from that integration formula is quite long, and more importantly it CONTAINS (x^2 + a^2)^(1/2), the very function being integrated, TWICE ... so it seems clear that this is not the right thing to do.


I tried using these parts:


x^2 and x * (x^2 + 4)^(1/2) dx


because this allows a seemingly nice substitution in the integrand dv,


w = x^2 + 4
dw = 2x dx


resulting in...


v = 1/2 * Integral[ w^(1/2) dw ]
v = 1/2 * 2/3 * w^(3/2)
v = 1/3 * (x^2 + 4)^(3/2)


So, Integral( du v ) = uv - Integral( v du )
= 1/3 * x^2 * (x^2 + 4)^(3/2) - Integral[ 2/3 x(x^2 + 4)^(3/2) dx ]


Now, finding Integral[ 2/3 x(x^2 + 4)^(3/2) dx ] ...


= 1/3 * Integral[ 2x(x^2 + 4)^(3/2) dx ]


Again using substitution ...

w = x^2 + 4
dw = 2x dx


= 1/3 * Integral[ w^(3/2) dw ]
= 1/3 * 2/5 * w^(5/2)
= 2/15 * (x^2 + 4)^(5/2)


which gives a final solution of ...


1/3 * x^2 * (x^2 + 4)^(3/2) - 2/15 * (x^2 + 4)^(5/2) + C


This seems, to me, like it should be correct, but I cannot find any way that this could be equal to the solution they have given.

Could anyone help me out here? Thank you!

 

[stapel]

Note: I'm sorry that I am using longhand text for the mathematical expressions, but my computer and this site don't seem to get along (i.e., when I type, only 75% of the letters actually show up), so I am writing this in Notepad.

Yeesh! That's a new problem. I've never heard of that one before. :shock:

\(\displaystyle \mbox{Evaluate, using any method: }\, \)\(\displaystyle \displaystyle \int\, \left(x^3\, \sqrt{\strut x^2\, +\, 4\,}\right)\, dx\)

Section 5.1 taught basic integration (polynomial, b*e^ax, etc.), Section 5.5 taught substitution, Section 5.6 taught integration by parts, and Section 5.7 gave a table of 26 formulae. The back of the book says that the solution is:

. . .\(\displaystyle \left(\dfrac{1}{15}\right)\, (3x^2\, -\, 8)\, \left(x^2\, +\, 4\right)^{{}^3/_2}\, +\, C\)

and that Sections 5.6 and 5.7 (by parts and tables) should be used.

Personally, I'd start by noticing that x² relates nicely, through derivatives, to x³. Then I'd try to do something with that. For instance:

. . . . .\(\displaystyle x^3\, =\, u\)

. . . . .\(\displaystyle 3x^2\, dx\, =\, du\)

But then we've got the "plus four" which isn't being helpful. So maybe we'll need to try something else....

...the result from that integration formula is quite long, and more importantly it CONTAINS (x^2 + a^2)^(1/2), the very function being integrated, TWICE ... so it seems clear that this is not the right thing to do.

Actually, sometimes that "looping" can be exactly what you need. In such a case, you name the part (usually an integral) being looped (that is, repeating itself) as "I", and then work with it almost like a variable itself. You end up with something you can complete, and then maybe divide by 2 or some such.

I tried using these parts:

. . . . .\(\displaystyle w\, =\, x^2\, +\, 4\)

. . . . .\(\displaystyle dw\, =\, 2x\, dx\)

resulting in...

I'm not sure I follow what you're doing here...? Let's try working "by parts". (I don't know what formulae you have in your tables. I'll be assuming something along the lines of this or this.) Since the square root integration is a pain, I'll start by assuming that this will be the part that we differentiate; otherwise, the thing gets only way messier.

. . . . .\(\displaystyle u\, =\, \sqrt{\strut x^2\, +\, 4\,}\)

. . . . .\(\displaystyle du\, =\, \dfrac{x}{\sqrt{\strut x^2\, +\, 4\,}}\, dx\)

Wait. Multiplying the derivative's numerator's x by x² would give us x³. Maybe there's a different substitution we could do...?

