Give the substitution, involving the trigonometric function of either tan(𝜃), sin(𝜃) or sec(𝜃), that would be used when solving these integrals. Be sure to use 𝜃 as the variable.

Question

a) ∫√x^2+10x+61 dx

x=

b) ∫√x^2+6x-55 dx

x=

Answer 1

For part (b), complete the square to get x² + 6x - 55 = x² + 6x + 9 - 9 - 55 = (x+3)² - 8².

Let secθ = (x+3)/8, then x+3 = 8secθ. So, x = 8secθ - 3 and dx = 8secθtanθdθ.

∫ √(x²-6x-55)dx = ∫ √[(x+3)² - 64] dx = ∫ √[64sec²θ - 64](8secθtanθ)dθ.

Recall that 1 + tan²θ = sec²θ. So, the integral can be rewritten as 64 ∫ tan²θsecθdθ =

64∫ (sec²θ - 1)secθdθ = 64 ∫sec³θdθ - 64 ∫secθdθ = 64 ∫sec³θdθ - 64 ln l secθ + tanθ l + C.

The remaining integral can be done using integration by parts.

Answer 2

√(x² + 10x + 61) = √(x² + 10x + 25 + 36) = √((x+5)² + 36) = √((36)((1/36)(x+5)² + 1) = 6√((1/36)(x+5)² + 1) =

6√([(1/6)(x+5)]² + 1)

Under the square root is a squared term plus 1 which fits the pattern of "tan²θ + 1 = sec²θ" so

tanθ = [(1/6)(x + 5)]

6tanθ = x + 5

x = 6tanθ - 5

Use the same exact process on the 2nd problem except consider sec²θ - 1 = tan²θ