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

Question

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

Ezra Adams · Accepted Answer

#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#

Thus, 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#

Fracturing the integral, we've got

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

which are equal, meaning that the integral is equal to

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

Alternatively, streamlining

#(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, to simplify using algebra,

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#

Christopher Allers · Answer

undefined To find the integral of $ \int \frac{x^2}{\sqrt{4 - x^2}} \, dx $, we can use the trigonometric substitution method. Let $ x = 2\sin(\theta) $, then $ dx = 2\cos(\theta) \, d\theta $. Substituting these into the integral, we get:

$$ \int \frac{(2\sin(\theta))^2}{\sqrt{4 - (2\sin(\theta))^2}} \cdot 2\cos(\theta) \, d\theta $$

Simplify this expression to:

$$ \int \frac{4\sin^2(\theta)}{\sqrt{4 - 4\sin^2(\theta)}} \cdot 2\cos(\theta) \, d\theta $$

$$ \int \frac{4\sin^2(\theta)}{\sqrt{4(1 - \sin^2(\theta))}} \cdot 2\cos(\theta) \, d\theta $$

$$ \int \frac{4\sin^2(\theta)}{\sqrt{4\cos^2(\theta)}} \cdot 2\cos(\theta) \, d\theta $$

$$ \int 2\sin^2(\theta) \cdot 2\cos(\theta) \, d\theta $$

$$ \int 4\sin^2(\theta)\cos(\theta) \, d\theta $$

Using the double angle identity for sine, $ \sin^2(\theta) = \frac{1}{2}(1 - \cos(2\theta)) $, we can rewrite the integral as:

$$ \int 4\left(\frac{1}{2}(1 - \cos(2\theta))\right)\cos(\theta) \, d\theta $$

$$ \int 2(1 - \cos(2\theta))\cos(\theta) \, d\theta $$

$$ \int (2\cos(\theta) - 2\cos(2\theta)\cos(\theta)) \, d\theta $$

$$ \int (2\cos(\theta) - \cos(2\theta)) \, d\theta $$

Integrate term by term:

$$ = 2\sin(\theta) - \frac{1}{2}\sin(2\theta) + C $$

Now, convert back to $ x $:

$$ = 2\sin^{-1}\left(\frac{x}{2}\right) - \frac{1}{2}\sin^{-1}\left(\frac{x^2}{2}\right) + C $$

So, the integral of $ \int \frac{x^2}{\sqrt{4 - x^2}} \, dx $ is $ 2\sin^{-1}\left(\frac{x}{2}\right) - \frac{1}{2}\sin^{-1}\left(\frac{x^2}{2}\right) + C $, where $ C $ is the constant of integration.