. . . . .\(\displaystyle u\, =\, x^2\, +\, 4\)

. . . . .\(\displaystyle u\, -\, 4\, =\, x^2\)

. . . . .\(\displaystyle du\, =\, 2x\, dx\)

. . . . .\(\displaystyle \dfrac{1}{2}\, du\, =\, x\, dx\)

. . . . .\(\displaystyle \sqrt{\strut x^2\, +\, 4\,}\, =\, \sqrt{\strut u\,}\)

. . . . .\(\displaystyle x^3\, \sqrt{\strut x^2\, +\, 4\,}\,dx\, =\, (x^2)\, \sqrt{\strut x^2\, +\, 4\,}\, (x\, dx)\, =\, (u\, -\, 4)\, \sqrt{\strut u\,}\, \dfrac{1}{2}\, du\)

Okay, that looks hopeful. Just polynomial stuff, right? So:

. . . . .\(\displaystyle \displaystyle \int\, \left(x^3\, \sqrt{\strut x^2\, +\, 4\,}\right)\, dx\, =\, \dfrac{1}{2}\, \int\, \bigg(\left(u\, -\, 4\right)\, u^{{}^1/_2}\bigg)\, du\, =\, \dfrac{1}{2}\, \int\, \left(\, u^{{}^3/_2}\, -\, 4u^{{}^1/_2}\, \right)\, du\)

Where does this lead?

 

[Deleted member 4993]

Integration Problem (pun intended) : Integral( x^3 * (x^2 + 4)^(1/2) dx )

\(\displaystyle \displaystyle{I = \int x^3 * \sqrt{x^2+4}dx \ = \ \int x^2 * \frac{1}{2}(2x)\sqrt{x^2+4}dx}\)


\(\displaystyle \displaystyle{I = x^2 * \frac{1}{3}[x^2+4]^{\frac{3}{2}} - \frac{1}{3}\int (2x) * [x^2+4]^{\frac{3}{2}}dx}\)

\(\displaystyle \displaystyle{3I = x^2 * [x^2+4]^{\frac{3}{2}} - \frac{2}{5} [x^2+4]^{\frac{5}{2}}}\)

\(\displaystyle \displaystyle{15I = [x^2+4]^{\frac{3}{2}}*[5x^2 - 2 x^2 - 8]}\)

and simplify.....

 

[eighsse]

Thanks!

Thank you very much, guys!

As for the typing issue, I have a lot of trouble with certain sites on this computer -- I have 30mbps internet, but my PC I believe is bogged down with who knows what kind of virus crap.


Anyway, so basically, what I did was correct, I just didn't take it the final couple of steps to the simplified form they gave in the back of the book. Really the only thing standing in my way was a silly mistake causing me to think that ( ... )^(3/2) cannot be easily factored out of ( ... )^(5/2) simply because 5/2 is not a multiple of 3/2. Obviously this is not true. A "duh" on me, but it was a little confusing because the back of the book indicated that substitution was not used, so I thought I was just all wrong.


Thank you for your thoughtful and prompt responses!

 

[Prove It]

Integral( x^3 * (x^2 + 4)^(1/2) dx )

\(\displaystyle \displaystyle \begin{align*} \int{x^3\,\left( x^2 + 4 \right) ^{\frac{1}{2}}\,\mathrm{d}x} \end{align*}\)

Let \(\displaystyle \displaystyle \begin{align*} x = 2\sinh{ \left( t \right) } \implies \mathrm{d}x = 2\cosh{ \left( t \right) } \,\mathrm{d}t \end{align*}\) and the integral becomes

