Question

Evaluate the integral

Answer 1

Evaluate the integral \[\int\limits \sqrt{1-4x^2}dx\] Using \[\frac{ u }{ 2 }\sqrt{a^2+u^2}-\frac{ a^2 }{ 2 }\ln (u+\sqrt{a^2+u^2})\]

Answer 2

Have you tried substituting \(\sin u=2x\) ?

Answer 3

this is what i got, idk if its right
let u= \[\sqrt{2x}\]
let a= 1 and
let \[u^2=4x^2\]
then \[\frac{ \sqrt{2x} }{ 2 }*\sqrt{1+4x^2}-\frac{ 1 }{ 2 } \ln(\sqrt{2x} + \sqrt{1+4x^2})\]=
\[\sqrt{1+4x^2}-\frac{ \sqrt{2} }{ 2 }\ln(\sqrt{2x}+\sqrt{1+4x^2})\]

Answer 4

no i didn't just because the equation i posted, is the form the "answer" will be in

Answer 5

he gives us a chart and says, compare the integral to the chart, and use the equation to find your answer

Answer 6

number 21

Answer 7

Fair enough. Personally, I'm not a fan of integral table formulas, but let's work with the given formula. Let's take the derivative to make sure this is the right one.
\[\frac{d}{du}\left[\frac{ u }{ 2 }\sqrt{a^2+u^2}-\frac{ a^2 }{ 2 }\ln (u+\sqrt{a^2+u^2})\right]\\
\quad\quad\quad\quad =\frac{1}{2}\sqrt{a^2+u^2}+\frac{u^2}{2\sqrt{a^2+u^2}}-\frac{a^2}{2}\frac{1+\dfrac{u}{\sqrt{a^2+u^2}}}{u+\sqrt{a^2+u^2}}\\
\quad\quad\quad\quad =\frac{a^2+u^2}{2\sqrt{a^2+u^2}}-\frac{a^2-u^2}{2\sqrt{a^2+u^2}}\\
\quad\quad\quad\quad =\frac{u^2}{\sqrt{a^2+u^2}}\]
This is not the same as the form of the given integral, however, which is more like \(\sqrt{a^2-u^2}\).

Answer 8

im not a fan either... ugh.... im not sure, guess i got lost here

Answer 9

The formulas on the bottom half of the page involve \(a^2\color{red}+u^2\). Check the section for those with \(a^2\color{blue}-u^2\) instead.

Answer 10

oh ok... let me look

Answer 11

would it be any of the ones with the du on top?

Answer 12

No, your integral takes on the form \(\int \sqrt{a^2-u^2}\,du\), so you should be looking for that form.

Answer 13

i think i found it on this other ref page...

Answer 14

jeez! these table integrals suck! number 30 looks better

Answer 15

Exactly, it's #30.

Answer 16

how do i know that its not number 39. where a and u have switched spots? because there isn't a zero?

Answer 17

\(a-b\) is fundamentally different from \(b-a\). #30 has \(a^2-u^2\), that is a constant minus \(u^2\), whereas #39 is the opposite, \(u^2\) minus a constant.

Answer 18

ok...the "constant" minus u^2....i understand that

Answer 19

I hate tables too. I don't see how using the tables is mathematically good for you.
and hey jlock

Answer 20

hey freckles!!! i am studying for the final... he is making it "comprehensive" so i got to restudy everything

Answer 21

have to

Answer 22

At any rate, it makes more sense to me to know where these formulas come from than it does being able to pick it out from a table, so I recommend deriving them using a general version of the trig sub I mentioned.
\[\int\sqrt{a^2-u^2}\,du\]
Set \(u=a\sin t\), then \(du=a\cos t\,dt\), and so
\[\begin{align*}\int\sqrt{a^2-u^2}\,du&=\int\sqrt{a^2-(a\sin t)^2}(a\cos t)\,dt\\\\&=a^2\int\sqrt{1-\sin^2t}\cos t\,dt\\\\
&=a^2\int \cos^2t\,dt\\\\
&=\frac{a^2}{2}\int (1+\cos2t)\,dt\\\\
&=\frac{a^2}{2}\left(t+\frac{1}{2}\sin2t\right)+C\\\\
&=\frac{a^2}{2}\left(\sin^{-1}\frac{u}{a}+\frac{u}{a}\times\frac{\sqrt{a^2-u^2}}{a}\right)+C\\\\
&=\frac{a^2}{2}\sin^{-1}\frac{u}{a}+\frac{u}{2}\sqrt{a^2-u^2}+C
\end{align*}\]

Answer 23

Skipped a step: I used the fact that \(\sin2t=2\sin t\cos t\), so \(\dfrac{1}{2}\sin2t=\sin t\cos t\).

Answer 24

wow... lots of identities! you make this look so easy

Answer 25

thank you as usual

Answer 26

You're welcome! Good luck