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# STEP Question Help watch

1. Anyone have any ideas on this one? (attached)

Thanks,

Vazzyb
Attached Images

2. I = int(0, pi/2) ln(sin(x)) dx

Let u=pi/2-x, then du=-dx. So:
I = int(pi/2, 0) ln(sin(pi/2 - u)) -du
= - int(pi/2, 0) ln(cos(u)) du
= int(0, pi/2) ln(cos(u)) du

This gives us the first equality. Now ln(sin(2u)) = ln(2sin(u)cos(u)) = ln(2) + ln(sin(u)) + ln(cos(u)). So:
I = int(0, pi/2) ln(sin(2u)) - ln(2) - ln(sin(u)) du
= int(0, pi/2) ln(sin(2u)) du - int(0, pi/2) ln(2) du - int(0, pi/2) ln(sin(u)) du
= int(0, pi/2) ln(sin(2u)) du - pi/2 ln(2) - I
=> 2I = int(0, pi/2) ln(sin(2u)) du - pi/2 ln(2)
=> I = (1/2) int(0, pi/2) ln(sin(2u)) du - pi/4 ln(2)

And that gives us the second equality. Now let's move on to the second part.

J = int(0, pi/2) ln(sin(2x)) dx

Let v=2x, then dv=2dx. So:
J = (1/2) int(0, pi) ln(sin(v)) dv

This gives us the first equality. Let's go a step further:
J = (1/2) int(0, pi) ln(sin(v)) dv
= (1/2) int(0, pi/2) ln(sin(v)) dv + (1/2) int(pi/2, pi) ln(sin(v)) dv

Let v=pi-u, then dv=-du. So:
int(pi/2, pi) ln(sin(v)) dv = int(pi/2, 0) ln(sin(pi-u)) -du
= - int(pi/2, 0) ln(sin(u)) du
= int(0, pi/2) ln(sin(u)) du

So let's sub this back into J:
J = (1/2) int(0, pi/2) ln(sin(v)) dv + (1/2) int(pi/2, pi) ln(sin(v)) dv
= (1/2) int(0, pi/2) ln(sin(v)) dv + (1/2) int(0, pi/2) ln(sin(u)) du
= int(0, pi/2) ln(sin(x)) dx
3. Thank you so much for that. Very helpful. Will rep tommorow.
4. No problem. Out of curiosity, which level of STEP is this?
5. Step 2.
6. Can I ask what year it was?
7. 1991 Step 2.

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