Need help w/ practice probs: integration methods

[roody__poo]

Hey everyone.

First, thanks for answering my previous question about little-oh and big-oh. It helped a lot!

Now, I've got some practice problems that I need help with. We were told to do as many out of 150 problems as we could get done in preparation for our upcoming final on Friday, and I have all but 9 of them done, and could use some help. I appreciate it a lot! These aren't being turned in for grades, but after doing 141 other problems, I'll be (rhymes with slammed, starts with 'D') if I don't get the remaining problems done. It's a matter of principal now! What I have come up with as my answers just look way wrong to me. I have underlined my answers.

a) Integrate using trig. substitution: dx/x^2 (x^2 + 4)^1/2 = 2sin^-1(x/2)-x(x^2+4)^1/2+C

b) Integrate (-8x^3)dx/(4x^2)- 1 = x-ln |2x-1|+C

c) Using the Trapeziodal Rule, Integrate: upper limit (pi)^1/2, lower limit 0, cos (x^2)dx; n=10 = 1.95643

d) Integrate by parts (x^1/2) lnx dx = ln2 - 3/2

e) Integrate (x-3)dx/(27x^3) - 8 = ln|3x-2| - 3ln|3x-2|+C

f) Integrate sin^2(3x)cos^2(4x)dx = 1/8(x-(1/2)sin2x+(1/3)sin^3(2x))

g) Integrate with upper limit infinity and lower limit 1; dx/1.000002 = 1

h) Integrate (((e^t/2)-4)^ln3) e^t/2 dt = ln(1+e^t)+C

i) Integrate dx/(9x^2 -12x-1)^1/2 = (1/2)sin^-1(2x/(3^1/2))+C

Any and all help would be GREATLY appreciated. I have wasted more paper trying to figure these out than is considered to be environmentally friendly! I can actually see my hair turning grey over these.

Thanks a lot!

 

[soroban]

Re: Need help with practice problems

Hello, roody_poo!

Thank you for the explanation . . . you have a commendable attitude!
Here are a couple of them . . .

a) Integrate using trig. substitution: \(\displaystyle \L\,\int\frac{dx}{x^2\sqrt{x^2 + 4}}\)

Let \(\displaystyle x\,=\,2\tan\theta\;\;\Rightarrow\;\;dx\,=\,2\sec^2\theta\,d\theta\)

\(\displaystyle \;\;\)and radical is: \(\displaystyle \,\sqrt{x^2\,+\,4}\:=\:\sqrt{4\tan^2\theta\,+\,4} \:=\:\sqrt{4(\tan^2\theta\,+\,1)} \:=\:\sqrt{4\sec^2\theta} \:=\:2\sec\theta\)


Substitute: \(\displaystyle \L\,\int\frac{2\sec^2\theta\,d\theta}{4\tan^2\theta\cdot2\sec^2\theta} \;=\;\frac{1}{4}\int\frac{d\theta}{\tan^2\theta} \;=\;\frac{1}{4}\int\cot^2\theta\,d\theta\)

We have: \(\displaystyle \L\,\frac{1}{4}\int(\csc^2\theta\,-\,1)\,d\theta \;= \;\frac{1}{4}\left(-\cot\theta\,-\,\theta)\,+\,C\)

Back-substitute: \(\displaystyle \L\,-\frac{1}{4}\left(\frac{2}{x} \,+\,\arctan\frac{x}{2}\right)\,+\,C\)

b) Integrate: \(\displaystyle \L\,\int\frac{-8x^3\,dx}{4x^2 \,-\,1}\)

Note that the numerator is of higher degree than the denominator.

\(\displaystyle \;\;\)We must use long division: \(\displaystyle \L\,\frac{-8x^3}{4x^2\,-\,1}\:=\:-2x\,-\,\frac{2x}{4x^2\,-\,1}\)

So we have: \(\displaystyle \L\,\int\left(-2x\,-\,\frac{2x}{4x^2\,-\,1}\right)\,dx \;= \;-2\int\left(x \,+\,\frac{x}{4x^2\,-\,1}\right)\,dx\)

Therefore: \(\displaystyle \L\,-2\left(\frac{1}{2}x^2 \,+\,\frac{1}{8}\ln\left|4x^2\,-\,1\right|\right)\,+\,C \;=\;-x^2\,-\,\frac{1}{4}\ln\left|4x^2\,-\,1\right|\,+\,C\)

 

[soroban]

Re: Need help with practice problems

Hello again, Roody!

