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integrate the following

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\[\int\limits_{0}^{\Pi/2} \sin ^{2}x \div (\sin ^{2}x+4\cos ^{2}x)\]
change cos^2 = 1 - sin^2
so you have for the denominator sin^2 + 4(1-sin^2) = sin^2 + 4 - 4 sin^2 -3sin^2 + 4

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Other answers:

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?
you dont need partional fractions
How will you do it by converting denominator to sin?
the denominator is -3sin^2 t + 4 , do long division
we gottta use the properties of definite integrals
shank, change cos^2 = 1 - sin^2
And then what?
long division
You're effectively doing the same thing only.
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.
ok you did trig substitution
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.
i see that, im asking about your t = tan x part
what sort of substitution is this called?
Yeah thats a trignometric substituion.
ok your solution is better, im not sure if i would go ahead with mine . not sure what do do next
you can divide top and bottom by cos^2 for my solution
shank, can i ask a simple question about kites
hmmm ok.

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