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PostPosted: Thu, 17 May 2012 23:49:55 UTC 
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I met a problem with calculating following integrals:
[0,pi] (x sinx)/(1+(cos x)^2)
[0,pi/2] [sqrt(sinx)]/[sqrt(sinx) + sqrt(cosx)]


if you could give me a tip with whitch I would be able to proceed with I'd be grateful.


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PostPosted: Fri, 18 May 2012 00:26:21 UTC 
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poniek wrote:
I met a problem with calculating following integrals:
[0,pi] (x sinx)/(1+(cos x)^2)
[0,pi/2] [sqrt(sinx)]/[sqrt(sinx) + sqrt(cosx)]


if you could give me a tip with whitch I would be able to proceed with I'd be grateful.


For the first one do a u-substitution. For the second I'd say the same, then multiply by the denominator's conjugate.

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PostPosted: Fri, 18 May 2012 06:25:35 UTC 
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thanks for the reply,
But have you done it? I've already tried with the substitution method but as there is also x witch must be changed into arc function it seems to doesn't work...


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PostPosted: Fri, 18 May 2012 06:43:06 UTC 
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Shadow wrote:
poniek wrote:
I met a problem with calculating following integrals:
[0,pi] (x sinx)/(1+(cos x)^2)
[0,pi/2] [sqrt(sinx)]/[sqrt(sinx) + sqrt(cosx)]


if you could give me a tip with whitch I would be able to proceed with I'd be grateful.


For the first one do a u-substitution. For the second I'd say the same, then multiply by the denominator's conjugate.


No, there is no need to rationalise the denominator in the second one if one notes the symmetry. For the first one, just use integration by parts and symmetry.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Fri, 18 May 2012 07:07:22 UTC 
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outermeasure wrote:
Shadow wrote:
poniek wrote:
I met a problem with calculating following integrals:
[0,pi] (x sinx)/(1+(cos x)^2)
[0,pi/2] [sqrt(sinx)]/[sqrt(sinx) + sqrt(cosx)]


if you could give me a tip with whitch I would be able to proceed with I'd be grateful.


For the first one do a u-substitution. For the second I'd say the same, then multiply by the denominator's conjugate.


No, there is no need to rationalise the denominator in the second one if one notes the symmetry. For the first one, just use integration by parts and symmetry.


I suppose, I was just thinking for antiderivative's sake.

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PostPosted: Fri, 18 May 2012 07:11:12 UTC 
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Shadow wrote:
I suppose, I was just thinking for antiderivative's sake.


They are ugly. The first one is a polylog-type expression, and the second is an elliptic integral.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Fri, 18 May 2012 08:51:51 UTC 
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outermeasure wrote:
Shadow wrote:
I suppose, I was just thinking for antiderivative's sake.


They are ugly. The first one is a polylog-type expression, and the second is an elliptic integral.


Ooh, blech.

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PostPosted: Sun, 20 May 2012 11:01:18 UTC 
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SYMMETRY!!!
thanks a million!


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PostPosted: Sun, 20 May 2012 15:10:26 UTC 
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For the firs case it works perfectly, but for the second one what substitution have you used? u=tan(x/2) ? - it complicates the integral and doesn't give a clear conclusion. It's also not so easy for u=sinx or u=sqrt(sinx) ....


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PostPosted: Sun, 20 May 2012 15:58:18 UTC 
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poniek wrote:
For the firs case it works perfectly, but for the second one what substitution have you used? u=tan(x/2) ? - it complicates the integral and doesn't give a clear conclusion. It's also not so easy for u=sinx or u=sqrt(sinx) ....


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

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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