# Also help with integral problem

1. Jun 16, 2008

### heyman123

I cant solve this exactly but who can help me with it.. I also need the primitive of sin x

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2. Jun 16, 2008

### arildno

If an exact value can be found for that definite integral, I suspect you should make a contour integral in the complex plane in order to crack it.

3. Jun 16, 2008

### heyman123

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. Jun 16, 2008

### snipez90

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).

Last edited: Jun 16, 2008
5. Jun 16, 2008

### dirk_mec1

What's wrong with one step IBP?

6. Jun 16, 2008

### snipez90

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.

Last edited: Jun 16, 2008
7. Jun 16, 2008

### Crosson

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

8. Jun 16, 2008

### snipez90

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?

Last edited: Jun 16, 2008
9. Jun 17, 2008

### JG89

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?

Last edited: Jun 17, 2008
10. Jun 17, 2008

### Defennder

11. Jun 17, 2008

### JG89

I have a question regarding to what I posted.

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. Jun 17, 2008

### snipez90

JG89, you won't be ignoring the other integral. You never have to evaluate an integral using the FTC in this problem except for once. At that point, you won't have to consider log(0). This is a hint that you're supposed to split up the integral and try to use various substitutions that will get you equal integrals.

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

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. Jun 17, 2008

### ice109

mathematica gives : $\pi \log(2)$

14. Jun 17, 2008

### Defennder

Yes, I agree it's a series of tricks. I just solved it yesterday and was amazed at the elegance of the solution. Dick's hint in the other thread proves to be very helpful.

15. Jun 18, 2008

### dirk_mec1

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

Use $$u=\frac{\pi}{2}-x$$ to find

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

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

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