The Method of Trigonometric Substitution

Main Idea The method helps dealing with integrals, where the integrand contains one of the following expressions: (where a and b are constants)

A simpler forms of the former expressions are the following ones:

To get rid of the root, we substitute sin θ, tanθ or secθ respectively dx becomes The radical becomes SubstitutionThe radical cos θ dθ cos θ x = sin θ sec 2 θ dθ sec θ x = tan θ sec θ tan θ dθ tan θ x = sec θ

To apply that to the general cases, we transfer the radical to a form similar to the respective simple form

The First Case We let: sin θ = (b/a)x → x = asinθ/b → dx = (a/b) cos θ dθ The radical becomes a cosθ

The Second Case We let: tan θ = (b/a)x → x = (a /b) tanθ → dx = (a/b) sec 2 θ dθ The radical becomes a sec θ

The Third Case We let: sec θ = (b/a) x → x = (a /b) secθ → dx = (a /b) sec θ tanθ dθ The radical becomes a tanθ

Examples

Example 1 The First case

We must write the answer in terms of x We have: sin θ = 3x/4 Hence θ = arcsin(3x/4) cos θ = √[1 - (3x/4) 2 ] = (1/4) √[16 - 9x 2 ] And so: sin2 θ = 2 sin θ cos θ = 2 (3x/4). (1/4). √[16 - 9x 2 ] = (3/8)x. √[16 - 9x 2 ]

A faster way to find cos θ is by using the triangle method, starting from the fact that sin θ = 3x / 4

Example 2 The second case

Let’s simplify the intrgrand (The expression inside the integral sign)

We must write the answer in terms of x We have: tan θ = 3x/5 Hence, sec θ = √[ 1 + (3x/5) 2 ] = √[ 1 + (9x 2 /25)] = 5 √[25 + 9x 2 ] & cos θ = 1/ 5 √[25 + 9x 2 ]

We could have also found sec θ and cos θ using the triangle method, starting from ths fact that tan θ = 3x / 5

If we wish, we can simplify the answer

Example 3 The Third case

We must write the answer in terms of x We have: sec θ = 2x/5 Hence, csc θ = 1 / √[1 –(5/2x) 2 ] = 2x / √[4x 2 –25] 2

Here is the triangle method to find csc θ, starting from the fact that sec θ = 2x / 5

Integrals Involving an Expression of the Form √(ax 2 +bx+c) We simply complete the square and rewrite the expression in one of the previous three forms

Example 1 X x + 16 = (x + 5) 2 – = (x + 5) = 9 [ ( ( x + 5) / 3 ) ] Let (x + 5 ) / 3 = sec θ → x = 3 secθ – 5 → dx = 3 secθ tanθ dθ

Example x x = 11 - [ x x ] = 11 - [ ( x + 5 ) 2 – 25 ] = 11 - ( x + 5 ) = 36 - ( x + 5 ) 2 = 36 [ 1 - ( (x+5) / 6 ) 2 ] Let ( x + 5 ) / 6 = sin θ → x = 6sinθ – 5 → dx = 6 cosθ dθ

Example 3 50x - 25x = - 25 (x 2 - 2x ) - 16 = - 25 [ (x - 1) 2 - 1] - 16 = 9 – 25 ( x – 1 ) 2 = 9[ 1 - ( 5(x -1) / 3 ) 2 ] Let (5/3 ) (x-1) = sin θ → x = (3 / 5) sinθ + 1 → dx = (3 / 5) cosθ