Beddy
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On the official solutions thread, the solution given for the 2nd part of question 6 is:

take y = pi - x.
=> dy = -dx. Substituting and noticing that sin x = sin(pi - x) gives that integral( xf(sin(x) ) = integral( (pi-x)f(sin x)).
We use this fact in evaluating I = integral(x/(1+sin x)). We get 2I = integral(pi/(1+sin x)). Hence I = pi/2.integral(1/(1+sin x)).

But if you make the substitution, you obtain integral(( pi -y) f(siny))dy. How can you just change the y's to x's???
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Zacken
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(Original post by Beddy)
On the official solutions thread, the solution given for the 2nd part of question 6 is:

take y = pi - x.
=> dy = -dx. Substituting and noticing that sin x = sin(pi - x) gives that integral( xf(sin(x) ) = integral( (pi-x)f(sin x)).
We use this fact in evaluating I = integral(x/(1+sin x)). We get 2I = integral(pi/(1+sin x)). Hence I = pi/2.integral(1/(1+sin x)).

But if you make the substitution, you obtain integral(( pi -y) f(siny))dy. How can you just change the y's to x's???
Because it's a dummy variable just like

\displaystyle \sum_{k=1}^n f(k) = \sum_{r=1}^n f(r) = \sum_{i=1}^n f(i) = \cdots

We have that

\displaystyle \int_{a}^{b} f(x) \, \mathrm{d}x = \int_{a}^{b} f(y) \, \mathrm{d}y = \int_{a}^{b} f(z) \, \mathrm{d}z = \cdots
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16Characters....
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(Original post by Beddy)
On the official solutions thread, the solution given for the 2nd part of question 6 is:

take y = pi - x.
=> dy = -dx. Substituting and noticing that sin x = sin(pi - x) gives that integral( xf(sin(x) ) = integral( (pi-x)f(sin x)).
We use this fact in evaluating I = integral(x/(1+sin x)). We get 2I = integral(pi/(1+sin x)). Hence I = pi/2.integral(1/(1+sin x)).

But if you make the substitution, you obtain integral(( pi -y) f(siny))dy. How can you just change the y's to x's???
Hi,

If you go into the Maths Exams subforum of the Maths subforum in the Maths, Science and Technology study help forum then you will see a STEP Prep thread 2016. You are likely to receive quicker answers if you post here.

I am not familiar with the question and typed mathematics like that is not the easiest to understand so please correct me if the below advice is not helpful.

Have they turned an integral of the form  \int g(y) dy into one of the form  \int g(x) dx ? If so they are completely the same integral. If you plotted graphs of z = g(y) and z = g(x) and found the areas between the curves and the y and x axes respectively you would get the same area.
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Zacken
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(Original post by Beddy)
On the official solutions thread, the solution given for the 2nd part of question 6 is:

take y = pi - x.
=> dy = -dx. Substituting and noticing that sin x = sin(pi - x) gives that integral( xf(sin(x) ) = integral( (pi-x)f(sin x)).
We use this fact in evaluating I = integral(x/(1+sin x)). We get 2I = integral(pi/(1+sin x)). Hence I = pi/2.integral(1/(1+sin x)).

But if you make the substitution, you obtain integral(( pi -y) f(siny))dy. How can you just change the y's to x's???
By the way, I'm not sure why you're posting this in the A-Levels forum, STEP isn't A-Levels... this would be much more relevant in the STEP Prep thread 2015/2016 or the maths forum.
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Zacken
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(Original post by 16Characters....)
x.
This happens all the time. (posting at the same time) :rofl:
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16Characters....
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(Original post by Zacken)
This happens all the time. (posting at the same time) :rofl:
Indeed it does :-) It's kind of coincidental that this thread was made today because earlier I couldn't be bothered with one of the differentiating trig functions examples my teacher was going through in class so I was playing around with changes of variable of the form  y = \pi - x on some trig integrals, though the only particularly interesting one was  \displaystyle \int_0^{\pi} x \sin x dx  = \pi.
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Zacken
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(Original post by 16Characters....)
Indeed it does :-) It's kind of coincidental that this thread was made today because earlier I couldn't be bothered with one of the differentiating trig functions examples my teacher was going through in class so I was playing around with changes of variable of the form  y = \pi - x on some trig integrals, though the only particularly interesting one was  \displaystyle \int_0^{\pi} x \sin x dx  = \pi.
I find it fascinating that that works on a whole class of integrals: \displaystyle \int_0^{\pi}xf(\sin x) \, \mathrm{d}x, the beauty of maths, eh?
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16Characters....
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(Original post by Zacken)
I find it fascinating that that works on a whole class of integrals: \displaystyle \int_0^{\pi}xf(\sin x) \, \mathrm{d}x, the beauty of maths, eh?
Aye
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