Are You Ready to Challenge Your Integral Solving Skills?

[Gib Z]

Awesome :) Let's try another :)

Edit: didnt see that post

 

[murshid_islam]

I used some trig substitution and got \int (\frac{2}{\sec \theta})^2 d\theta. Lost from there..

you were correct. it comes out to be \int \frac{4}{\sec^{2} \theta} d\theta which is equal to \int 4\cos^{2} \theta d\theta. now use the identity 2\cos^{2}\theta = 1 + \cos(2\theta)

 

[Gib Z]

I did integration by parts a few times, I got I=\frac{\sin^2 x}{2} - x^2 \cos x :D I think that's right.

Note: This ones for x^2 sin x dx.

 

[Gib Z]

you were correct. it comes out to be \int \frac{4}{\sec^{2} \theta} d\theta which is equal to \int 4\cos^{2} \theta d\theta. now use the identity 2\cos^{2}\theta = 1 + \cos(2\theta)

I've already got that one, but knowing how to do it that way would be good. I understand that identity, but can't see how that will help me..

Edit: Wait gimme a min, i think i see it..

Edit 2: I Get 2\theta + \sin 2\theta Then what do i do..

 

[murshid_islam]

And now for something a little different;

I = \int x^2\sin x\;dx

does the answer come out to be -x^2\cos^{2}x + 2x\sin x + 2\cos x + C?

 

[Hootenanny]

does the answer come out to be -x^2\cos^{2}x + 2x\sin x + 2\cos x + C?

I did integration by parts a few times, I got I=\frac{\sin^2 x}{2} - x^2 \cos x :D I think that's right.

Note: This ones for x^2 sin x dx.

Neither is correct, but murshid is closer than Gib.

 

[Gib Z]

Umm Ok Ok, Problem 1. I can't see what to do with murshid_islams method, and When I found the derivative of my solution to I, for x^2 sin x dx, I get (2x + cos x)sin x, that doesn't work..

Edit: Ok i was wrong lol, Ill try it again

 

[murshid_islam]

Neither is correct, but murshid is closer than Gib.

lol. the answer i came up with is
-x^2\cos x + 2x\sin x + 2\cos x + C
i think this is correct.

I Get 2\theta + \sin 2\theta Then what do i do..

you used the substitution x = \frac{1}{2}\tan\theta
from this, you get, \tan\theta = 2x and from this, you get \theta = \arctan(2x) and \sin\theta = \frac{2x}{\sqrt{4x^2+1}} and \cos\theta = \frac{1}{\sqrt{4x^2+1}}.

and you also know that \sin 2\theta = 2\sin\theta\cos\theta = \frac{4x}{4x^2+1}

so your answer will be
2\theta + \sin2\theta = 2\arctan(2x) + \frac{4x}{4x^2+1}

 

[Hootenanny]

lol. the answer i came up with is
-x^2\cos x + 2x\sin x + 2\cos x + C
i think this is correct.

This is indeed correct.

 

[Gib Z]

Ok, I made a few mistakes the first time, I got I = x^2 + 2x \sin x + \cos x + C, but when I derive it it doesn't work again :'(

EDIT: AWW KILL ME! Sorry Guys, I got the same answer and Murshid, Its just that during my working I wrote down x^2 -\cos x...It was ment to me the multiplication of these 2, or -x^2 (cos x), but I took it as x^2 minus cos x...:'( sorry guys

 

[Gib Z]

you used the substitution x = \frac{1}{2}\tan\theta
from this, you get, \tan\theta = 2x and from this, you get \theta = \arctan(2x) and \sin\theta = \frac{2x}{\sqrt{4x^2+1}} and \cos\theta = \frac{1}{\sqrt{4x^2+1}}. and you also know that \sin 2\theta = 2\sin\theta\cos\theta

Sorry, I'm afraid you may be a bit confused. I used the substitution x=1/2 tan theta on Hootenanys approach, When I get to where you were showing me, I was doing my own trigonometric substitution that was not x=1/2 tan theta. SOrry

 

[murshid_islam]

which substitution did you use? you can get to \int \frac{4}{\sec^{2}\theta} by using x = \frac{1}{2}\tan\theta

anyway, which substitution did you use?

I feel this is simpler than sqrt(tanx). Putting tan(x/2) = z and writing

sin(x) = 2tan(x/2)/[1+{tan(x/2)}^2]

the result easily follows.

i substituted \sin(x) = \frac{2\tan(\frac{x}{2})}{1+\tan^{2}(\frac{x}{2})} and z = \tan(\frac{x}{2}) into \int \frac{dx}{2+sinx} and got \int \frac{dz}{z^2+z+1}

now what? shall i break the denominator into partial fractions? or is there an easier method?

 

[ssd]

i substituted \sin(x) = \frac{2\tan(\frac{x}{2})}{1+\tan^{2}(\frac{x}{2})} and z = \tan(\frac{x}{2}) into \int \frac{dx}{2+sinx} and got \int \frac{dz}{z^2+z+1}

now what? shall i break the denominator into partial fractions? or is there an easier method?

Write z^2+z+1= (z+1/2)^2 +3/4, put z+1/2 =u. Then the integral is of the form
du/(u^2+a^2), a=sqrt(3)/2. The integral results in (1/a)tan_1(u/a) +c, where by tan_1(u) I mean {tan inverse(u)}.

 

[Gib Z]

murshid, sorry about that :) turns out my trig substitution was tan theta =2x :) thanks

 

[Gib Z]

Heys guy, can someone post another one? The last few that have been posted were good.

 

[murshid_islam]

try these:

\int_{0}^{\infty}\frac{\sin x}{x}dx

\int_{0}^{\infty}\exp\left(-x^2\right)dx

 

[Gib Z]

Ahh I've seen those before...I know the answers, just not how to get there...

 

[murshid_islam]

Ahh I've seen those before...I know the answers, just not how to get there...

well then, you know the answers. try to get to the answers.

some hints on the first one:

let f(s) = \int_{0}^{\infty}\frac{\sin x}{x}\exp(-sx)dx
now, after finding f(s), you just need to find f(0) to get the inegral you want. to find f(s):

f'(s) = \frac{d}{ds}\int_{0}^{\infty}\frac{\sin x}{x}\exp(-sx)dx = \frac{-1}{1+s^2}

now,
f'(s) = \frac{-1}{1+s^2}
f(s) = -\tan^{-1}s + C\ldots\ldots\ldots(1)

since f(s) = \int_{0}^{\infty}\frac{\sin x}{x}\exp(-sx)dx, we have f(\infty) = 0

therefore from (1),
C = \frac{\pi}{2}

f(s) = -\tan^{-1}s + \frac{\pi}{2}
f(0) = \frac{\pi}{2}

 

[Gib Z]

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

 

[acm]

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

 

[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 =<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.