I dearly love new and challenging problems . . . here are a few more.

d) Integrate by parts: \(\displaystyle \L\,\int x^{\frac{1}{2}}\,\ln x\,dx\)

Let: \(\displaystyle u\,=\,\ln x\;\;\;\;dv\,=\,x^{\frac{1}{2}}\,dx\)

Then: \(\displaystyle du\,=\,\frac{dx}{x}\;\;\;v\,=\,\frac{2}{3}x^{\frac{3}{2}}\)


We have: \(\displaystyle \L\,\frac{2}{3}x^{\frac{3}{2}}\,\ln x \,- \,\int\left(\frac{2}{3}x^{\frac{3}{2}}\right)\left(\frac{dx}{x}\right) \;= \;\frac{2}{3}x^{\frac{3}{2}}\,\ln x\,-\,\frac{2}{3}\int x^{\frac{1}{2}}\,dx\)

Therefore: \(\displaystyle \L\,\frac{2}{3}x^{\frac{3}{2}}\,\ln x \,-\,\frac{4}{9}x^{\frac{3}{2}}\,+\,C\;=\;\frac{2}{9}x^{\frac{3}{2}}\,(3\,\ln x \,- \,2)\,+\,C\)

h) Integrate: \(\displaystyle \L\, (e^{\frac{t}{2}}\,-\,4)^{\ln3}\,e^{\frac{t}{2}}\,dt\)

This one looked scary . . . then I realized that \(\displaystyle \ln3\) is just a constant.

Let \(\displaystyle u\,=\,e^{\frac{t}{2}}\;\;\Rightarrow\;\;du\,=\,\frac{1}{2}e^{\frac{t}{2}}\,dt\;\;\Rightarrow\;\;dt\,=\,\frac{2\,du}{e^{\frac{t}{2}}}\)

Substitute: \(\displaystyle \L\,\int(u\,-\,4)^{\ln3}\,e^{\frac{t}{2}}\,\left(\frac{2\,du}{e^{\frac{t}{2}}\right) \;= \;2\int(u\,-\,4)^{\ln3}\,du\)

And we have: \(\displaystyle \L\,\frac{2}{\ln 3\,+\,1}(u\,-\,4)^{\ln3\,+\,1}\,+\,C\)

Back-substitute: \(\displaystyle \L\,\frac{2}{\ln3\,+\,1}\left(e^{\frac{t}{2}}\,-\,4\right)^{\ln3\,+\,1} \,+\,C\)

i) Integrate: \(\displaystyle \L\,\int\frac{dx}{\sqrt{9x^2\,-\,2x\,-\,1}}\)

This one is truly messy . . . (well, the way I did it)

Complete the square: \(\displaystyle \,9x^2\,-\,12x\,-\,1\;=\;9\left(x^2\,-\frac{4}{3}x\,-\,\frac{1}{9}\right)\;=\;9\left(x^2\,-\,\frac{4}{3}x\,+\,\frac{4}{9}\,-\,\frac{4}{9}\,-\,\frac{1}{9}\right)\)

\(\displaystyle \;\;\;=\;9\left[\left(x\,-\,\frac{2}{3}\right)^2\,-\,\frac{5}{9}\right] \;= \;9\left(x\,-\,\frac{2}{3}\right)^2\,-\,5\)

We have: \(\displaystyle \L\,\int \frac{dx}{\sqrt{9\left(x\,-\,\frac{2}{3}\right)^2\,-\,5}}\)

Let: \(\displaystyle 3\left(x\,-\,\frac{2}{3}\right)\,=\,\sqrt{5}\,\sec\theta\;\;\Rightarrow\;\;dx\,=\,\frac{\sqrt{5}}{3}\,\sec\theta\tan\theta\,d\theta\)
\(\displaystyle \;\;\)and the radical becomes: \(\displaystyle \,\sqrt{5}\,\tan\theta\)

