# Evaluating the indefinite integral $\int\frac{x^2}{\sqrt{4-x^2}}dx$

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?

• @AhmedHussein oops missed that. What identity allows you to simplify like that? – qbert Jun 8 '16 at 17:49
• $\sin(2y) = 2\sin(y)\cos(y)$, 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. – Spine Feast Jun 8 '16 at 17:57
• @DepeHb Got it! Thank you – qbert Jun 8 '16 at 18:02

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.

• Thank you! I will edit. Any tips on identities I can use to simplify the final expression? – qbert Jun 8 '16 at 17:53
• @qbert You might like to use the identity $2 \arcsin x= \arcsin (2x\sqrt{1-x^2})$ – SchrodingersCat Jun 8 '16 at 17:57

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;

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