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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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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Integration by substitution

Sometimes it is difficult to integrate a function using the simple formulas of integrals. We have to introduce a second variable to make the integration easier.

Example 1

Calculate ∫(x+1)⁵dx

If it was the integral of x⁵ it would be easy to calculate. We can introduce another variable to make the integral easy to calculate. If we make x + 1 equal to u we will just have to calculate the integral of u. So let's write u = x+1

Then our integral becomes ∫u⁵dx. We calculate the integral as if it was the integral of x⁵.

 ∫u⁵dx = u⁵⁺¹/5+1 + C = 1/6 u⁶ + C

Let's substitute u we get  ∫u⁵dx = 1/6 (x+1)⁶ + C

Example 2

Calculate ∫√4x+3dx

Let's write the expression as a power:

 ∫√4x+3dx = ∫(4x+3)¹/² dx

Let's write u = 4x+3 du = 4dx then dx = du/4. Let's substitute 4x+3 and dx

∫√4x+3dx = ∫u¹/².du/4 = 1/4∫u¹/²du = 1/4(u¹/²+¹/¹/²+¹) + C = 1/4(u³/²/3/2) + C = (1/4)(2/3)u³/² + C = 1/6u³/² + C = 1/6√(4x+3)³ + C

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Using derivative rules to find the integrals of some basic functions

Integral of the power function

Let f be f(x) = xⁿ its antiderivative is F(x) = 1/n+1x^n+1+C (Read F(x) equals  1 over n+1 times x exponent n+1 plus the constant of integration C) . You can rewrite the expression by utilizing a horizontal bar for the fraction 1 over n+1. You can also rewrite the expression x^n+1 by putting n+1 besides x and above. The rewriting will make the expression clearer.

 The antiderivative of f(x) being the indefinite integral, we can write ∫f(x)dx =  1/n+1.x^n+1+C (Read sum f(x) dx equals 1 over n+1 times  x exponent n+1 plus the constant of integration C).

In order to prove this, we have to demonstrate that the derivative of  1/n+1.x^n+1+C is xⁿ.

Let's calculate the derivative of F(x) = 1/n+1.x^n+1+C 
F'(x) = n+1/n+1x^n+1-1 (Read n+1 over n+1 times x exponent n+1-1. By simplifying and reducing F'(x) = xⁿ

Integral of the inverse function

The integral of the inverse function f(x) = 1/x is ∫1/xdx = lnx +C.

Proof

The derivative of F(x) = lnx is F'(x) = 1/x

Integral of the exponential function

The integral of the exponential function f(x) = e^x is ∫e^xdx = e^x + C. The term e^x reads e exponential x.

Proof

The derivative of F(x) = e^x is F'(x) = e^x.

Rules of integrable functions

If f and g are integrable functions and C is a constant, then

∫[f(x)+g(x)] = ∫f(x)dx+∫g(x)dx

∫[f(x)-g(x)] = ∫f(x)dx-∫g(x)dx

∫kf(x)dx = k∫f(x)dx

Example

Solution

Using the basic rules of integration we have:

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The formula of the antiderivative of a function simply explained

Let's consider the function f(x) = x² +3 its derivative is f '(x) = 2x. The function f '(x) = 2x is the derivative of the function f(x) = x² + 3. The function f(x) = x² +3 is the antiderivative of the function
f '(x) = 2x. The antiderivative is the function from which the derivative was calculated.

If we consider the function f(x) = 2x it can be considered as the derivative of another function F. This function which is F(x) = x² + 3 is the antiderivative of the function f. Actually, there are several functions whose derivative is f(x) = 2x. Some examples of these functions are F(x) = x²+1, F(x) = x²-3, F(x) = x² + 6. There is a bunch of functions whose derivative is f '(x) = 2x. We can't list them all. For this reason we represent all them by the function F(x) = x²+C.

Notation and notion of indefinite integral

The process of finding the antiderivative of a function is called antidifferentiation or integration. If F is the antiderivative of a function f we write ∫f(x)dx = F(x) +C. The left side of this equation is called integral of f(x). The function f(x) is called the integrand. The letter C is the constant of integration. The symbol dx means that the function is integrated with respect to x.

Formula to calculate the antiderivative of the function power

Let's retake the function f(x) = 2x. Its antiderivative is F(x) = x² + C or F(x) = 2/2x² + C.  Let's consider another function g(x) = 3x. Its antiderivative can be G(x) = 3/2x²+C. For the function h(x) = x² its antiderivative is H(x) = 1/3x³+C. For the function i(x) =5 x³ its antiderivative is I(x) = 5/4x⁴ + C

By observing each of these functions and their antiderivative, two points draw our attention.

1) The exponent of x in the antiderivative function is one degree more than the exponent of x in the given function.
2) The coefficient of x in the antiderivative function is obtained by dividing the coefficient of x in the given function by the exponent of x in the antiderivative function.

Let's come back to the examples above. In the function antiderivative F(x) the exponent of x is 2 and the exponent of x in the function f is 1. The coefficient of x 2/2 in the function F(x) is equal to the coefficient 2 of x in the function f(x) divided by the exponent 2 of x in the antiderivative function.

 In the function antiderivative G(x) the exponent of x is 2 and the exponent of x in the function g(x) is 1. The coefficient 3/2 of x in the function G(x) is equal to the coefficient 3 of x in the function f(x)   divided by the exponent 2 of x in the function G(x). The same is true for the functions H(x) and I(x).

