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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Integrating the power of secants and tangents

In this post you are going to learn how to integrate the power of secant and tangent. There are some formulas that allow to calculate these powers. Before to learn these formulas let's learn first the formulas to integrate the secant and tangent.

Formulas to integrate the function tangent and secant

You learn first the expression of tangent since if you know its integral it is easy to know the integral of secant. The integral of each of these functions has the function ln, sec.

The integral of tangent is an expression of secant

∫tanx = ln❘secx❘ + C

The integral of secant is an expression of sec and tangent. We just add tangent between the absolute bars.from the integral of tanx

∫secx = ln❘secx + tanx❘ + C

N.B. These formulas can be demonstrated but the demonstration is not done here.

Formulas to integrate the power of tangent and secant

Integral of the power of secant

The integral of the power of secant is made of two terms. The first term is a quotient. The second term is the product of a constant by the integral of secant. We are going to learn the formula according to the following steps:

1. The numerator of the quotient and the and the second part of the integral are the same. It is obtained by decreasing the exponent of secx by 2.

∫secⁿ x = secⁿ⁻²x +  ∫secⁿ⁻²x

2. The denominator of the first quotient is obtained by taking only the exponent n and decreasing it by 1

∫secⁿ x = secⁿ⁻²x/n-1 +  ∫secⁿ⁻²x

3. In the first term we take the quotient of the exponent (n-2) of secx and the denominator (n-1). We then obtain the quotient before the integral

∫secⁿ x = secⁿ⁻²x/n-1 + n-2/n-1  ∫secⁿ⁻²x. This is the formula of the integral of the power of secant. We just learn how to memorize  the formula.

Example. Calculate the integral  ∫sec³x

We apply the formula ∫secⁿ x = secⁿ⁻²x/n-1 + n-2/n-1  ∫secⁿ⁻²x.

We substitute n by 3:

  ∫sec³x =  sec³⁻²x/3-1 + 3-2/3-1  ∫sec³⁻²x.

            =   secx/2 + 1/2 ∫secx
         
            =  1/2secx +ln ❘secx + tanx❘ + C

Integral of the power of tangent

The integral of the power of tangent is given by the formula: ∫ tanⁿxdx = tanⁿ⁻ⁱx/n-1- ∫tanⁿ⁻²xdx

 Example. Solve ∫tan⁵xdx

∫tan⁵xdx = tan⁴x/4 - ∫ tan³xdx

 Let's calculate ∫ tan³xdx:

∫ tan³xdx = tan²x/2 - ∫ tanxdx
                                 
              =    tan²x/2 - ln|secx| + C
              =    tan²x/2 + ln|cosx| + C

Let's substitute  tan³xdx:

∫tan⁵xdx = tan⁴x/4 - tan²x/2 - ln|cosx| + C

Practice

Solve:

1) ∫ sec⁵xdx
2)  ∫tan³xdx

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Integrating the product of the power of sine by the power of cosine

Integrating a function of the product of sine by cosine: f(x) = sin^mxcos^nx ( read sine  exponent m x by cos exponent n x)

Method

1. If m is odd let u = cosx
2. If n is odd let u = sinx
3. If m and n are even use identities to reduce the power of sine and cosine

Example I

Solve ∫sin³xcos⁴xdx

 Since m is odd let u = cox then du =- sinxdx  dx = -du/sinx

Let's substitute cox and dx in the integral

∫sin³xcos⁴xdx = ∫sin³xu⁴.-du/sinx

Let's simplify by sinx:

∫sin³xcos⁴xdx = - ∫sin²xu⁴du

In order to have the integral as a function of u let's express sin²x as an expression of cosx

sin²x = 1-cos²x = 1-u²

Let's substitute sin²x

∫sin³xcos⁴xdx = -∫(1-u²)u⁴du

                     = -∫(u⁴-u⁶)du

                     = -(u⁵/5-u⁷/7 +C

                     = -u⁵/5-u⁷/7 +C

                     =  -cos⁵x/5-cos⁷x/7 +C ( by substituting u by cosx)

Example II

Solve ∫sin²xcos²xdx.

We have m and n even. We use identities to reduce power.

 sin²xcos²xdx. = (1 - cos2x)/2.(1 + cos2x)/2 = 1 - cos²2x/4 = sin²2x/4

Let's reduce the power of sin²2x

 sin²2x = (1 - cos4x)/2

 sin²2x/4 = (1 - cos4x)/8 = 1/8 - 1/8cos4x

 ∫sin²xcos²xdx = ∫(1/8 - 1/8cos4x)dx

                      =1/8 ∫dx - 1/8 ∫cos4xdx

                      = 1/8x - 1/8sin4x + C

Practice

Solve

1) ∫sin⁴xcos³x
2) ∫sin⁴xcos⁴x
.

Integration of trigonometric functions involving the power of sine or the power of cosine

Objective:

To compute the integral of a function involving the power of sine or the power of cosine

Method

In order to solve the integral of the power of sine or the power of cosine we have to reduce their power. We use the following formulas:

sin²x  = 1/2 (1-cos2x )
cos²x = 1/2 ( 1 + cos2x )

Example I

Solve   ∫ sin²xdx

Let's substitute sin²x by  1/2 ( 1-cos2x )

∫sin²xdx = ∫ [1/2 ( 1-cos2x )] dx
         
            = 1/2 ∫ ( 1-cos2x ) dx
         
            = 1/2 [ ∫ dx - ∫ cos2x dx ]

 Let's solve ∫ cos2x dx

Let's use the substitution method by writing u = 2x

Then du = 2xdx dx = du/2

∫ cos2x dx =  ∫ cosu.du/2
                = 1/2 ∫ cosudu

                = 1/2 sinu + C

                =  1/2 sin2x + C

Let's solve the expression between brackets:

 ∫sin²xdx = 1/2 ( x - 1/2 sin2x ) + C

             =  1/2 x - 1/4 sin2x + C
  
Example II   Evaluate ∫ cos⁴x

The technique used to solve  the integral of the power of cosine and the power of sine consists in reducing the power of these functions.

Let's first reduce the power of cos⁴x
cos⁴x = ( cos²x )²

Let's reduce the power of cos²x by using the formula cos²x = 1/2 ( 1 + cos2x )

cos⁴x = [ 1/2 ( 1 + cos2x ) ]²

         = 1/4 ( 1 + 2cos2x + cos²2x )

         = 1/4 + 1/2cos2x + 1/4cos²2x

Let's reduce the power of cos²2x:

cos²x = 1/2 ( 1 + cos2x ). By analogy  cos²2x = 1/2 ( 1 + cos4x ). Then 1/4cos²2x = 1/8 + 1/8cos4x

Therefore cos⁴x = 1/4 + 1/2cos2x + 1/8 + 1/8cos4x

                 cos⁴x = 3/8 + 1/2cos2x + 1/8cos4x
               
∫cos⁴xdx = ∫( 3/8 + 1/2cos2x + 1/8cos4x )dx
             
              = 3/8∫dx + 1/2∫ cos2xdx + 1/8∫ cos4xdx
           
In the example above we found ∫cos2xdx = 1/2 sin2x + C. By analogy ∫cos4xdx = 1/4 sin4x + C

  ∫cos⁴xdx = 3/8x + 1/2 ( 1/2 sin2x ) + 1/8 ( 1/4 sin4x ) + C 
             
                 = 3/8x + 1/4 sin2x  + 1/32 sin4x  + C

Practice 

Evaluate:

1) ∫cos²xdx
2) ∫sin⁴xdx

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