Fundamental Integrals involving algebraic trigonometric

Integration Formulae

Integration formula is applied for finding the integration of functions, algebraic expression, logarithmic and exponential functions, trigonometric ratios, and inverse trigonometric ratios. The integration of the functions will result in the original function for which derivatives were obtained. This sort of integration formula is used to determine the anti-derivative of the function. 

The integration formulas have been defined as the following six sets of formulae. In general terms, integration refers to the method of combining the part to find the complete function. The formula encompasses the basic integration formula, product of function, integration of trigonometric functions, inverse trigonometric functions, and some advanced set of integration formulas. Therefore, integration is known as the inverse operation of differentiation, and the basic integration formula is f’(x).dx = f(x) + C

Basic Integration Formulas

By applying the fundamental theorems of integral, we will find some generalized results called the integration formula in indefinite integration.

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• ∫ x^n.dx = x(n + 1)/(n + 1)+ C
• ∫ e^x.dx = e^x + C
• ∫ e^x[f(x) + f'(x)].dx = e^x.f(x) + C
• ∫ 1.dx = x + C
• ∫1/x.dx = log|x| + C
• ∫ a^x.dx = a^x /log a+ C

Integration of Trigonometric functions

As we have studied that integration refers to finding the integration, there are some essential results formulas that need to be remembered for making quick calculations. When we find trigonometric functions, we can simplify them to rewrite them as functions that are intangible. Some of the essential results of trigonometric and inverse trigonometric functions are listed below. 

• ∫ cosx.dx = sinx + C
• ∫ sinx.dx = -cosx + C
• secx.dx = log|secx + tanx| + C 
• ∫ cosecx.dx = log|cosecx – cotx| + C
• ∫ sec^2x.dx = tanx + C
• ∫ cosecx.cotx.dx = -cosecx + C 
• ∫ tanx.dx =log|secx| + C 
• ∫ cosec^2x.dx = -cotx + C 
• ∫ secx.tanx.dx = sec x + C 
• ∫ cotx.dx = log|sinx| + C 

Integration Formulas of Inverse Trigonometric Functions:

• ∫1/√(1 – x^2).dx = sin-1x + C
• ∫ /1(1 – x^2).dx = -cos-1x + C
• ∫ 1/(1 +x^2 ).dx = -cot-1x + C
• ∫ 1/x√(x^2 – 1).dx = sec-1x + C
• ∫ 1/x√(x^2 – 1).dx = -cosec-1 x + C

Advanced Integration Formulas

• ∫1/(x^2 – a^2).dx = 1/2a.log|(x – a)(x + a| + C 
• ∫ 1/(a^2 – x^2).dx =1/2a.log|(a + x)(a – x)| + C
• ∫√(a^2 – x^2).dx = 1/2.x.√(a^2 – x^2).dx + a^2/2.sin-1 x/a + C
• ∫1/√(x^2 + a^2 ).dx = log|x + √(x^2 + a^2)| + C 
• ∫ √(x^2 + a^2 ).dx =1/2.x.√(x^2 + a^2) + a^2/2. log|x + √(x^2 + a^2 )| + C
• 1/√(x^2 – a^2)dx = log|x +√(x^2 – a^2)| + C
• ∫ √(x^2 – a^2).dx =1/2.x.√(x^2 – a^2)-a^2/2 log|x + √(x^2 – a^2)| + C 
• ∫1/(x^2 + a^2).dx = 1/a.tan^-1x/a + C
• ∫1/√(a^2 – x^2).dx = sin^-1 x/a + C

Different Integration Formulas

There are three types of integration methods available, and each is implemented with its unique purpose for finding the integrations and known as the standard integrals formulas. Each has its properties of integration.

Integration by parts formula:

If you will find the given function as the product of two functions, you have to apply integration by part formula to find the integral formula. The integration formula by using integration by parts can be written as

∫ f(x).g(x) = f(x).∫g(x).dx -∫ (∫g(x).dx.f'(x)).dx + c

Integration by substitution formula:

In case if a given function is a function of another function, you have to apply an integration formula for substitution. The integration formula using the integration of functions by substitution can be written as I = ∫ f(x) dx = ∫ f(g(t)) g'(t) dt

Integration by partial fractions formula:

When we need to determine the integral of an improper function i.e. (x)/Q(x), wherein the degree of P(x) < that of Q(x), it is required to use integration by partial fractions. If the fraction is split using the partial fraction in the form P(x)/Q(x) = T(x) + P (x)/ Q(x), where T(x) is a polynomial in x and P (x)/ Q(x) is a proper rational function.

