Question

Find the integral of (4-x^2)^(-3/2) dx from 0 to 1. (Answer: sqrt(3)/12) (I got [-2(4-x^2)^(-1/2)] from 0 to 1 but that's not the answer.)

Answer 1

Is there a easier way to do this?

Answer 2

You could of used Wolfram. That's easier method.

Answer 3

I thought I did change the limits @Loser66... you could change the bottom one to 0.

@Idealist, have you learned about trig substitution yet?

Answer 4

I think I did.

Answer 5

Let me try one more time... lol

Answer 6

OHHHH! He said your method is unfamiliar and implied that your method was dumb!! Buuuuuuuuurn!!!

Answer 7

\[\int_0^1\left(4-x^2\right)^{-3/2}~dx\]
This integral is with respect to \(x\), which changes from 0 to 1. If we let \(x=2\sin u\), then \(u=\arcsin \dfrac{x}{2}\), which changes from \(\arcsin 0=0\) to \(\arcsin \dfrac{1}{2}=\dfrac{\pi}{6}\).
\[\int_0^{\pi/6}(4-4\sin^2u)^{-3/2}~2\cos u~du\]

Answer 8

So, @Idealist, the procedure here is to reduce to a more easily integrable expression:
\[\frac{1}{4}\int_0^{\pi/6}(1-\sin^2u)^{-3/2}~\cos u~du\\
\frac{1}{4}\int_0^{\pi/6}(\cos^2u)^{-3/2}~\cos u~du\\
\frac{1}{4}\int_0^{\pi/6}\sec^2u~du\\
\]

Answer 9

lmao

Answer 10

abbot

Answer 11

And then you integrate, right?

Answer 12

Yup

Answer 13

Wow, I like this method more.

Answer 14

@SithsAndGiggles Thanks for giving us the easiest way to get the answer.

Answer 15

This is the hardest integral that I've seen!

Answer 16

hahaa

Answer 17

Don't laugh.

Answer 18

<3

Answer 19

@Idealist sweetie, ignore him

Answer 20

Hahaha