Question

Find or evaluate the integral. (Complete the square, if necessary.) $$ \int \frac{2}{\sqrt{-x^{2}+4 x}} d x $$

Short answer

The evaluated integral is \(2 \sin^{-1}((x-2)/2) + C\).

Step-by-step solution

Completing The Square

The aim is to make \( -x^2 + 4x \) into a square. The general form will be \( -(x-a)^2 \). The term 'a' is half of the coefficient of x which gives \( -((x - 2)^2 - 4) \). Thus the integral becomes \( \int \frac{2}{\sqrt{-(x - 2)^2 + 4}} dx \)

Substitution

Let \( u = x - 2 \). Then \( du = dx \). The integral becomes \( \int \frac{2}{\sqrt{-u^2 + 4}} du \)

Simplify And Integrate

\(\int \frac{2}{\sqrt{-u^2 + 4}} du = 2 \int \frac{1}{\sqrt{4 - u^2}} du \). This integral is a standard form and can be represented as \( \sin^{-1}(u/2) \). So, \( 2 \int \frac{1}{\sqrt{4 - u^2}} du = 2 \sin^{-1}(u/2) + C \), where C is the constant of integration.

Re-substitute

Substitute back \( u = x - 2 \) to get the final answer. This gives \( 2 \sin^{-1}((x-2)/2) + C \)