Answer to Question #222886 in Calculus for cayyy

Question

Consider the following integral:

I =  ∫ 1 / (√2x²-x²) dx

(a) complete the square of f(x) = 2x²-x²

(b) use (a) together with the method of trigonometric substitution to determine the integral.

Answer 1

$\displaystyle (a)\\ \textsf{Completing the square}\\ 2x - x^2 = -(-2x + x^2) = -(1 - 2x + x^2) + 1 = -(x - 1)^2 + 1\\ (b)\\ \textsf{Under the substitution}\,\, x - 1 = \sin\theta\\ \begin{aligned} \int \frac{1}{\sqrt{2x - x^2}}\mathrm{d}x &= \int \frac{1}{\sqrt{1 - (x - 1)^2}}\mathrm{d}x \\&= \int \frac{1}{\sqrt{1 - \sin^2{\theta}}}\cos{\theta}\mathrm{d}\theta \\&= \int \frac{1}{\sqrt{\cos^2{\theta}}}\cos{\theta}\mathrm{d}\theta \\&= \int \mathrm{d}\theta = \theta + C \\&= \arcsin(x - 1) + C \end{aligned}$