Are You Ready to Challenge Your Integral Solving Skills?

[Gib Z]

Ahh but when I but the integral into mathematica I get csc x - cot x, which is what I sort of got the other way...and when I derive tan x/2 i don't get the same thing..

 

[cristo]

Well, I get tan(x/2) and so does the integrator. Try expressing cosecx - cotx in terms of x/2 using the double angle formulae.

 

[Gib Z]

Maybe there's something wrong with the page I go to:

http://www.calc101.com/webMathematica/integrals.jsp

Another then I guess, If i got this one..

 

[Gib Z]

>.< Another Integral I meant. Somebody?

 

[tehno]

Hey guys I was just wondering If you guys could post over a few indefinite integrals for me to solve, because I seem to find some of the integrals on these forums easy, and others I am completely lost, I just want to see where I am up to so I can start learning from there, my learning of calculus has been independent, so it hasnt been structured very well...

Just post maybe a few at a time, with varying difficulty and I'll try to do them. I would prefer to get advice and do them on this thread rather than learn from a link if you guys don't mind...

Here are few more examples for you and Tom1992.
Enjoy evaluating:
\int \sqrt{x^6+2x^4-x^2}dx

\int \sqrt{x^4-2x^3+2x^2}dx

\int \frac{arcsin(e^x)}{e^x}dx

 

[Schrodinger's Dog]

The bottom one I'd integrate arcsin (e^x) with e^x where x=a then substitute.

Then use 1(du)/e^x to integrate the problem. Then use the fact that 1(du)/e^x is the same as log^x(du) or -e^-x. And erm how now? Am I anywhere near? Sorry my integration needs work

How good am I: not very one day

 

[Gib Z]

Is the arcsin one ln (1-\sqrt{1-u}) - \frac{arcsin u}{u} where u = e^x? I was swear I was right, but when i check it it says I am slightly off >.<

 

[quasar987]

Here's one we could not do last week. But all the people I was with (including me) all suck at integrals so I don't know if it's really hard. In either case, if you can do it, I'd be interested to see how:

\int\sqrt{-x^2+2x+17}dx

 

[Schrodinger's Dog]

Here's one we could not do last week. But all the people I was with (including me) all suck at integrals so I don't know if it's really hard. In either case, if you can do it, I'd be interested to see how:

\int \sqrt {-x^2+2x+17}dx

\int \frac {1}{2}\sqrt -x^2+2x+17 =<br /> <br /> \frac {1}{2} \int \sqrt {-x^2+2x+17}

er maybe not

pass?

Punching above my weight I think.

I'll give it to someone who can if you like?

mathcalc101 says it's soluble, but it looks a real nightmare.

-18 sin^{-1} \frac {(x-1)}{(3\sqrt 2)} where did that come from?

 

[J77]

\int\sqrt{-x^2+2x+17}dx=\int\sqrt{(3\sqrt{2})^2-(1-x)^2}dx

let y=1-x\implies-\int\sqrt{(3\sqrt{2})^2-y^2}dy

which equals

-\frac{y}{2}\sqrt{18-y^2}-9\arcsin\frac{y}{3\sqrt{2}}

how good am i?



ps: sd's solution above is wrong.

 

[dextercioby]

A correct solution to an integration involving square roots should always include an analysis of the sign of the quadatic under the square root sign.

 

[Schrodinger's Dog]

\int\sqrt{-x^2+2x+17}dx=\int\sqrt{(3\sqrt{2})^2-(1-x)^2}dx

let y=1-x\implies-\int\sqrt{(3\sqrt{2})^2-y^2}dy

which equals

-\frac{y}{2}\sqrt{18-y^2}-9\arcsin\frac{y}{3\sqrt{2}}

how good am i?



ps: sd's solution above is wrong.

no really, NS Sherlock.

I copied that off mathcad what I mean to say was it's (du) -18 sin^{-1} \frac {(x-1)}{(3\sqrt 2)} - (du)

that's what I meant sorry :/.

I'll copy the answer off mathcalc, but it's probably equivalent anyway although it is much different.

\int\sqrt{-x^2+2x+17}dx= \frac{1}{2} (\sqrt -x^2+2x+17x)+--18 sin^{-1} \frac {(x-1)}{(3\sqrt 2)} - \sqrt -x^2+17x+17)+C

Actually looking at your solution? If it's equivalent I'm not seeing it, if you could explain

I need a good lesson, my integration is not great atm.

 

[J77]

A correct solution to an integration involving square roots should always include an analysis of the sign of the quadatic under the square root sign.

pedant

OK - x\in[1-\frac12\sqrt{72},1+\frac12\sqrt{72}]

 

[J77]

no really, NS Sherlock.

