Integration Question (1 Viewer)

gigapuddi · Apr 21, 2012

Stuck on this one:

<img src="http://latex.codecogs.com/gif.latex?\int_{0}^{\frac{a}{2}} x^{2}\sqrt{a^{2}-x^{2}}" title="\int_{0}^{\frac{a}{2}} x^{2}\sqrt{a^{2}-x^{2}}" />

Tried by parts in different ways and special formula (f(a) = f(a-x)).

Online integral form is crazy and i do not know how to get that far.

Thanks.

nightweaver066 · Apr 21, 2012

$\int_{0}^{\frac{a}{2}} x^{2}\sqrt{a^{2}-x^{2}}dx$

$$Let $ x = asin\theta$

$dx = acos\theta d\theta$

$x = 0, \theta = 0$

$x = \frac{a}{2}, \theta = \frac{\pi}{6}$

$\therefore \int a^2sin^2\theta \sqrt{a^2 - a^2sin^2\theta} acos\theta d\theta$

$= \int a^4 sin^2\theta cos^2\theta d\theta$

$= \frac{a^4}{4} \int sin^2 2\theta d\theta$

$= \frac{a^4}{8} \int 1 - cos4\theta d\theta$

$= \frac{a^4}{8} [\theta - \frac{sin4\theta}{4}]^{\frac{\pi}{6}}_0$

$= \frac{a^4}{8} (\frac{\pi}{6} - \frac{\sqrt{3}}{8})$

Omitted the limits when typing the integral because its tiresome to type.

RealiseNothing · Apr 21, 2012

nightweaver066 said:
$\int_{0}^{\frac{a}{2}} x^{2}\sqrt{a^{2}-x^{2}}dx$

$$Let $ x = asin\theta$

$dx = acos\theta d\theta$

$x = 0, \theta = 0$

$x = \frac{a}{2}, \theta = \frac{\pi}{6}$

$\therefore \int a^2sin^2\theta \sqrt{a^2 - a^2sin^2\theta} acos\theta d\theta$

$= \int a^4 sin^2\theta cos^2\theta d\theta$

$= \frac{a^4}{4} \int sin^2 2\theta d\theta$

$= \frac{a^4}{8} \int 1 - cos4\theta d\theta$

$= \frac{a^4}{8} [\theta - \frac{sin4\theta}{4}]^{\frac{\pi}{6}}_0$

$= \frac{a^4}{8} (\frac{\pi}{6} - \frac{\sqrt{3}}{8})$

Omitted the limits when typing the integral because its tiresome to type.

When you do trigonometric substitution, are you allowed to change the limits, because you are going from algebraic terms to trigonometric terms?

I had a question similar to this in my half yearly, and changed the limits so it was in terms of theta. But afterwards I got worried if I could or not.

When my teacher taught us trigonometric substitution, to find theta after integrating, he used a triangle to put it back into terms of 'x'. So I was wondering if just changing limits would still be right.

nightweaver066 · Apr 21, 2012

RealiseNothing said:
When you do trigonometric substitution, are you allowed to change the limits, because you are going from algebraic terms to trigonometric terms?

I had a question similar to this in my half yearly, and changed the limits so it was in terms of theta. But afterwards I got worried if I could or not.

When my teacher taught us trigonmetric substitution, to find theta after integrating, he used a triangle to put it back into terms of 'x'. So I was wondering if just changing limits would still be right.

Of course lol. Whenever you substitute, you have to change everything in terms of that substitution (the dx -> d(theta), the original expression -> new expression, limits -> new limits)

You only need to use the triangle when going back to the original variable.

RealiseNothing · Apr 21, 2012

nightweaver066 said:
Of course lol. Whenever you substitute, you have to change everything in terms of that substitution (the dx -> d(theta), the original expression -> new expression, limits -> new limits)

You only need to use the triangle when going back to the original variable.

Ok good, I was worried lol.

The reason I thought you couldn't for a trigonometric substitution is that the answer in my half yearly was something like:

$1 + \frac{\pi}{2}$

So it made me doubt that you could go from algebraic to trigonometry.

Integration Question (1 Viewer)

gigapuddi

New Member

nightweaver066

Well-Known Member

RealiseNothing

what is that?It is Cowpea

nightweaver066

Well-Known Member

RealiseNothing

what is that?It is Cowpea

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