Question

Please how do you solve the Integral of x^2/sqrt(4-x^2) using trig substitution, below I will show how far I've gotten b4 I get stuck

Answer 1

i get to the answer of 2(theta)-sin(2theta)+C

then, I know that if x=sin(theta), theta should equal x/2. But, I have a note in my notes that says something about inverse sin (x/2)....
help!!

Answer 2

\[\int\limits(x^2)/\sqrt{4-x^2} \]

Answer 3

I didn't take many notes so far on this problem, but did you obtain this as well?

\[\Large 4\int \sin^2\alpha d\alpha \]

Answer 4

yes

Answer 5

I got to the answer in terms of theta
2&-sin2% +C

Answer 6

Good, then you used the double angular formula I assume.

Answer 7

yes

Answer 8

I don't know how to back sub for theta

Answer 9

Ahh, okay. I see where your problem is I guess.
Keep in mind that:

\[\Large 2\alpha-\sin(2\alpha)+C=2\alpha-2\sin\alpha\cos\alpha+C \]

Answer 10

why do you change sin 2& with 2sin&cos&. I know that is a sub again, but why do you need to do it?

Answer 11

You will see in a minute. As you've figured out yourself, you need to use trigonometric back-substitution. To work this out yourself, I recommend you using a right triangle:

remember that:

\[\Large x=2\sin\alpha \longrightarrow \sin\alpha=\frac{x}{2}=\frac{opposite}{hypotenuse} \]