Integral problem

Question

Solve step by step. If possible, without using trigonometric substitution.

Answer 1

The integration expression is ʃ√(1 - 2x - x²)dx.

Write the quadratic expression in the complete square form.

= 1 - 2x - x².

= 1 - (2x + x²).

= 1 - (x² + 2x + 1 - 1).

= 1 - (x+1)² + 1.

= 2 - (x + 1)².

Let (x + 1) = u ⇒ dx = du.

= 2 - u².

ʃ√(1 - 2x - x²)dx = ʃ√[(√2)² - u²]du.

Apply formula: ʃ√(a² - u²)du = (1/2)[a²sin^-1(u/a) + u√(a² - u²)] + C.

= (1/2)[2sin^-1(u/√2) + u√(2 - u²)] + C

= (1/2)[2sin^-1{(x+1)/√2} + (x+1)√{2 - (x + 1)²}] + C      [Since (x + 1) = u]

ʃ√(1 - 2x - x²)dx = (1/2)[2sin^-1{(x+1)/√2} + (x+1)√{2 - (x + 1)²}] + C