Are You Ready to Challenge Your Integral Solving Skills?

  • Thread starter Gib Z
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In summary, the individual tried to do an integrative problem involving tan x, but was not able to get the answer correct. They substituted for dx and got an incorrect answer.
  • #36
Gib Z said:
Thats really good, thanks for that. For the partial fractions one, I actually managed to get A and C before, but I could only reduce it to B+D=0, I couldn't acutally get the values themselves. You could explain how you got that one?
Even if I did get B and D, What steps did you do to get to the next result? I have no idea...sorry about that.

Ok when u multiply the partial fraction identity by their common denominator, u get [tex]u^2 = (A+C)u^3 + (B+D+(C-A)\sqrt{2})u^2 + (A+C+(D-B)\sqrt{2})u + B + D [/tex]. Compare coefficients of descending powers of u, u get A+C =0, B+D+(C-A)[tex]\sqrt{2}[/tex]=1, A+C+(D-B)[tex]\sqrt{2}[/tex]=0 and B+D=0. Sub. eqn 1 into eqn 3, u get (D-B)[tex]\sqrt{2}[/tex]=0. Thus D-B must be 0. B=D=0. (shown!)
For the next part, u use the values of A,B,C and D to express the integral as partial fractions. The resultant integral is [tex]\int \frac{-\frac{u}{2\sqrt{2}}}{u^2+u\sqrt{2}+1} + \frac{\frac{u}{2\sqrt{2}}}{u^2-u\sqrt{2}+1} du[/tex]. U can factor out the constant [tex]\frac{1}{2\sqrt{2}}[/tex]. After that, try to make the top a derivative of the bottom into the form f'(u)/f(u) where the result is in terms of ln function. For the denominator, u need to complete the square and u get integral of the form [tex]\int \frac{du}{u^2+a^2}[/tex] where the result is in terms of [tex]\tan^{-1}[/tex] function. Rationalise all the terms with surds at the bottom and use identity ln a - ln b= ln(a/b) to combine the 2 ln terms together. U'll eventually get the answer in thread #33.
 
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  • #37
Gib Z said:
As for ssd's let's z= sin x - cos x...i can't see how to manipulate your expression for that to help me..
Write sqrt(tanx) = sqrt(sinx)/sqrt(cosx) and similarly write cotx. Simplify sqrt(tanx)+sqrt(cotx)... check that the numerator = dz/dx=(sinx+cosx). Write the denominator as mentioned in reply #17.
 
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  • #38
Hmm ok I've managed to do it the u^2=tan x method, thanks guys. But i want to learn ssd's method, it looks simpler and shorter, and seeing as he posted the question I think he might know how to do it the best, no offense guys. ssd: When I make the numberator of the first part dz/dx, i get the 2nd parts numerator being z, I can't work out how to finish this off.
 
  • #39
Gib Z said:
Hmm ok I've managed to do it the u^2=tan x method, thanks guys. But i want to learn ssd's method, it looks simpler and shorter, and seeing as he posted the question I think he might know how to do it the best, no offense guys. ssd: When I make the numberator of the first part dz/dx, i get the 2nd parts numerator being z, I can't work out how to finish this off.

Do the two parts separately.

Put u= sinx+cosx in the second part. You get -du/dx=sinx-cosx= the numerator of the second part. Use 2sinx.cosx = (sinx+cosx)^2 -1.
You appear right in the sense that I feel its the simplest and shortest method when the answer is not known beforehand.
 
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  • #40
Hey, I know its been a while since I've posted on this thread, sorry about that guys. Ok well, Could someone give me another integral? Maybe one abit easier than that sqrt tanx
 
  • #41
No worries. Okay a simpler one for you; Evaluate the following integral

[tex]\int\frac{8\; dx}{\left(4x^2+1\right)^2}[/tex]
 
  • #42
I used some trig substitution and got [tex]\int (\frac{2}{\sec \theta})^2 d\theta[/tex]. Lost from there..
 
  • #43
Which substitution did you use? I would recommend the following one;

[tex]x:=\frac{1}{2}\tan\theta[/tex]
 
  • #44
[tex]\int \frac{8dx}{\sec^2 \theta}[/tex]...still lost
 
  • #45
O wait, I get [tex]\int 4 d\theta[/tex]...I don't Understand what to do now..the answers 4theta? What..
 
  • #46
Gib Z said:
O wait, I get [tex]\int 4 d\theta[/tex]
Correct!
Gib Z said:
...I don't Understand what to do now..the answers 4theta? What..
You now have to do the integration with respect to theta. What happens when you integrate a number with respect to something?
 
  • #47
Yes, and now express theta in terms of the initial variable, x. Add the integration constant.

Daniel.
 
  • #48
So is the solution [tex]4\arctan 2x[/tex]...Hey wait, i derived it, it works :), anuda one?

Edit: Forgot the +C
 
  • #49
Looks good to me.
 