Substitute: \(\displaystyle \L\,\int\frac{\frac{\sqrt{5}}{3}\,\sec\theta\tan\theta\,d\theta}{\sqrt{5}\,\tan\theta} \;= \;\frac{1}{3}\int\sec\theta\,d\theta \;=\;\frac{1}{3}\,\ln|\sec\theta\,+\,\tan\theta|\,+\,C\)

Back-substitute: \(\displaystyle \L\,\frac{1}{3}\left|\frac{3\left(x\,-\,\frac{2}{3}\right)}{\sqrt{5}}\,+\,\frac{\sqrt{9x^2\,-\,12x\,-\,1}}{\sqrt{5}}\right|\,+\,C\)

. . . \(\displaystyle \L=\;\frac{1}{3}\,\ln\left|\frac{3x\,-\,2\,+\,\sqrt{9x^2\,-12x\,-\,1}}{\sqrt{5}}\right|\,+\,C\)


. . . \(\displaystyle \L=\:\frac{1}{3}\,\ln|3x\,-\,2\,+\,\sqrt{9x^2\,-\,12x\,-\,1}|\,+\,C\)

I'll let someone else explain where the \(\displaystyle \sqrt{5}\) went.
I need a nap . . .

I can actually see my hair turning grey over these.

Problems like these can cause ulcers, loss of hair, insomnia, and loss of short-term memory.
(Did I mention baldness?)

 

[roody__poo]

Thank you for the help!

I appreciate it a lot!

If you get a chance to look at those other problems, I'll check back in periodically. If not, thanks for all the help. I feel like an idiot not seeing how to do a couple of those, but such is the life of the math student!

Thanks again!

 

[galactus]

You certainly don't have to feel like an idiot. These integrals are rather tough. They're good ones to give you practice and to see how innovative you can get. There's always more than one way to do them.


Here's a stab at e:

First, partial fractions:

\(\displaystyle \L\\\frac{x-3}{27x^{3}-8}=\frac{x-3}{(3x-2)(9x^{2}+6x+4)}\)

\(\displaystyle \L\\\frac{A}{3x-2}+\frac{Bx+C}{9x^{2}+6x+4}=x-3\)

Skipping ahead throughthe algebra, the solutions are:

\(\displaystyle \L\\A=\frac{-7}{36};\;\ B=\frac{7}{12};\;\ C=\frac{10}{9}\)

Then we have:

\(\displaystyle \L\\\int\frac{-7}{36(3x-2)}dx+\int\frac{7}{12(9x^{2}+6x+4)}dx+\int\frac{10}{9(9x^{2}+6x+4)}dx\)

Now, the first part is easy by comparison.

The 2nd part is the same as the 3rd except for the constant it's multiplied by. So, I will step through the second part and leave the 1st and 3rd to you. Okey-doke?.

You must use some imaginative substitutions to do integrals like these.

\(\displaystyle \L\\\frac{7}{12}\int\frac{1}{9x^{2}+6x+4}dx\)

Let \(\displaystyle \L\\u=\frac{3x+1}{3};\;\ du=dx;\;\ x=\frac{3u-1}{3}\)

When you do the subbing you wind up with:

\(\displaystyle \L\\\int\frac{1}{9u^{2}+3}du=\int\frac{1}{3(3u^{2}+1)}du=\frac{1}{3}\int\frac{1}{3u^{2}+1}du\)

If you remember, the derivative of arctan is \(\displaystyle \L\\\frac{1}{x^{2}+1}\)

Let \(\displaystyle \L\\w=tan^{-1}(\sqrt{3}u);\;\ dw=\frac{\sqrt{3}}{3u^{2}+1}du;\;\ \frac{1}{\sqrt{3}}dw=\frac{1}{3u^{2}+1}du\)

So, after all that you get something simple(it's about time!):

\(\displaystyle \L\\\frac{\sqrt{3}}{9}\int{dw}\)

\(\displaystyle \L\\\frac{\sqrt{3}}{9}w\)

Resub:

\(\displaystyle \L\\\frac{\sqrt{3}}{9}tan^{-1}(\sqrt{3}u)\)

Resub again:

\(\displaystyle \L\\\frac{\sqrt{3}}{9}tan^{-1}(\sqrt{3}(\frac{3x+1}{3}))\)

Now, don't forget aboput the 7/12 waiting back in the wings. You thought I'd forget, didn't you?.

