Proving the Integral of sqrt(a^2-x^2) without a Prefix

[_wolfgang_]

Homework Statement

Prove that \int\sqrt{}9-x^2 dx

=\frac{9\Theta}{2}+\frac{9sin2\Theta}{4}+c

given that x=3sin\Theta

Homework Equations

The Attempt at a Solution

\int\sqrt{}9-x^2 dx

=\frac{(9-x^2)^{1.5}}{10x}

=\frac{(-x^2)(\sqrt{-x^2+9}+9\sqrt{(-x^2)+9}}{10x}

=\frac{-x(x^2\sqrt{(-x^2)+9}-9(\sqrt{(-x^2)+9}}{10}

=\frac{-3sin\Theta(9(sin^2)\Theta\sqrt{9(sin^2)\Theta-9}-9\sqrt{9(sin^2)+9}}{10}


Im not to sure if I am going in the right direction if i am not guidance would be appreciated

 

[vela]

Homework Statement

Prove that

\int\sqrt{9-x^2}dx=\frac{9\Theta}{2}+\frac{9\sin2\Theta}{4}+c

given that x=3\sin\Theta.

Homework Equations

The Attempt at a Solution

\int\sqrt{}9-x^2 dx

=\frac{(9-x^2)^{1.5}}{10x}

=\frac{(-x^2)(\sqrt{-x^2+9}+9\sqrt{(-x^2)+9}}{10x}

=\frac{-x(x^2\sqrt{(-x^2)+9}-9(\sqrt{(-x^2)+9}}{10}

=\frac{-3sin\Theta(9(sin^2)\Theta\sqrt{9(sin^2)\Theta-9}-9\sqrt{9(sin^2)+9}}{10}


I'm not to sure if I'm going in the right direction. If I am not, guidance would be appreciated.

You're not going in the right direction. Your very first step is wrong. The power rule only applies when the integrand is of the form xⁿdx where n \ne -1. It doesn't apply when you have some function of x taken to a power, as you do in this case. Also, I have no idea where that 10x in the denominator came from.

Use the substitution given and rewrite the integral in terms of \theta first.

 

[_wolfgang_]

ok so if i do...

\int\sqrt{9-9sin^2\Theta}
=\int3-3sin\Theta
=3\Theta+3cos\Theta

is it going in the right direction now?? thanks for the help

 

[vela]

Marginally better. First of all,

\sqrt{a^2-b^2} \ne a-b

Second, you forgot the dx and then didn't write it in terms of d\theta. Finally, not that it really matters, you didn't integrate the first term correctly.

I would suggest you review your textbook on the topic of trig substitutions. There's probably a similar example you could use as a template for solving this problem.

 

[HallsofIvy]

\sqrt{9- 9 sin^2(\theta)}= 3\sqrt{1- sin^2(\theta)}= 3\sqrt{cos^2(\theta)}

 

[_wolfgang_]

okay i was finally able to prove it!

\int\sqrt{a^2-x^2} dx
=\int a^2-a^2sin^2\Theta acos\Theta d\Theta
=\int\sqrt{a^2(1-sin^2\Theta} acos\Theta d\Theta
=\int\sqrt{a^2cos^2\Theta} acos\Theta d\Theta
=\int a^2cos^2\Theta d\Theta

=\frac{a^2}{2}(\Theta+sin\Thetacos\Theta)

i know i definatly skipped a couple of steps in integrating cos^2