\(\displaystyle \displaystyle \begin{align*} \int{ x^3\,\left( x^2 + 4 \right) ^{\frac{1}{2}} \,\mathrm{d}x} &= \int{ \left[ 2\sinh{(t)} \right] ^3\,\left\{ \left[ 2\sinh{(t)} \right] ^2 + 4 \right\} ^{\frac{1}{2}}\,2\cosh{(t)}\,\mathrm{d}t } \\ &= 2\int{ 8\sinh^3{\left( t \right) } \, \left[ 4\sinh^2{ (t)} + 4 \right]^{\frac{1}{2}} \,\cosh{(t)} \,\mathrm{d}t } \\ &= 16\int{ \sinh^3{(t)} \,\left\{ 4\,\left[ \sinh^2{(t)} + 1 \right] \right\} ^{\frac{1}{2}} \,\cosh{(t)}\,\mathrm{d}t } \\ &= 16\int{ \sinh^3{(t)} \, \left[ 4\cosh^2{(t)} \right] ^{\frac{1}{2}}\,\cosh{(t)}\,\mathrm{d}t } \\ &= 16\int{ \sinh^3{(t)}\,\left[ 2\cosh{(t)} \right]\,\cosh{(t)}\,\mathrm{d}t } \\ &= 32\int{ \cosh^2{(t)}\sinh^3{(t)}\,\mathrm{d}t } \\ &= 32\int{ \cosh^2{(t)}\sinh^2{(t)}\sinh{(t)}\,\mathrm{d}t } \\ &= 32\int{ \cosh^2{(t)}\,\left[ \cosh^2{(t)} - 1 \right] \,\sinh{(t)} \,\mathrm{d}t } \\ &= 32\int{ \left[ \cosh^4{(t)} - \cosh^2{(t)} \right]\,\sinh{(t)}\,\mathrm{d}t } \\ &= 32\int{ \left( u^4 - u^2 \right) \,\mathrm{d}u } \textrm{ after substituting } u = \cosh{(t)} \implies \mathrm{d}u = \sinh{(t)}\,\mathrm{d}t \\ &= 32\,\left( \frac{u^5}{5} - \frac{u^3}{3} \right) + C \\ &= 32\,\left[ \frac{\cosh^5{(t)}}{5} - \frac{\cosh^3{(t)}}{3} \right] + C \\ &= 32 \,\left\{ \frac{\left[ \cosh^2{(t)} \right] ^{\frac{5}{2}}}{5} - \frac{\left[ \cosh^2{(t)} \right] ^{\frac{3}{2}}}{3} \right\} + C \\ &= 32\,\left\{ \frac{\left[ 1 + \sinh^2{(t)} \right] ^{\frac{5}{2}} }{5} - \frac{\left[ 1 + \sinh^2{(t)} \right] ^{\frac{3}{2}}}{3} \right\} + C \\ &= 32\,\left\{ \frac{\left[ 1 + \left( \frac{x}{2} \right) ^2 \right] ^{\frac{5}{2}}}{5} - \frac{\left[ 1 +\left( \frac{x}{2} \right) ^2 \right] ^{\frac{3}{2}}}{3} \right\} + C \\ &= 32\,\left[ \frac{\left( \frac{4 + x^2}{4} \right) ^{\frac{5}{2}}}{5} - \frac{\left( \frac{4 + x^2}{4} \right) ^{\frac{3}{2}}}{3} \right] + C \\ &= 32\,\left[ \frac{\frac{\left( 4 + x^2 \right) ^{\frac{5}{2}}}{32}}{5} - \frac{\frac{\left( 4 + x^2 \right) ^{\frac{3}{2}}}{8}}{3} \right] + C \\ &= \frac{\left( 4 + x^2 \right) ^{\frac{5}{2}}}{5} - \frac{4\,\left( 4 + x^2 \right) ^{\frac{3}{2}}}{3} + C \\ &= \left( 4 + x^2 \right) ^{\frac{3}{2}} \left[ \frac{4 + x^2}{5} - \frac{4}{3} \right] + C \\ &= \left( 4 + x^2 \right) ^{\frac{3}{2}} \, \left( \frac{12 + 3\,x^2}{15} - \frac{20}{15} \right) + C \\ &= \left( 4 + x^2 \right) ^{\frac{3}{2}}\,\left( \frac{3\,x^2 - 8}{15} \right) + C \end{align*}\)