That gives:

\(\displaystyle \L\\\frac{7\sqrt{3}}{108}tan^{-1}(\sqrt{3}(\frac{3x+1}{3}))\)

I hope there isn't a mistake somewhere:roll:

Like I said, paet 3 is essentially the same. The first part is just the 'ln' thing.

 

[roody__poo]

I feel like an idiot because it always looks so easy when someone else does it. It's like "Ergh! Why didn't I see that?"

Again, thanks for the help!

If anyone wants a stab at c), f), or g); feel free. All the help I've gotten has been most helpful, and I can't say "I appreciate it" enough times. I really can't.

 

[galactus]

part c is a Fresnel integral. Not doable by elementary means. That's why a sum is used.

I ran it through Maple:

I get .6628372275

For f, try the substitutions \(\displaystyle sin^{2}(3x)=\frac{1}{2}(1-cos(6x))\)

and \(\displaystyle cos^{2}(4x)=\frac{1}{2}(1+cos(8x))\)

 

[roody__poo]

How do you do that??

It looks so childishly easy. I've looked at every answer that you and the user soroban have given me, and I end up hitting my head on the desk thinking to myself " Well, duh. Why didn't I think of that? It's right in front of me!"

I hate to ask this of anyone, but if anyone wants to do f) and g), I will be eternally grateful. I keep trying them, and I'm getting different answers every time I try them. All I want is to make it to summer break, which starts for me on Thursday at 1:30pm and it isn't soon enough!

As far as galactus and soroban are concerned, all I need is an address (e-mail works too) and you are getting a Christmas card! Maybe even a Thanksgiving card too!

Much thanks for saving what little sanity I have left. (I have to be crazy to be doing this kind of math!)

 

[soroban]

Re: Need help with practice problems

Hello, roody__poo!

Is there a typo in question (e)?
It seems much too long for a single problem . . .

e) Integrate: \(\displaystyle \L\,\frac{x\,-\,3}{27x^3\,-\, 8}\,dx\)

From your answer, it looks like you thought that: \(\displaystyle \,27x^3\,-\,9\:=\3x\,-\,2)^3\) . . . not true!

Basically, I did it the same that Galactus did . . .

Using Partial Fractions: \(\displaystyle \L\,\frac{x\,-\,3}{(3x\,-\,2)(9x^2\,+\,6x\,+\,9)}\;=\;\frac{A}{3x\,-\,2}\,+\,\frac{Bx\,+\,C}{9x^2\,+\,6x\,+\,4}\)

\(\displaystyle \;\;\)we find that: \(\displaystyle \,A\,=\,-\frac{7}{36},\;B\,=\,\frac{7}{12},\;C\,=\,\frac{10}{9}\)


So we have: \(\displaystyle \L\,\frac{1}{36}\left[\int\frac{-7}{3x\,-\,2}\,dx \;+\,\int\frac{21x\,+\,40}{9x^2\,+\,6x\,+\,4}\,dx\right]\)

The first integral is of the form \(\displaystyle \frac{du}{u}\) . . . a log.


Here's where my approach diverges . . . I was taught differently.

The second integral would also be \(\displaystyle \frac{du}{u}\,\) if the numerator was \(\displaystyle 18x\,+\,6\)
Can we get it? . . . well, yes . . . sort of . . .

Some Olympic-level gymnsatics and we have: \(\displaystyle \,21x\,+\,40\;=\;\frac{7}{6}(18x\,+\,6)\,+\,33\)

So the second integral becomes two integrals:
\(\displaystyle \L\;\;\;\frac{7}{6}\int\frac{18x\,+\,6}{9x^2\,+\,6x\,+\,4}\,dx \,+\,33\int\frac{dx}{9x^2\,+\,6x\,+\,4}\)


The last integral can be hammered into an \(\displaystyle \arctan\) form by completing the square.

