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I think everything I have done is kosher, but unless I am missing an identity it is a different answer than the online quiz and wolfram alpha give.

I tried to use the trig substitution $$ x=2\sin(\theta)\Rightarrow dx=2\cos(\theta)$$

Which yields $$\int\frac{x^2}{\sqrt{4-x^2}}dx=\int\frac{4\sin^2(\theta)}{2\sqrt{1-\sin^2{\theta}}}2\cos(\theta)d\theta=4\int \sin^2(\theta)d\theta\\ =2\int (1-\cos(2\theta))d\theta=2\theta-\sin(2\theta)$$

By the half angle formula. Then since $x=2\sin(\theta)\Rightarrow \theta=\arcsin(x/2)$ this gives a final answer of

$$\int\frac{x^2}{\sqrt{4-x^2}}dx=2\arcsin(x/2)-\sin(2\arcsin(x/2))+c $$

Is this right? If not, where did I go wrong?

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  • $\begingroup$ @AhmedHussein oops missed that. What identity allows you to simplify like that? $\endgroup$ Jun 8, 2016 at 17:49
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    $\begingroup$ $\sin(2y) = 2\sin(y)\cos(y)$, then to simplify $\cos(\arcsin(x/2))$ draw a right triangle with appropriate lengths so that an angle is given by $\arcsin(x/2)$, calculate the cosine of that angle. $\endgroup$ Jun 8, 2016 at 17:57
  • $\begingroup$ @DepeHb Got it! Thank you $\endgroup$ Jun 8, 2016 at 18:02

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There are $3$ mistakes that you have made:

  1. Missed the $d\theta$ in the first line.

$$ x=2\sin(\theta)\Rightarrow dx=2\cos(\theta)$$

  1. Then you did the integration in the following step as:

$$=2\int (1-\cos(2\theta))d\theta=2\theta-2\sin(2\theta)$$

But actually, this is wrong. What you should have done is:

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

  1. And finally you missed the constant of integration "c"... :P

Hope this helps.

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  • $\begingroup$ Thank you! I will edit. Any tips on identities I can use to simplify the final expression? $\endgroup$ Jun 8, 2016 at 17:53
  • $\begingroup$ @qbert You might like to use the identity $2 \arcsin x= \arcsin (2x\sqrt{1-x^2})$ $\endgroup$ Jun 8, 2016 at 17:57
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for $\displaystyle\int\dfrac{x^2}{\sqrt{4-x^2}}dx$ ;

Integral by parts;

Let be $\quad du=\dfrac{-x}{\sqrt{4-x^2}}dx$

$u=\sqrt{4-x^2}$

And $\quad(-x)=v\longrightarrow -dx=dv$

$\displaystyle\int\dfrac{x^2}{\sqrt{4-x^2}}dx=-x.\sqrt{4-x^2}+\displaystyle\int\sqrt{4-x^2}dx$

And, for $\quad\displaystyle\int\sqrt{4-x^2}dx$;

$x=2\sin a$

$\arcsin \frac{x}{2}=a$

$dx=2.\cos a.da$

$\quad\displaystyle\int\sqrt{4-x^2}dx=4\displaystyle\int \cos^2a \;da=2\displaystyle\int \dfrac{\cos 2a+1}{1} \;da=\cos 2a+2a+C=\cos (2.(\arcsin \frac{x}{2}))+2.\arcsin \frac{x}{2}+C$

And all of Integral'll be;

$\boxed{\boxed{\displaystyle\int\dfrac{x^2}{\sqrt{4-x^2}}dx=-x.\sqrt{4-x^2}+\cos (2.(\arcsin \frac{x}{2}))+2.\arcsin \frac{x}{2}+C}}$

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$$ \begin{aligned} \int \frac{x^{2}}{\sqrt{4-x^{2}}} d x &=-\int x d \sqrt{4-x^{2}} \\ &=-x \sqrt{4-x^{2}}+\int \sqrt{4-x^{2}} d x \\ &=-x \sqrt{4-x^{2}}+\frac{x \sqrt{4-x^{2}}}{2}+2 \sin ^{-1}\left(\frac{x}{2}\right)+C\\&= -\frac{x \sqrt{4-x^{2}}}{2}+2 \sin ^{-1}\left(\frac{x}{2}\right)+C \end{aligned} $$

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