How do you find the integral of $int ( x^2 / sqrt(4 - x^2) ) dx$ ?

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Answer 1 · 2015-11-21T17:52:38

$I = (2x)/(4 - x^2) + 1/4*ln|(2 + x)/sqrt(4 - x^2)| + c$

Let's say that $x = 2sin(t)$ , then $dx = 2cos(t) dt$

So we have

$int(sin^2(t)/cos^2(t)*dt/(2cos(t))) = 1/2int(tan^2(t)*sec(t)dt)$

However we know that $tan^2(x) = sec^2(x) - 1$ , so we can say it's

$int(sec^3(t) - sec(t))dt$

Breaking the integral, we have

$intsec^3(t)dt - intsec(t)dt$

Which are both standard integrals, so we have the integral equal

$(sec(t)tan(t))/2 + ln|sec(t) + tan(t)|/2 - ln|sec(t) + tan(t)| + c$

Or, simplifying

$(sec(t)tan(t) -ln|sec(t)+tan(t)|)/2 + c$

$int(sin^2(t)/cos^2(t)*dt/(2cos(t))) = 1/2int(tan^2(t)*sec(t)dt) = (sec(t)tan(t) -ln|sec(t)+tan(t)|)/4 + c$

But the original integral was in terms of $x$ , and since

$x = 2sin(t)$ , we have that $t = arcsin(x/2)$ so substituting that we have

$int ( x^2 / sqrt(4 - x^2) ) dx = (sec(t)tan(t) -ln|sec(t)+tan(t)|)/4 + c$

$I = int ( x^2 / sqrt(4 - x^2) ) dx = (sec(arcsin(x/2))tan(arcsin(x/2)) -ln|sec(arcsin(x/2))+tan(arcsin(x/2))|)/4 + c$

Or, using some algebra to simplify,

We know that $sin(arcsin(x/2)) = x/2$ and that $cos(arcsin(x/2)) = sqrt(1 - x^2/4)$ , so we have

$sec(arcsin(x/2)) = 1/cos(arcsin(x/2)) = 1/sqrt(1 - x^2/4) = 2/sqrt(4 - x^2)$

$tan(arcsin(x/2)) = x/(2sqrt(1 - x^2/4)) = x/sqrt(4 - x^2)$

$I = (2x)/(4 - x^2) + 1/4*ln|(2 + x)/sqrt(4 - x^2)| + c$

How do you find the integral of int ( x^2 / sqrt(4 - x^2) ) dxint ( x^2 / sqrt(4 - x^2) ) dx?