Question

Trig sub?

Answer 1

\[\int\limits_{}^{} \frac{x^2}{\sqrt{4-x^2}}dx\]
\[4-x^2=4-4\sin^2 \theta=4cos^2\theta\]
\(x=2sin\theta\)
\(dx=2cos\theta\)
\[=\int\limits_{}^{} \large \frac{2sin^2\theta}{\sqrt{4cos^2\theta}}*2cos\theta ~d\theta \]
\[=\int\limits_{}^{} \large \frac{2sin^2\theta}{2cos\theta}*2cos\theta ~d\theta\]
\[=\int\limits_{}^{} 2sin^2\theta ~d\theta \]
\[=\theta-\int\limits_{}^{} cos(2\theta)~d\theta \]
\[=\theta-\frac{1}{2}sin(2\theta)+c\]
trig identity for sin(2theta)
\[=\theta-sin(\theta)cos(\theta)+c\]
Substitute the x expressions for theta, sine, and cosine
\(x=2sin\theta\)
\(\theta=\arcsin(\frac{x}{2})\)
\(sin(\theta)=x/2\)
\(cos(\theta)=adj/2\)
\(x^2+a^2=2^2\)
\(a=\sqrt(4-x^2)\)
\(cosine=\frac{\sqrt(4-x^2)}{2}\)

\[=arcsin(\frac{x}{2})-(\frac{x}{2})(\frac{\sqrt(4-x^2)}{2})+c\]

Answer 2

answer is
\[2*arcsin(\frac{x}{2})-\frac{x\sqrt{4-x^2}}{2}+c\]

Answer 3

but my answer was
\[arcsin(\frac{x}{2})-\frac{x}{2}*\frac{\sqrt{4-x^2}}{2}+c\]
\[arcsin(\frac{x}{2})-\frac{x\sqrt{4-x^2}}{4}+c\]

Answer 4

@Kainui @iambatman @ganeshie8

Answer 5

Looks like when you first plugged it in you put 2sin^2 theta instead of 4sin^2 theta on top where you were looking at x^2.

Answer 6

Thanks @Kainui !

Answer 7

Yeah awesome job on the trig sub, it's just funny how you do all the hard part correct and it ends up being something silly like that haha.

Answer 8

Yeah lol xD

Answer 9

i have tons of others to do so yeah lol
But thanks again

Answer 10

Yeah, I'll be here off and on, feel free to tag me if you ever need me to check your work again.

Answer 11

I will start working on them now, and sure i'll tag you whenever you're on :)

See ya, few more qs and i'm off to bed