Trignometric Substiution Problem

[Archimedes II]

Homework Statement

$\displaystyle \int _0 ^a 4x^2 \sqrt{a^2-x^2} dx$

Homework Equations

$\cos^2 x =1 - \sin^2 x$

The Attempt at a Solution

Substuting

Substituting $x=a \sin(\theta)$

$dx=a\cos( \theta) d\theta$

x=0
asin$\theta$=0
sin$\theta$=0
$\theta$=0

$\displaystyle \int_{0}^{a} 4x^2 \sqrt{a^2-x^2} dx =$

$\displaystyle \int_{0}^{\pi/2} 4(a\sin\theta)^2 \sqrt{a^2-(a\sin\theta)^2} a\cos\theta d\theta=$

$\displaystyle \int_{0}^{\pi/2} 4a^2sin^2\theta \sqrt{a^2-a^2sin^2\theta}a\cos\theta d\theta=$

$\displaystyle \int_{0}^{\pi/2} 4a^2sin^2\theta a \sqrt{1-sin^2\theta}a\cos\theta d\theta=$

$\displaystyle \int_{0}^{\pi/2} 4a^2sin^2\theta a cos\theta a\cos\theta d\theta=$

$\displaystyle 4a^4\int_{0}^{\pi/2} sin \theta cos^2 \theta d\theta=$

U substitution

$u=cos\theta$

$-du= sin\theta d\theta$

$\displaystyle 4a^4\int u^2 du=$
When I put in the bounds it yields from 1 to 0. What did I do wrong

 

[Archimedes II]

So the code is messed up basically

Integral from 0 to a [4x^2 squareroot(a^2 - x^2)]dx =

4a^4 integral from 0 to pi/2 [sin theta cos^2 theta] dtheta

I'm stuck trying to change the bounds with a u substitution

 

[SammyS]

Homework Statement

$\displaystyle \int _0 ^a 4x^2 \sqrt{a^2-x^2} dx$

Homework Equations

$\cos^2 x =1 - \sin^2 x$

The Attempt at a Solution

Substituting $x=a \sin(\theta)$

$dx=a\cos( \theta) d\theta$

New Bounds

x=a
a sin$\theta$=a
sin$\theta$=1
$\theta$=$\pi$/2x=0
asin$\theta$=0
sin$\theta$=0
$\theta$=0

$\displaystyle \int_{0}^{a} 4x^2 \sqrt{a^2-x^2} dx =$

$\displaystyle \int_{0}^{\pi/2} 4(a\sin\theta)^2 \sqrt{a^2-(a\sin\theta)^2}\ \, a\cos\theta d\theta=$

I fixed your LaTeX code up to this point.

$\int from 0 to \pi/2 4(a^2sin^2\theta)\sgrt{a^2-(a^2sin^2\theta)} acos\theta d\theta$=

$\int from 0 to \pi/2 4(a^2sin^2\theta) a \sgrt{(1- sin\theta)} acos\theta d\theta$=

$\int from 0 to \pi/2 4(a^2sin^2\theta) a sos\theta acos\theta d\theta$=

$4a^4\int form 0 to \pi/2 sin \theta cos^2 \theta d\theta$

Let u= $cos/theta$

thus -du=$sin/theta d\theta$

My problems is when I set the new bounds I get from 1 to 0 not 0 to 1. What am I doing wrong?

Here is the Code for some of the lines I fixed:

[itex]\displaystyle \int_{0}^{a} 4x^2 \sqrt{a^2-x^2} dx =[/itex]

[itex]\displaystyle \int_{0}^{\pi/2}  4(a\sin\theta)^2 \sqrt{a^2-(a\sin\theta)^2} a\cos\theta d\theta=[/itex]

The "\displaystyle" modifier allows the integral symbol to be larger than otherwise.

I'll let you edit the Original Post .

 

[Archimedes II]

Ok, I fixed it thank you very much. I didn't know how to put a bound into the intergral code.

 

[SammyS]

...

$\displaystyle 4a^4\int_{0}^{\pi/2} sin \theta cos^2 \theta d\theta=$

U substitution

$u=cos\theta$

$-du= sin\theta d\theta$

When I put in the bounds it yields from 1 to 0. What did I do wrong

So you get

$\displaystyle -4a^4\int_{1}^{0} u^2\,du$

Right?

 

[Archimedes II]

So you get

$\displaystyle -4a^4\int_{1}^{0} u^2\,du$

Right?

Yes but the how can the bound be from a higher number to a smaller one.

 

[SammyS]

Yes but the how can the bound be from a higher number to a smaller one.

What does switching the order do to the integral?

 

[Archimedes II]

What does switching the order do to the integral?

I solved it I forgot to square the sin at the end. I did't need to use a u substitution.

Any regardless if you have that bound it yields a negative answer.

Thanks for helping though.

The answer is

$\pi a^4/4$

 

[SammyS]

I solved it I forgot to square the sin at the end. I didn't need to use a u substitution.

Any regardless if you have that bound it yields a negative answer.

Thanks for helping though.

The answer is

$\pi a^4/4$

Yes. It works out just fine either way.

Switching the limits of integration switches the sign. That's good because it got rid of that leading negative sign.

... and you learned a bit more of LaTeX coding .