Are You Ready to Challenge Your Integral Solving Skills?

[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!