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- Thread starter heyman123
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arildno

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i ll manage to fix it i think but only I need to get the primitive of sin x.

And thats the thing that i cant find, and not know how.

- #4

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Hint: try to decompose the integral into two integrals. one should have limits of integration from 0 to pi/2

exploit symmetry and manipulate the limits of integration to solve the problem. these identities might also help: sin(x) = sin(pi - x), 2sin(x)cos(x) = sin(2x).

exploit symmetry and manipulate the limits of integration to solve the problem. these identities might also help: sin(x) = sin(pi - x), 2sin(x)cos(x) = sin(2x).

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- #5

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Hint: try to decompose the integral into two integrals. one should have limits of integration from 0 to pi/2

exploit symmetry and manipulate the limits of integration to solve the problem. these identities might also help: sin(x) = sin(pi - x), 2sin(x)cos(x) = sin(2x).

What's wrong with one step IBP?

- #6

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How would you attack this by integrating by parts? The integrand is not log(x)sin(x). Then again I've never computed any IBP recursion formulas but I consider IBP rather blunt and inelegant. But with the composition of functions, substitution should be a priority technique. This is pretty much the same problem as an old putnam question (1953 I think) and it certainly doesn't require anything beyond some trig and basic calculus.

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Use the substitution x = ArcSin(u) and then use integration by parts.

- #8

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Interesting idea, but then wouldn't your new limits of integration both be 0?

Also, heyman, does log signify the logarithm base ten or natural logarithm?

Also, heyman, does log signify the logarithm base ten or natural logarithm?

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Am I missing something or is this integral impossible to evaluate because then sin0 = 0 and then you'd be taking the log of 0, which is impossible?

And even if you split the integrals into two separate ones there will still be a 0 in one of the limits of integration?

And even if you split the integrals into two separate ones there will still be a 0 in one of the limits of integration?

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- #10

Defennder

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Apparently this has been double-posted here:

https://www.physicsforums.com/showthread.php?t=240629

https://www.physicsforums.com/showthread.php?t=240629

- #11

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If you split up the integral into two other integrals, only one of the integrals will have 0 in the limit of integration. Does this mean that the integral without the 0 in its limits of integration can be evaluated as the final answer? Just ignoring the other integral?

- #12

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Here is a useful identity/substitution: [tex]\int_0^a f(x)\, dx = \int_0^a f(a-x)\, dx[/tex]

You can prove it by substituting for a-x on the RHS. As Dick mentioned in the other thread, this integral is mainly a series of tricks.

Defennder, I think it's the natural log but that really shouldn't affect the problem except for a change of functions in the final answer.

- #13

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mathematica gives : [itex]\pi \log(2)[/itex]

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Defennder

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- #15

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[tex]

\Psi=\int_0^{\pi} \ln{\sin{x}} \ dx= 2 \int_0^{\frac{\pi}{2}} \ln{\sin{x}} \ dx

[/tex]

Use [tex]

u=\frac{\pi}{2}-x

[/tex] to find

[tex]

\Psi=2 \int_0^{\frac{\pi}{2}} \ln{\cos{x}} \ dx

[/tex]

[tex]

\Psi=\int_0^{\frac{\pi}{2}} \ln{\frac{1}{2} \sin{2x}} \ dx=\int_0^{\frac{\pi}{2}} \ln{ \sin{2x}}-\int_0^{\frac{\pi}{2}} \ln{2} \ dx=\frac{1}{2} \Psi-\int_0^{\frac{\pi}{2}} \ln{2} \ dx

[/tex]

[tex]

\Psi=-2 \int_0^{\frac{\pi}{2}} \ln{2} \ dx

[/tex]

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