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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