From these observations, we can deduct a rule. In order to find the antiderivative of f((x) = xⁿ add 1 to the exponent and divide the coefficient by the new exponent,

The formula that generalizes this procedure is as follows:

If f(x) = kxⁿ its derivative is given by F(x) = k/n+1xⁿ⁺¹ + C.

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Derivative of logarithmic functions

In this post I'll show some techniques to remember the formulas for logarithmic  functions. I'll do some examples and leave some exercises to practice.

Derivative of logarithmic functions 

Derivative of log_bx

d/dx (log_bx) = 1/xlnb
To remember this formula let's apply the following technique:
1) Multiply the number of which we calculate the logarithm by the natural logarithm of the base. The number here is x and the base is b. Therefore we have xlnb
2) Take the inverse of this product. The inverse of the product is 1/xlnx

Derivative of lnx

d/dx(lnx) = 1/x
The derivative of the logarithm of any number is equal to the inverse of this number.

Derivative of log_bu

Since log_bu is a composite function its derivative is given by d/dx(log_bu) = d/du(log_bu).du/dx

d/dx(log_bu) = 1/ulnnb.du/dx

Rule: The derivative of the logarithm of a composite function is equal to its derivative with respect to the new variable (u) multiplied by the derivative of the new variable (u) with respect to x.

Derivative of lnu 

Since u is a composite function we have d/dx(lnu) = d/du(lnu).du/dx
                                                                              = i/u.du/dx

Rule: The derivative of the natural logarithm of a composite function u is equal to the inverse of the function multiplied by its derivative with respect to x

Example 1, Calculate the derivative of y = x³log₅2x

The derivative of y is y" = (x³log₅2x)'

Let's apply the product rule:
Y' = (x³)'(log₅2x) + x³(log₅2x)'
The derivative of x³ is obvious. Let's calculate the derivative  of log₅2x
Let's write u = 2x we have (log₅u)' = d/du(log₅u),du/dx
                                                     = i/uln5.u'
                                                    = 1/2x.ln5.(2x)'
                                                   = 1/2x.ln5.2
                                                   = 1/xln5
let's go back to the derivative of y we have:
y' = 3x²log₅2x + x³.1/xlnx
   = 3x²log₅2x+x²/lnx

Example 2.  Calculate the derivative of y = ln(2x²-4x+3)

Let's write u = 2x²-4x+3
We have y = lnu
Then dy/dx = d/dx(lnu)
Since lnu is a composite function then dy/dx = d/du(lnu).du/dx
                                                                      = 1/u(4x-4)
Substitute u: dy/dx = (1/2x³-4x+3).(4x-4)
dy/dx = 4x-4/2x²-4x+3
           = 4(x-1)/2x³-4x+3

Practice

Calculate the derivative of the following functions;
1.log₅(2x+5)
2. 5/log(x+4)
3. ln(sinx)

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Derivative computations

The formula lim f(x)-f(x+h)/h when  x→h that defines the derivative of a function f implies tedious calculations to calculate the derivative of some types of functions and combinations of functions..

Therefore some formulas have been established to determine the derivatives of a combination of functions and some specific types of functions.

The formulas for the constant function and the power functions are called respectively constant rule and power rule. The formulas for the sum, product and quotient of functions are called respectively addition rule, product rule and quotient rule. The derivative of a composition of 2 functions f and g is called the chain rule. It is an extension of the power rule The trigonometric, logarithmic and exponential functions have their specific formula.

The derivative of an implicit function is called implicit differentiation.

It is essential to memorize the formulas. Otherwise, it would be difficult to calculate the derivatives of these particular functions. Today we are going to limiting ourselves to the learning of the basic formulas: constant, power, sum, product and quotient rule.

Derivative of a constant

The derivative of the function constant is 0. If f(x) = c the derivative of f(x) is 0. We write:  f′(x) = 0.

The Power rule

The derivative of the function power defined by f(x) =  xⁿ is equal to n multiplied by x to the power of n-1. The formula is .f’(x) = nx^n-1

Derivative of the product of a constant by a function

The derivative of the product of a function by a constant is equal to the product of the constant by the derivative of the function power.

If  f(x) = axⁿ its derivative is f’(x) = ax^n-1


Derivative of the function f(x) = x

The derivative of the function f(x) = x can be calculated using the formula for the derivative of the function power. In order to use this formula we have to write f(x) = x as the function power. We write f(x) = x as f(x) = x¹ 
By applying the formula for the function power we obtain f’(x) = x^1-1 
 f’(x) = x⁰ ⇒ f’(x) = 1 

Derivative of a sum of functions

If f. g. h,;;; are differentiable for any value of x of their domain the derivative of the sum of these functions is f’+g’+h’+ ....

Derivative of the product of 2 functions

If f and g are differentiable for any value x of their domain the derivative of the product f.g is fg’+gf’

Derivative of the quotient of 2 functions

If f and g are differentiable for any value of their domain the derivative of the quotient f/g is (f∕g)’ = f’g-gf’∕g²

These formulas have to be demonstrated and the learners have to do some exercises to apply them. If anyone is interested in learning more subscribe to these courses via this link Free Introductory Calculus Course and Complete Calculus Course

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