Antiderivative of square root function

[cemar.]

1. find the antiderivative of sqrt(81-x^2) in the bounds 0 to 9/2

okay so the first thing i did was sin substitution.
where I made x = 9sin(t) dx = 9cos(t) where t is theta.
then after some manipulating i got that that new integral was 9cos^2(t)
i then used the double angle formula and made it equal to 9/2 * (1 + cos(2t))
I then found the antiderivative of this integral to be 9/2 * (t + sin(2t)/2)
Using the identity that sin(2t) = 2sin(t)cos(t) the antiderivative became
(9/2) (t + cos(t)sin(t))
I then changed the bounds of the integral.
When x goes from 0 to 9/2, theta goes from 0 to pi/6
i then solved the integral to get 4.304
The actual answer is 38.74 so i think I am completely missing the boat here.

 

[cemar.]

Ps any tips you might have would really help me out it doesn't have to be too in detail I am usually pretty good at this stuff. Thank you!

 

[cemar.]

okay so it was actually a really dumb sneaky mistake:
after all the subtituting and manipulating was done instead of 9cos^2(t) its actually supposed to be 81cos^2(t) then follow my steps from there and its right!

 

[rohanprabhu]

It is a standard integral [i.e. anti-derivatives of a general form of problems]:

$$\int \sqrt{a^2 - x^2}dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} sin^{-1}\left(\frac{x}{a}\right) + C$$

 

[cemar.]

thats really cool! I've never seen that formula before!

 

[HallsofIvy]

It is, honestly, better to know HOW to do the integral, the way you did, rather than remember (or even look up) a formula.