Integration problems: Intriguing Integrals

[BioCore]

Homework Statement

\intx^3\sqrt{}a^2-x^2dx

Homework Equations

The Attempt at a Solution

I came this far with my solution:

a^5\sqrt{}sin^3\Theta cos^2\Thetad\Theta
a^5\sqrt{}sin^3\Theta-sin^5\Thetad\Theta

The answer given to me is:

-1/15(a^2 - x^2)^3/2 (2a^2 + 3x^2) + C

It seems that they used sine theta, and not cos theta. I have currently tried to do that but am not sure how I would get rid of the extra sin\Theta.

Thanks for the help.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[rock.freak667]

\int x^3 \sqrt{a^2-x^2}dx

let x=acos\theta \Rightarrow dx=-asin\theta

\int x^3 \sqrt{a^2-x^2}dx \equiv \int (acos\theta)^3\sqrt{a^2-a^2cos^2\theta} \times -asin\theta d\theta

\int - a^4cos^3\theta \sqrt{a^2(1-cos^2\theta)} \times -asin\theta d\theta

-a^5 \int cos^3\theta sin^2\theta d\theta

recall that sin^2\theta +cos^2\theta=1

and substitute for sin^2\theta

 

[BioCore]

\int x^3 \sqrt{a^2-x^2}dx

let x=acos\theta \Rightarrow dx=-asin\theta

\int x^3 \sqrt{a^2-x^2}dx \equiv \int (acos\theta)^3\sqrt{a^2-a^2cos^2\theta} \times -asin\theta d\theta

\int - a^4cos^3\theta \sqrt{a^2(1-cos^2\theta)} \times -asin\theta d\theta

-a^5 \int cos^3\theta sin^2\theta d\theta

recall that sin^2\theta +cos^2\theta=1

and substitute for sin^2\theta

I would assume that using your solution I would have the right answer. The only problem is that in the solutions my Professor gave us, he has an intermediate step before the final solution such as this:

a^5 \int sin^3\theta cos^2\theta d\theta

Also in explaining to us Trig Substitution he placed \sqrt{a^2 - x^2} as the bottom portion of the right triangle. Thus he stated that

x = asin\Theta
dx = acos\Theta
\sqrt{a^2 - x^2} = acos\theta

 

[rock.freak667]

I would assume that using your solution I would have the right answer. The only problem is that in the solutions my Professor gave us, he has an intermediate step before the final solution such as this:

a^5 \int sin^3\theta cos^2\theta d\theta

Also in explaining to us Trig Substitution he placed \sqrt{a^2 - x^2} as the bottom portion of the right triangle. Thus he stated that

x = asin\Theta
dx = acos\Theta
\sqrt{a^2 - x^2} = acos\theta

Your teacher used x=sin\theta, I used x=cos\theta.
So in his triangle, sin\theta=\frac{x}{a} which would make the adjacent side,\sqrt{a^2-x^2}. At the end of that integral you would get terms involving cos\theta

In my substitution, the opposite side would be \sqrt{a^2-x^2} and at the end of my integral I would have terms involving sin\theta

So in theory it should all work out the same.

 

[BioCore]

Ok following your steps:

-a^5 \int \sqrt{1 - sin^2\theta} (1 - sin^2\theta) (sin^2\theta)d\theta

= -a^5 \int (1 - sin^2\theta)^3/2 (sin^2\theta)d\theta

= -a^5 \int sin^3\theta - sin^6\theta d\theta

= -a^5 [1/4sin^4\theta + 1/7sin^7\theta ] +C

Now I know that I should substitute for the sine function, but is this so far correct?