# Use trigonometric substitution to verify

1. Nov 24, 2008

### celeramo

1. The problem statement, all variables and given/known data
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

2. Relevant equations

$$\sqrt{a^2-x^2}$$

x=asin($$\theta$$)

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

3. 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.
:)

Last edited by a moderator: Nov 25, 2008
2. Nov 24, 2008

### 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

3. Nov 25, 2008

### celeramo

Thanks a lot.