Use trigonometric substitution to verify

[celeramo]

Homework Statement

Using trigonometric substitution, verify that
\int \sqrt{a^2[-t^2}dt (INTEGRAL FROM 0 TO \pi)=(1/2)a^2sin^{-1}(x/a)+(1/2)x\sqrt{a^2-x^2

Sorry it doesn't seem to want to allow me to place superscript inside a square root. but inside the first sq. root is a²-t² and the second sq. root contains a²-x²

Homework Equations

\sqrt{a^2-x^2}

x=asin(\theta)

-(\pi/2) \leq \theta \leq (\pi/2)

The Attempt at a Solution

replacing x with asin\theta in the original integral
\int\sqrt{a^2-(asin^2\theta}dt
multiply and factor out. Change 1-sin^2\theta to cos^2\theta
replace dt with dt=(-a)cos\theta d\theta because t=asin\theta
End up with
\int (a)(cos\theta)(-a)(cos\theta)d\theta
Combine and get
-\int a²cos²\theta d\theta

At this point I considered pulling the a² out in front since, unless I'm confused, it represents a constant and changing cos²\theta to 1-sin²\theta but that hasn't seemed to get me any close. Any help would be much appreciated or if I've made a mistake already identifying that for me would be excellent. Please and thank you all.
:)

 

[Dick]

No, changing cos^2(theta) to sin^2(theta) doesn't help much. But using a double angle formula does. Change it to (1+cos(2*theta))/2

 

[celeramo]

Thanks a lot.