Question

With integral 0 to a, evaluate (3a^2 -x^2)/(2a(sqrt(2a^2-x^2)) using the substitution x = a(sqrt2)sin(theta)

Answer 1

\[\Large \int\limits_0^a\frac{3a^2-x^2}{2a \sqrt{2a^2-x^2}}\;dx\]Here are the pieces we will use for our substitution:\[\Large \color{orangered}{x\quad=\quad a \sqrt2 \sin \theta}\]\[\Large \color{#CC0033}{dx\quad=\quad a \sqrt2 \cos \theta\;d\theta}\]

Answer 2

\[\Large \int\limits\limits_{x=0}^a\frac{3a^2-\color{orangered}{x}^2}{2a \sqrt{2a^2-\color{orangered}{x}^2}}\;\color{#CC0033}{dx}\]Understand `where` we'll be plugging these in?
It's a bit of work to simplify it down, shouldn't be too bad though.
Confused on anything yet?

Answer 3

Use the comments section here to chat :D not the private message box lol

Answer 4

So making the substitution gives us:\[\Large \int\limits \frac{3a^2-(\color{orangered}{a \sqrt2 \sin \theta})^2}{2a\sqrt{2a^2-(\color{orangered}{a\sqrt2 \sin \theta})^2}}\color{#CC0033}{a\sqrt2 \cos \theta\;d\theta}\]

Answer 5

Which simplifies to:\[\Large \int\limits\limits \frac{3a^2-2a^2 \sin^2\theta}{2a\sqrt{2a^2-2a^2\sin^2\theta}}a\sqrt2 \cos \theta\;d\theta\]Factoring a 2a^2 out of each term under the root, taking it out of the root.\[\Large \int\limits\limits\limits \frac{3a^2-2a^2 \sin^2\theta}{2a(a \sqrt2)\sqrt{1-\sin^2\theta}}a\sqrt2 \cos \theta\;d\theta\]

Answer 6

Understand how the portion under the root will simplify now?
We need to use our trig identity,\[\Large \sin^2\theta+\cos^2\theta=1\qquad\to\qquad \color{royalblue}{1-\sin^2\theta =\cos^2\theta}\]

Answer 7

Zepdrix the typing is all the place. However you typed in your first response was much clearer

Answer 8

Oh you don't like the black font? :) lol
Too much math all jumbled up?

Answer 9

Maybe I'm misunderstanding you.. :o

Answer 10

it is all jumbled with a lot of words and numbers

Answer 11

Ohh I see.
Are you using Internet Explorer by chance?

I think the Plugin for the math stuff won't work in your on IE :(

Answer 12

yes I am using internet explorer

Answer 13

Ok let's try using the drawing tool then.