\(\displaystyle \;\;\)The denominator is: \(\displaystyle \,9\left(x^2\,+\,\frac{2}{3}x\,+\,\frac{4}{9}\right) \;=\;9\left(x^2\,+\,\frac{2}{3}x\,+\,\frac{1}{9}\,+\,\frac{4}{9}\,-\,\frac{1}{9}\right) \;=\;9\left[\left(x\,+\,\frac{1}{3}\right)^2\,+\,\frac{1}{3}\right]\)

The last integral becomes: \(\displaystyle \L\,\frac{11}{3}\int\frac{dx}{\left(x\,+\,\frac{1}{3}\right)^2\,+\,\frac{1}{3}}\)

\(\displaystyle \;\;\)And we have the \(\displaystyle \arctan\) form with: \(\displaystyle \,u\:=\:x\,+\,\frac{1}{3}\) and \(\displaystyle a\,=\,\frac{1}{\sqrt{3}}\)

 

[roody__poo]

That could be why I was getting a funky answer.

Since you seem to be a mathematical genius, which I am not even close to being, would you mind looking at these last two of mine:

f) Integrate sin^2(3x)cos^2(4x)dx = 1/8(x-(1/2)sin2x+(1/3)sin^3(2x))

g) Integrate with upper limit infinity and lower limit 1; dx/1.000002 = 1

I keep taking them to our teacher's work study, and I keep getting told that my answers are wrong. The work study can't help us, because she is only in differentials and doesn't know what integration is. Is peace of mind too much to ask for these days? I swear, if our final is something UBER easy, I'm going to be so dissatisfied. The pain and suffering I've gone through with these 150 problems is intolerable!

Thanks again soroban and galactus!

PS - By the by, you wouldn't be interested in taking my 2 hour final later this week, would you? ;-) Tons of fuuuuuuuuun!! No? Well, thought I'd offer anyways!

 

[galactus]

What the heck. Let's try old part f.

If we rewrite it like mentioned:

\(\displaystyle \L\\sin^{2}(3x)=\frac{1}{2}(1-cos(6x))\;\ \text{and}\;\ cos^{2}(4x)=\frac{1}{2}(1+cos(8x))\)

So, we have:

\(\displaystyle \L\\\frac{1}{4}\int{(1-cos(6x))(1+cos(8x))dx\)

Expand:

\(\displaystyle \L\\\frac{1}{4}\left[1+cos(8x)-cos(6x)-cos(6x)cos(8x)\right]\)

Integrate:

\(\displaystyle \L\\\frac{1}{4}\left[\int{dx}+\int{cos(8x)}dx-\int{cos(6x)}dx-\int{cos(6x)cos(8x)}dx\right]\)

The first three parts are a breeze:

\(\displaystyle \L\\\int{dx}=x;\;\ \int{cos(8x)}dx=\frac{sin(8x)}{8};\;\ \int{cos(6x)}dx=\frac{sin(6x)}{6}\)

Now, the last part is the trickiest:

But there's a trick I use when what is inside the parenthese of the cos are different.

You can rewrite them as:

\(\displaystyle \L\\cos(6x)cos(8x)=\frac{cos(2x)}{2}+\frac{cos(14x)}{2}\)

Subtract what's inside the cos to get the first part and add to get the second part.

8x-6x=2x and 8x+6x=14x. See?.

Now, we have:

\(\displaystyle \L\\\frac{1}{2}\int{cos(2x)}dx+\frac{1}{2}\int{cos(14x)}dx\)

Use u-substitution on them:

\(\displaystyle \L\\u=2x\;\ \text{and}\;\ \frac{du}{2}=dx\)

You should be able to finish and put it all together now.

 

[roody__poo]

On all of these so far, I've been like "Why didn't I see that?" On that one, part f), I'm glad I asked for help, because I was waaaaaaay off.

Thanks a lot!