Are You Ready to Challenge Your Integral Solving Skills?

[murshid_islam]

Aww come on, at least gimme a hint...

i have edited previous post and given some hints.

Exp(x^2) = e^(x^2)?
If so Int (Exp(x^2) ) = Infinity.

sorry that was a typo. i have edited that post now.

Gib Z, hint for the second one (don't look at all of it until you have tried):

let $$I = \int_{0}^{\infty}\exp(-x^2)dx = \int_{0}^{\infty}\exp(-y^2)dy$$

now,
$$I^2 = \int_{0}^{\infty}\int_{0}^{\infty}\exp(-x^2-y^2)dxdy$$

now converting into polar coordinates, you get,
$$I^2 = \int_{0}^{\frac{\pi}{2}}\int_{0}^{\infty}\exp(-r^2)rdrd\theta = \frac{\pi}{4}$$

$$I = \frac{\sqrt{\pi}}{2}$$

 

[Gib Z]

Ok, Thanks, The first one your hint pretty much did it for me :) And for the 2nd one, I tried, I failed, and then I saw your hint. I haven't done polar coordinates, Nor did I realize Multiplying integrals resulted in a double integral. Sorry about that.

Maybe a little bit easier one perhaps :P?

 

[Gib Z]

Anyone? Come on lol

 

[cristo]

Here's a simple one: $$\int\frac{2x}{x^2+2x+1}dx$$

 

[andytoh]

I'm just curious. Doesn't practising with integrals so much only improves your computational skills? Personally, I don't try to solve math "problems" that any computer can do.

 

[cristo]

Does one have a computer in an exam?

 

[Gib Z]

ok, I just split it up with partial fractions, then some simple straight forward Integration, I get it down to
$$\int \frac{2x}{x^2+2x+1} dx= 2\int \frac{1}{x+1} dx -2\int \frac{1}{(x+1)^2}= 2 (log_e (x+1) + \frac {1}{x+1})$$, which I differentiated to check and I got it right.

As to andytoh, I can't even begin to imagine why you would think a computer should do mathematics is place of a human...How can we develop more advanced techniques without knowing the basics? If we can't develop the techniques, who will programme the computers to know it?

Computers should only be used for tedious calculations one already knows how to perform, not for anything else. Not to mention, Computer assisted proofs are ugly.

 

[Gib Z]

Ok maybe try a tiny bit more advanced one this time?

 

[Gib Z]

Come on! Somebody! I am up for the challenge~!

 

[cristo]

Ok. This is a bit harder: $$\int\frac{1-\cos x}{\sin^2x}dx$$

 

[Gib Z]

I got to $$\int \frac{1-\cos x}{sin^2 x} dx = (\int \frac{1}{sin^2 x} dx) + \csc x$$ then i started to cry...

 

[Gib Z]

Hmph I kind of happened to notice that $$\int \frac{1}{sin^2 x} dx$$ was $$-\cot x$$, but that's probably cheating huh..

 

[cristo]

Hint: use the substitution u=x/2.

 

[d_leet]

Here's a slightly challenging one, it isn't too difficult, but not really simple either.

$$\int sec^3 x \ dx$$

 

[Gib Z]

Ahh ok with cristos hint $$\sin^2 x = \frac{4t^2}{t^4+2t^2+1}$$...I'll see what I can do from there...

I can't see d_leet's one...

Edit: Can't do cristos either now...

 

[d_leet]

For mine try splitting it into secant times secant squared and then using an identity. It still isn't really a direct computation from there though.

 

[Gib Z]

I Already tried that, I got $$\int (\tan^2 x +1)\sec x dx$$ then couldn't do it...

 

[cristo]

Ahh ok with cristos hint $$\sin^2 x = \frac{4t^2}{t^4+2t^2+1}$$...I'll see what I can do from there...

I can't see d_leet's one...

Edit: Can't do cristos either now...

OK, for mine, making the substitution x=2u, dx=2 du:$$\int\frac{1-\cos x}{\sin^2x}dx=2\int\frac{1-\cos2u}{\sin^22u}du$$ Now, can you simplify this using expressions for cos(2u) and sin(2u)?

 

[Gib Z]

$$2\int\frac{1-\cos2u}{\sin^22u}du=2\int \frac{1-(2\cos^2 u -1)}{4\sin^2 u \cos^2 u} du= 2\int \frac{2-2\cos^2 u}{4\sin^2 u \cos^2 u}= \int \frac{1}{\cos^2 u} du=\int \sec^2 u du= \tan u + C = \tan (\frac{x}{2}) + C$$...What did I do wrong...

 

[cristo]

What did I do wrong...

Nothing-- that's correct!

 

[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 independant, 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:

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

$$\int \frac {1}{2}\sqrt -x^2+2x+17 = \frac {1}{2} \int \sqrt {-x^2+2x+17}$$

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