This MIT Integration Bee Integral Seems Tricky

Ajouté : 2026-06-01

[00:00:00]Okay, this integral looks a little scary especially for the denominator. It looks to be going to zero at both end points and log of x is going to negative infinity near x is zero. But then again, if you rewrite this denominator as x parentheses 1 - x inside of the square root, then everything seems to be easier to work with. So for the denominator, especially inside of the square root, x - x², [applause] this is the same as x parentheses 1 - x.

[00:00:28]And let me just call this integral as the I and rewrite this right. [applause] So integral I is the same as integral from 0 to 1 of flow of x that over denominator is the same as square root of x parenthesis 1 - x then we have dx and every time you have square root of x * 1 - x we have famous substitution x is sin square theta right. So let me just substitute x as sin^ square theta.

[00:01:03]This will work because 1 - x will turn into cosine square theta and square root simplify. So using this substitution, right?

[00:01:13]We can just talk about derivative, right? So dx is then going to be 2 * [applause] sin theta cossine theta d theta and at the same time 1 - x is the same as 1 - sin square theta which is cossine square theta.

[00:01:40]So that is why now square root of x parenthesis 1 - x this is the same as square root of sin square theta * cossine^ square theta.

[00:01:54]Then think about the interval right because your lower bound is now zero and upper bound is now one. But then again when your theta is between zero and pi / 2 both s and cosine are non- negative. So that is why the square root of sin^ square theta * cosine^ square theta. This is just the same as sin theta * cossine [applause] theta.

[00:02:26]Okay. So using this let's just talk about the lower bound of the upper bound because it will be changing right. So for the lower bound when x is okay zero then your theta is also zero upper bound of the x is now one. So when x is equal to 1 theta is going to [applause] pi / 2.

[00:02:51]Okay, then let's just rewrite this integral I using the theta, right?

[00:02:56]>> [applause] >> So using this integral I is the same as 10 integral from 0 to pi / 2 [applause] then that of log of sin [applause] square theta that divided by sin theta [applause] time cossine theta then dx was 2 * sin theta cosine theta d theta Right? So that times 2 * sin theta * cossine theta and we have d theta. So that a lot of terms got cancelceled out.

[00:03:36]So we can cancel sin theta out cossine theta out and pulling this two outside of this integral. This is same as 2 * integral from 0 to pi / 2 of just the log of sine square theta. And then we have d theta and in our bound s is greater than zero.

[00:04:07]So that's why we can just rewrite this as uh 2 * integral from 0 to pi / 2 and at that times we can pull this two out 2 * log of sin [applause] theta d theta and we can also pull this two outside. So eventually we have four * integral from 0 to pi / 2 of log of sin theta [applause] d theta And we can now evaluate this integral.

[00:04:44]And let me call this integral as say capital letter a. Right?

[00:04:50]Okay. So once again your capital letter a [applause] is integral from 0 to pi / 2 of log of sin [applause] theta d theta.

[00:05:03]And once again in our bounds we can use symmetry. So we can simply replace theta to p<unk> / 2 minus theta.

[00:05:13]And it seems like the lower bound at the upper bound will change because when theta is going to zero then the lower bound is now pi / 2 and when your theta is going to pi / 2 then the upper bound seems to be zero. So seems like the lower bound at the upper bound is now changing from pi / 2 to zero. But then again if you get derivative of this whole expression with respect to theta it becomes negative d theta. So we can just pull the negative sign outside switching lower bound and the upper bound then this a is just the same as integral from still zero to pi / 2 and at the same time if you consider the s of pi / 2 minus theta right this is the same as cosine of theta. So that is why the same expression a is the same as integral from 0 to pi / 2 of log of cossine theta and then we have d theta.

[00:06:10]So both of these integrals are the same as just the a.

[00:06:14]So that is why we can just add them up to have 2 * a. Right? So if you add them up we will have 2 a.

[00:06:23]2 a is going to be then the same as since we have the same lower bound at the upper bound from 0 to pi / 2 we should have log okay and we can just add them up let me just make a bracket log of sin theta plus [applause] log of cosine theta and we have d theta using log property we can just Add these two loes up by multiplying sin theta and cosine theta. Right? So it is the same as then integral from 0 to pi / 2 that of log sin theta time cossine theta and then we have d theta.

[00:07:13]Okay. It's a time to use double angle identity, right? Which is about sin theta * cossine theta. Using double angle identity, this is the same as 1 / 2 * sin of 2 theta.

[00:07:38]Okay. So using this we can rewrite this as now then. So 2 a is still and it's going to be the same as integral from 0 to pi / 2. Okay. Then that of now log of 1 [applause] / 2 * s of 2 theta and we have d theta. It seems like we can split this integral into two integrals, right? So we can rewrite this integral as integral from 0 to pi / 2 of log of sin [applause] 2 theta and d theta and log of 1 / 2 this is going to be now negative log of two. So that's why minus integral from 0 to pi / 2 that of log of 2 and then we have d theta second integral is now easy to evaluate right. So second integral.

[00:08:38]Okay, this seems to be pi / 2 * log of two. [applause] And the first integral using symmetry is just the same as a. So we can easily check by substituting say like u as 2 * theta. Then using the same symmetry we can easily check that this integral is just the same as integral a. So that is why what we have is going to be now 2 a is the same as a and then minus pi [applause] / 2 * log of 2.

[00:09:17]So a is then going to be the same as<unk> / 2 * low of two.

[00:09:28]But then again integral I that we looking for I was the same as 4 * integral A right so knowing how I was 4 * A. So the answer to this question is 4 * since A was<unk> / 2 * log of 2. So I has written the same as 4 *<unk> / 2 * log of 2. So the answer for this question is -2<unk>i * log of 2.

[00:09:59]This is the answer for the question.

[00:10:03]Okay, it's a pretty interesting integral from MIT integration B. How amazing.