Question

Calculus II: Can someone explain this problem to me? integral of sqrt 169-x^2 using trig substitutions.

Answer 1

The correct answer is \[169/2 \sin ^{-1} (x/13) + x/2 \sqrt{169-x ^{2}} + C\] My question is, how do you get X/2 in the answer?

Answer 2

note that 169 = 13^2

use substitution
\[x = 13\sin u\]
\[dx = 13 \cos u du\]
\[\rightarrow 169 \int\limits_{}^{}\cos^{2} u du\]
\[ = 169 (\frac{\sin u \cos u}{2} + \frac{u}{2}) +C\]
substitute back in for x where
\[u = \sin^{-1} (x/13)\]
\[\sin u = \frac{x}{13}\]
\[ \cos u = \sqrt{1-(x/13)^{2}} = \frac{\sqrt{169-x^{2}}}{13}\]