Examples using Trig Substitutions

Example involving (x² –a²) ^p

Evaluate the integral I = ∫ dx / [ (ax)² – b² ] ^3/2 (here p = 3/2)

First let u = ax, then du = a dx and dx = (1/a) du (u is an intermediate variable)

The integral becomes I = (1/a) ∫ du / [ u² – b² ] ^3/2

In this case use the “trig substitution” u = sec θ , so du = sec θ tan θ dθ

and [ u² – b² ] = b² sec² θ – b² = b² tan² θ so

I = (1/a) ∫ [ ( b sec θ tan θ dθ ) / (b³ tan³ θ) ] dθ = [ (1/a b²) ∫ sec θ / tan² θ ] dθ

or I = (1/a b²) ∫ cos² θ / sin² θ cos θ ] dθ = (1/a b²) ∫ cos θ /sin² θ ] dθ

Now let w = sin θ dw = cos θ and the integral becomes

I = (1/a b²) ∫ dw / w² = (1/a b²) ( -1/w) + C = ( - 1 /a b²) ( 1/sin θ ) + C

or I = (- 1/a b²) csc θ + C

Now express the result in terms of the original variable x.

Start with the triangle involving the intermediate variable, u. Recall u = sec θ .

The relevant triangle is:

Therefore sec θ = u / b and csc θ = u / √ ( u² – b² ) and also recall u = ax

So I = (- 1/a b²) ax /√ [ (ax)² – b² ] + C or finally

I = (- 1/ b²) { x /√ [ (ax)² – b² ] } + C (result)

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