NS Watson

 

[Schrodinger's Dog]

NS Watson

 

[murshid_islam]

Nor did I realize Multiplying integrals resulted in a double integral.

as far as i know, multiplying integrals does not always result in a double integral. it does when the two single integrals are of variables that are independent of each other. did i put it correctly? someone?

 

[Gib Z]

I'm kinda scratchy on my Calculus, I am reviewing my Vector Calculus right now, which I am Glad to say I remember most of. I've never encountered something like that double integral method of solving integrals, like the one murshid_islam gave...Maybe Someone could gimme pointers on those?

 

[quasar987]

I think the general idea is that if we can say that* I=\sqrt{I^2}, then Fubini's theorem permits us to interchange the order of integration at will, and then we can make a change of variables, such as r(x,y)=\sqrt{x^2+y^2}, to put the integral in a form that can be evaluated using the FTC.

*This implies knowing that I, if it exists, is real and positive.

 

[tehno]

Is the arcsin one ln (1-\sqrt{1-u}) - \frac{arcsin u}{u} where u = e^x? I was swear I was right, but when i check it it says I am slightly off >.<

You're right that you are wrong about your solution for that integral.
Here's the most important hint for solving it:
Write it as \int e^{-x}\cdot arcsin \ e^x \ dx and do it by a method of partial integration.

 

[Gib Z]

RAWR I silly mistaked! I did it the same as my first attempt, but on that time Near the end, I had ln ( (1+root1-e^2x)/ e^x), and I simplyfied that to
ln 1+root 1-e^2x - ln e^x, and I simplified the last part to 1 instead of x...shoot me...

The Right solution is

\int \frac{\arcsin (e^x)}{e^x} dx = x - ( \frac{arcsin (e^x)}{e^x} + log_e ( 1+\sqrt{1-e^{2x}} ) )

Yay

 

[Warr]

I'm kinda scratchy on my Calculus, I am reviewing my Vector Calculus right now, which I am Glad to say I remember most of. I've never encountered something like that double integral method of solving integrals, like the one murshid_islam gave...Maybe Someone could gimme pointers on those?

Double integrals just mean that more than one variable can change. You wouldn't have double integrals for functions of one variable. The method of solving double integrals (and higher order integrals for that matter) depends mainly on choosing the order of integration (which variable to integrate first), and the implied limits of integration based on this choice. Any calc textbook that goes into multivariable calc. should have them.

 

[Gib Z]

Ahh yes right, The book I am doing my vector calculus in goes through those lol.

Funnily enough the titles - Multivariable Calculus by McCALLUM, HUGHES-HALLETT, GLEASON, ET AL.

What ever that means...yea well the vector bits are on page 87, while the double integrals are further..Ill try to get there lol

 

[Gib Z]

Umm, okie the double integrals I am working on. Someone give me a single variable one again :)

 

[murshid_islam]

try this: \int_{0}^{1}\exp\left(x^2\right)dx

 

[Gib Z]

I don't think that can be done in terms of elementary functions >.< All I could tell from finding its derivative is that it grows really quickly, so the area is probably large for larger limits of integration.

1.4626 is what I get from some simpsons rule. That doesn't look familiar to me >.<

I tried repeated integration by parts, not working.

 

[murshid_islam]

try expanding \exp(x^2) into a series and then integrate term by term. after that put the limits and you should get a series that looks like \sum_{k=0}^{\infty}\frac{1}{(2k+1)k!}

 

[Gib Z]

O thats, I did that and it becomes the series you stated, but how can i evaluate that? I know it will converge, just not what to >.<

 

[murshid_islam]

can anybody help me on this?

\int_{0}^{\pi}\frac{\theta\sin\theta}{1 + \cos^4\theta}d\theta

 

[ssd]

\int_{0}^{\pi}\frac{\theta\sin\theta}{1 + \cos^4\theta}d\theta<br /> =\int_{0}^{\pi}\frac{\pi\sin\theta}{1 + \cos^4\theta}d\theta<br /> - \int_{0}^{\pi}\frac{\theta\sin\theta}{1 + \cos^4\theta}d\theta<br />
Move the 2nd integral to left to get 2I, I=the original integral.
For the first integral put z=cos\theta

 

[murshid_islam]

how did you get this:

\int_{0}^{\pi}\frac{\theta\sin\theta}{1 + \cos^4\theta}d\theta=\int_{0}^{\pi}\frac{\pi\sin\theta}{1 + \cos^4\theta}d\theta- \int_{0}^{\pi}\frac{\theta\sin\theta}{1 + \cos^4\theta}d\theta