Help with integral with a radical

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knottb04

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I’ve been having trouble solving this integral. I’ve tried u-substitution, but that gives me an extra y in du that isn’t in the integral. I then tried a trig substitution by substituting sin u for 4y, but I couldn’t change the limits of integration because I couldn’t insert 0.5 (the upper limit) into my u-substitution since it’s not part of the range. Do you have any suggestions on how I should go about solving this?

Thanks in advance.
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[imath]4y = \sin v[/imath] will not get you from [imath]\sqrt{1-4y^2}[/imath] to [imath]\sqrt{1-\sin^2v}[/imath]
 
Did you verify that sqrt(cos2u) = cos u?
You also didn't finish. Please do so.
 
Steven G said:
Did you verify that sqrt(cos2u) = cos u?
You also didn't finish. Please do so.
Click to expand...
I believe that I’ve finished it. My result matches the answer I got when I plugged the whole integral into my calculator.
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sqrt(coś2x) is not necessarily cos x

Just like sqrt(x2) is not always x.
Just consider sqrt((-5)2). That doesn't equal -5. It equals 5. sqrt(x2)=|x|.
sqrt(coś2x) = |cosx|.
In your problem, you replaced sqrt(coś2x) with cosx. Why not -cosx? Please explain.

Also, try using some equal signs.
 
Since the area that I am integrating is positive (from 0 to pi/2), don’t I only need the positive square root of cosx, thereby negating the need for the absolute value symbols?

Also, thanks for the note about the equals signs.
 
 
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