Useful integral #1
∫ 0 2 π ln ( sin ( x ) ) d x
If the value of the integral above equals to − A π ln ( B ) for positive integers A and B , find the value of A B .
Try more here! .
The answer is 4.
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3 solutions
@Raj Rajput You wrote all the limits and properties at side. Thats amazing!!Well i did the same too!
Thanks a lot! The solution was very lengthy, that's why I didn't upload. U saved my time. Thanks a lot. Do try my other problems.
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The question had already been posted .
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Can u provide me a link??
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@Parth Lohomi – Actually it is a bit different. I can't waste the efforts of Syed and Raj. So just leave it. I know that both lead to the same answer in same methods.
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@Aditya Kumar – in 0 to π /2 ∫ ln ( sin ( x ) ) = ∫ ln ( cos ( x ) ) so there exactly same , but let it be , atleast one can see the difference of sin and cos!
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@Parth Lohomi – Yes that's what I meant to say. I entered the answer in that problem in 2s
Thanks and yes i will try :) :)
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I wouldn't if I were you. Just reading his solution made me pass out.
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@Pi Han Goh – Why what happened?? It uses basics in a complicated manner.
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@Aditya Kumar – Really? I would like to see you evaluate H(1/12) manually.
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@Pi Han Goh – is there something that i have made wrong @Pi Han Goh , i thought solution is perfect :P
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@Raj Rajput – pi han was talking about me. One of my solutions wasn't as expected. There's no problem with your solution.
There is a mistake in your solution,
You have wrote that period of
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is 2
π
∫π20ln(sin(x))dx Let's assume the following (eq.1*): I=∫π20ln(sin(x))dx
And we will need this property: ∫a0f(x)dx=∫a0f(a−x)dx
Therefor we can now calculate I: ∫π20ln(sin(π2−x))dx
There is literally a cofunction identity that says: sin(π2−x)=cos(x)
So we can arrive at: ∫π20ln(cos(x))dx
Now by combining eq.1 and the line above, we embark on a long journey: 2I=∫π20( ln(sin(x))+ln(cos(x)) )dx
=∫π20( ln(sin(x)cos(x)) )dx
We will use the following identity: sin(2x)=2sin(x)cos(x)
And get: ∫π20ln(sin(2x)2)dx
Then the log property: ln(AB)=ln(A)−ln(B)
So we get: ∫π20( ln(sin(2x))−ln(2) )dx
Now we split up the equation using simple multiplication: ∫π20ln(sin(2x))dx−∫π20ln(2)dx
We can integrate the right side (ln(2)) first because that is the easiest: ∫π20ln(sin(2x))dx−ln(2)(x)π20
∫π20ln(sin(2x))dx−ln(2)(π2)
Now we are going to use U substitution and should also recognize the symmetry of the sin function about pi/2 (sin(pi/2)=1; think a bell curve): (12)∫π0ln(sin(u))duu=2x−ln(2)(π2)
Note the integral changed, and we can eliminate the .5 that was just added by noticing the following: ∫π0=2∫π20
Thus we can change our integral to: (12)(2)∫π20ln(sin(u))duu=2x−ln(2)(π2)
Since we know from earlier: I=∫π20ln(sin(x))dx
We can say: 2I=I−(π2)ln(2)
Which is reduced to: I=−(π2)ln(2)
Ahah! we have finally arrived at our answer:
∫π20ln(sin(x))dx=−π2ln(2)
∫π20ln(sin(x))dx Let's assume the following (eq.1*): I=∫π20ln(sin(x))dx
And we will need this property: ∫a0f(x)dx=∫a0f(a−x)dx
Therefor we can now calculate I: ∫π20ln(sin(π2−x))dx
There is literally a cofunction identity that says: sin(π2−x)=cos(x)
So we can arrive at: ∫π20ln(cos(x))dx
Now by combining eq.1 and the line above, we embark on a long journey: 2I=∫π20( ln(sin(x))+ln(cos(x)) )dx
=∫π20( ln(sin(x)cos(x)) )dx
We will use the following identity: sin(2x)=2sin(x)cos(x)
And get: ∫π20ln(sin(2x)2)dx
Then the log property: ln(AB)=ln(A)−ln(B)
So we get: ∫π20( ln(sin(2x))−ln(2) )dx
Now we split up the equation using simple multiplication: ∫π20ln(sin(2x))dx−∫π20ln(2)dx
We can integrate the right side (ln(2)) first because that is the easiest: ∫π20ln(sin(2x))dx−ln(2)(x)π20
∫π20ln(sin(2x))dx−ln(2)(π2)
Now we are going to use U substitution and should also recognize the symmetry of the sin function about pi/2 (sin(pi/2)=1; think a bell curve): (12)∫π0ln(sin(u))duu=2x−ln(2)(π2)
Note the integral changed, and we can eliminate the .5 that was just added by noticing the following: ∫π0=2∫π20
Thus we can change our integral to: (12)(2)∫π20ln(sin(u))duu=2x−ln(2)(π2)
Since we know from earlier: I=∫π20ln(sin(x))dx
We can say: 2I=I−(π2)ln(2)
Which is reduced to: I=−(π2)ln(2)
Ahah! we have finally arrived at our answer:
∫π20ln(sin(x))dx=−π2ln(2)
