How Good Am I at Integrals?

[LRJ85]

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 $$u^2 = (A+C)u^3 + (B+D+(C-A)\sqrt{2})u^2 + (A+C+(D-B)\sqrt{2})u + B + D$$. Compare coefficients of descending powers of u, u get A+C =0, B+D+(C-A)$$\sqrt{2}$$=1, A+C+(D-B)$$\sqrt{2}$$=0 and B+D=0. Sub. eqn 1 into eqn 3, u get (D-B)$$\sqrt{2}$$=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 $$\int \frac{-\frac{u}{2\sqrt{2}}}{u^2+u\sqrt{2}+1} + \frac{\frac{u}{2\sqrt{2}}}{u^2-u\sqrt{2}+1} du$$. U can factor out the constant $$\frac{1}{2\sqrt{2}}$$. 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 $$\int \frac{du}{u^2+a^2}$$ where the result is in terms of $$\tan^{-1}$$ 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.

 

[ssd]

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.

 

[Gib Z]

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.

 

[ssd]

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.

 

[Gib Z]

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

 

[Hootenanny]

No worries. Okay a simpler one for you; Evaluate the following integral

$$\int\frac{8\; dx}{\left(4x^2+1\right)^2}$$

 

[Gib Z]

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

 

[Hootenanny]

Which substitution did you use? I would recommend the following one;

$$x:=\frac{1}{2}\tan\theta$$

 

[Gib Z]

$$\int \frac{8dx}{\sec^2 \theta}$$...still lost

 

[Gib Z]

O wait, I get $$\int 4 d\theta$$...I don't Understand what to do now..the answers 4theta? What..

 

[Hootenanny]

O wait, I get $$\int 4 d\theta$$

Correct!

...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?

 

[dextercioby]

Yes, and now express theta in terms of the initial variable, x. Add the integration constant.

Daniel.

 

[Gib Z]

So is the solution $$4\arctan 2x$$...Hey wait, i derived it, it works :), anuda one?

Edit: Forgot the +C

 

[Hootenanny]

Looks good to me.

 

[Hootenanny]

And now for something a little different;

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

 

[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.