  • #50
And now for something a little different;

[tex]I = \int x^2\sin x\;dx[/tex]
 
  • #51
Awesome :) Let's try another :)

Edit: didnt see that post
 
  • #52
Gib Z said:
I used some trig substitution and got [tex]\int (\frac{2}{\sec \theta})^2 d\theta[/tex]. Lost from there..
you were correct. it comes out to be [tex]\int \frac{4}{\sec^{2} \theta} d\theta[/tex] which is equal to [tex]\int 4\cos^{2} \theta d\theta[/tex]. now use the identity [tex]2\cos^{2}\theta = 1 + \cos(2\theta)[/tex]
 
  • #53
I did integration by parts a few times, I got [tex]I=\frac{\sin^2 x}{2} - x^2 \cos x[/tex] :D I think that's right.

Note: This ones for x^2 sin x dx.
 
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  • #54
murshid_islam said:
you were correct. it comes out to be [tex]\int \frac{4}{\sec^{2} \theta} d\theta[/tex] which is equal to [tex]\int 4\cos^{2} \theta d\theta[/tex]. now use the identity [tex]2\cos^{2}\theta = 1 + \cos(2\theta)[/tex]

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 [tex] 2\theta + \sin 2\theta[/tex] Then what do i do..
 
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  • #55
Hootenanny said:
And now for something a little different;

[tex]I = \int x^2\sin x\;dx[/tex]
does the answer come out to be [tex]-x^2\cos^{2}x + 2x\sin x + 2\cos x + C[/tex]?
 
  • #56
murshid_islam said:
does the answer come out to be [tex]-x^2\cos^{2}x + 2x\sin x + 2\cos x + C[/tex]?

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

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

Neither is correct, but murshid is closer than Gib.
 
  • #57
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
 
  • #58
Hootenanny said:
Neither is correct, but murshid is closer than Gib.
lol. the answer i came up with is
[tex]-x^2\cos x + 2x\sin x + 2\cos x + C[/tex]
i think this is correct.

Gib Z said:
I Get [tex]2\theta + \sin 2\theta[/tex] Then what do i do..
you used the substitution [tex]x = \frac{1}{2}\tan\theta[/tex]
from this, you get, [tex]\tan\theta = 2x[/tex] and from this, you get [tex]\theta = \arctan(2x)[/tex] and [tex]\sin\theta = \frac{2x}{\sqrt{4x^2+1}}[/tex] and [tex]\cos\theta = \frac{1}{\sqrt{4x^2+1}}[/tex].

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

so your answer will be
[tex]2\theta + \sin2\theta = 2\arctan(2x) + \frac{4x}{4x^2+1}[/tex]
 
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  • #59
murshid_islam said:
lol. the answer i came up with is
[tex]-x^2\cos x + 2x\sin x + 2\cos x + C[/tex]
i think this is correct.
This is indeed correct.
 
  • #60
Ok, I made a few mistakes the first time, I got [tex]I = x^2 + 2x \sin x + \cos x + C[/tex], 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 [tex]x^2 -\cos x[/tex]...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
 
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  • #61
murshid_islam said:
you used the substitution [tex]x = \frac{1}{2}\tan\theta[/tex]
from this, you get, [tex]\tan\theta = 2x[/tex] and from this, you get [tex]\theta = \arctan(2x)[/tex] and [tex]\sin\theta = \frac{2x}{\sqrt{4x^2+1}}[/tex] and [tex]\cos\theta = \frac{1}{\sqrt{4x^2+1}}[/tex]. and you also know that [tex]\sin 2\theta = 2\sin\theta\cos\theta[/tex]

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
 
  • #62
which substitution did you use? you can get to [tex]\int \frac{4}{\sec^{2}\theta}[/tex] by using [tex]x = \frac{1}{2}\tan\theta[/tex]

anyway, which substitution did you use?

ssd said:
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 [tex]\sin(x) = \frac{2\tan(\frac{x}{2})}{1+\tan^{2}(\frac{x}{2})}[/tex] and [tex]z = \tan(\frac{x}{2})[/tex] into [tex]\int \frac{dx}{2+sinx}[/tex] and got [tex]\int \frac{dz}{z^2+z+1}[/tex]

now what? shall i break the denominator into partial fractions? or is there an easier method?
 
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  • #63
murshid_islam said:
i substituted [tex]\sin(x) = \frac{2\tan(\frac{x}{2})}{1+\tan^{2}(\frac{x}{2})}[/tex] and [tex]z = \tan(\frac{x}{2})[/tex] into [tex]\int \frac{dx}{2+sinx}[/tex] and got [tex]\int \frac{dz}{z^2+z+1}[/tex]

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)}.
 
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  • #64
murshid, sorry about that :) turns out my trig substitution was tan theta =2x :) thanks
 
  • #65
Heys guy, can someone post another one? The last few that have been posted were good.
 
  • #66
try these:

[tex]\int_{0}^{\infty}\frac{\sin x}{x}dx[/tex]

[tex]\int_{0}^{\infty}\exp\left(-x^2\right)dx[/tex]
 
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  • #67
Ahh I've seen those before...I know the answers, just not how to get there...
 
  • #68
Gib Z said:
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. :wink:

some hints on the first one:

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

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

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

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

therefore from (1),
[tex]C = \frac{\pi}{2}[/tex]

[tex]f(s) = -\tan^{-1}s + \frac{\pi}{2}[/tex]
[tex]f(0) = \frac{\pi}{2}[/tex]
 
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  • #69
Aww come on, at least gimme a hint...
 
  • #70
Exp(x^2) = e^(x^2)?
If so Int (Exp(x^2) ) = Infinity.
 

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