Integrating an expression with radical by using trigonometric substitutions

In order to integrate integrals containing the radical expressions ⎷a²-x², ⎷a² + x², ⎷x² - a², we put these expressions in trigonometric form. In order to find the equivalent trigonometric expressions, we make each of the expressions the side of a right triangle.

We start by drawing three right triangles whose  one side is one of these expressions : ⎷a²-x², ⎷a² + x², ⎷x² - a²





Case 1 ⎷a²-x²

In this case we consider the middle triangle whose hypotenuse is a and the vertical side of the right angle is x. The horizontal side of the right angle is the expression containing the radical.

In considering the relations in a right triangle we write x = asinθ. In this case we use the identity 1-sin²θ = cos²θ

Case 2 ⎷a² + x²

In this case we consider the first right triangle whose hypotenuse is  ⎷a² + x²

In a right triangle the tangent is equal to the ratio of the opposite side to the adjacent side. Therefore tanθ = x/a then x = atanθ. We use the identity  1 + tan² = sec²θ

Case 3 ⎷x² - a²

We consider the third triangle where the vertical side of the right angle is ⎷x² - a². In a right triangle a side of a right angle is equal to the product of the hypotenuse by the cosine of the adjacent acute angle. Therefore  a = xcosθ then x = a/cosθ x = a.1/cosθ x= asecθ the relation used is sec²θ-1 = tan²θ

Example I Evaluate ∫dx/x²⎷4-x². 

Pay attention to the fraction bar. Read  ∫dx over x²⎷4-x². The radical affects the whole expression 4-x². The software I am using doesn't give me the horizontal bar of the radical sign.

Here we have case 1 ⎷a²-x². We have ⎷4-x² = √2² - x² then   a = 2. Therefore x = asinθ  x = 2sinθ. The identity used is  1-sin²θ = cos²θ . For x = asinθ we have dx/dA = 2cosθ. Then dx = 2cosθdθ.

Let's transform x²⎷4-x²:
 
x²⎷4-x² = 4sin²θ√4-4 4sin²θ

             =  4sin²θ√4(1-sin²θ)

            =  4sin²θ√4cos²θ

            =   4sin²θ2cosθ

            =   8sin²θcosθ

Let's substitute dx and x²⎷4-x in the original expression. we have;

∫dx/x²⎷4-x² = ∫2cosθdθ/ 8sin²θcosθ

                   =  ∫dθ/ 4sin²θ

                   = 1/4∫1/sin²θdθ

                   =1/4∫1/csc²θdθ
                  = -1/4cotθ + C

Let's express cotθ in function of x. Using the right triangle we can write: tanθ = x/ ⎷a²-x². then cotθ =   ⎷a²-x²/x

Let's substitute a by 2 we have  tanθ = x/ ⎷4-x². Then cotθ = ⎷4-x²/x.

Finally we have ∫dx/x²⎷4-x² = -1/4 ⎷4-x²/x + C

Example2  Evaluate ∫√x²-3/x dx

Here we have ⎷a²-x² we use the relation x = asecθ
Let's find a:

√x²-3 = √x²-(√3)². Let's note that the radical affect x²-(√3)². We have a = √3. then x = √3secA

dx = √3tanθsecθdθ

Let's substitute x and dx:

∫√x²-3/x dx = ∫√(√3secθ)²-3/√3secθ.√3tanθsecθdθ

                  = ∫√3sec²θ-3/√3secθ.√3tanθsecθdθ

                  = ∫√3sec²θ-3/√3secθ.√3secθtanθdθ (rearranging the expression to have √3secθ)

                 =   ∫√3sec²θ-3tanθdθ (simplifying by √3secθ)

                 =   ∫⇃3(sec²θ-1)tanθdθ

                 =   ∫√3 tan²θtanθdθ

                 =    ∫√3 tanθtanθdθ

                 =     ∫√3tan²θdθ

Let's calculate  ∫√3tan²θdθ

Let's apply the formula ∫tanⁿx =  tanⁿ⁻¹x/n-1 - ∫tanⁿ⁻²xdx

 ∫√3tan²θdθ = √3∫tan²θdθ

                     =  √3(tanθ -∫dθ)

                     =   √3(tanθ - θ) + C

 Let's calculate tan using the third rectangle triangle:
 tanθ = √x² - a²/a = √x²-3/√3. We use the rule that in a right triangle the tangent of an acute angle is equal to the ratio of the opposite side to the adjacent side. We can also say that A = tan⁻¹( √x²-3/√3) 
  ∫√x²-3/x dx =  √3[√x²-3/√3 - tan⁻¹(√x²-3/√3)] + C
                     =  √x²-3 - √3 tan⁻¹(√x²-3/√3) + C

Practice   Evaluate ∫ dx/x²√x² + 1

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