fortheloveofscience
integrate the following
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fortheloveofscience
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\[\int\limits_{0}^{\Pi/2} \sin ^{2}x \div (\sin ^{2}x+4\cos ^{2}x)\]
perl
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change cos^2 = 1 - sin^2
perl
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so you have for the denominator
sin^2 + 4(1-sin^2) =
sin^2 + 4 - 4 sin^2
-3sin^2 + 4
shankvee\
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I think this is aprtial fractions.
If you take cos^2 x common from denominator you have
tan^2 x/(tan^2 x+4) Put tan x=t.
you'll get dx=dt/sec^2 x= dt/(1+t^2).
So you have t^2/((t^2+4)(t^2+1)). Do you know partial fractions?
perl
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you dont need partional fractions
shankvee\
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How will you do it by converting denominator to sin?
perl
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the denominator is -3sin^2 t + 4 , do long division
fortheloveofscience
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we gottta use the properties of definite integrals
perl
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shank, change cos^2 = 1 - sin^2
shankvee\
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And then what?
perl
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long division
shankvee\
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?
perl
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|dw:1328867815825:dw|
shankvee\
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You're effectively doing the same thing only.
perl
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|dw:1328867933423:dw|
shankvee\
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By doing by partial fractions it reduces to the same thing.
Only it is in terms of t^2 so easier to integrate. You directly get tan inverse form which is easy.
perl
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ok you did trig substitution
shankvee\
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For the second term you divide by cos ^2 and then you get tan form and then you put that as t to integrate, You get that directly by partial fractions in terms of t.
perl
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i see that, im asking about your t = tan x part
perl
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what sort of substitution is this called?
shankvee\
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Yeah thats a trignometric substituion.
perl
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ok your solution is better, im not sure if i would go ahead with mine . not sure what do do next
perl
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you can divide top and bottom by cos^2 for my solution
perl
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shank, can i ask a simple question about kites
shankvee\
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hmmm ok.