Examples of Integral formula 

Example 1: Find the value of ∫ (7x+ 23)/ (x+4)^2 .dx
Solution:

(9x+ 25)/ (x+3)^2 is a rational function. 

Using the partial fraction decomposition, we have (9x+ 25)/(x+3)^2 = A/(x+3) + B/(x+3)^2 

While considering the LCD, we get

(7x+ 23)/(x+4)^2 = [A(x+4) +B]/(x+4)^2 

Equating the numerator, we get

7x+ 23 = A(x+4) +B

Finding the value of B when x = -4, we get B = -5

Now finding value of A when x = 0, we get A = 7

Therefore, the partial fraction is decomposed as 7/(x+4) -5/(x+4)^2

Now we have to find the integral of 7/(x+4) -5/(x+4)^2 . 

∫ [7/(x+4)] dx – ∫ -5(x+4)^2 .dx = 7 ln(x+4) – 5 /(x+4) +C

Answer: ∫(7x+ 23)/ (x+4)^2 .dx = 7 ln(x+4) – 5 /(x+4) +C

Example 2. Find the value of ∫ Cosec X (Cosecx – Cotx).dx using the integration formula.

Solution: 

∫Cosec X (Cosecx – Cotx).dx

= ∫Cosec2x.dx – ∫ Cotx.Cosecx.dx

By implementing the basic trigonometric integration of function formulas,

= -Cotx – (-Cosecx)

= -Cotx + Cosecx

= Cosecx – Cotx + C 

Answer: ∫Cosecx (Cosecx – Cotx).dx = Cosecx – Cotx + C

Example 3: Use the integration formula and find the value of ∫(3 + 5Cosx)/Sin2x.dx
Solution:

∫(3 + 5Cosx)/Sin2x.dx

= ∫3/Sin2x. dx +∫5Cosx/Sin2x.dx

= ∫3Cosec2x.dx + ∫5Cotx.Cosecx.dx

=3∫Cosec2x.dx + 5∫Cotx.Cosecx.dx

By implementing the trigonometric formula, we will get the result as 

=3(-Cotx) + 4(-Cosecx)

= -3Cotx – 5Cosecx + C

Example 4. Integrate sin4x cos3x with respect to x

Solution: 

Sinx cosy=1/2[sin(x+y) +sin (x−y)] sin⁡ x cos⁡y=1/2[sin⁡(x+y) +sin⁡ (x−y)]

Form this identities sin4x cos3x=12(sin7x+sinx) sin⁡4xcos⁡3x=1/2(sin⁡7x+sin⁡x)

Therefore,

∫(sin4xcos3x) dx=∫1/2(sin7x+sinx) dx∫ (sin⁡4xcos⁡3x) dx=∫1/2(sin⁡7x+sin⁡x) dx

From the above equation we have:

∫1/2(sin7x+sinx)dx=1/2∫ (sin7x+sinx)dx∫1/2(sin⁡7x+sin⁡x) dx=1/2∫(sin⁡7x+sin⁡x) dx (ii)

As we know that in accordance to properties of integration, the integral of adding two function is equivalent to addition of integral of the given function it can be therefore been written as 

∫[f(x) +g(x)] dx=∫f(x).dx+∫g(x).dx∫[f(x) +g(x)] dx=∫f(x).dx+∫g(x).dx

So, now equation 2 can be rewritten as:

1/2∫ (sin7x) +12∫ (sinx) dx1/2∫ (sin⁡7x) +1/2∫ (sin⁡x) dx

= −cos7x/ 14+−cosx/ 2 +C

Definite Integration Formula

Definite integration is that integration that has a pre-existing value of limits which means it makes the final value of the integral definite 

= G (b) – G (a)

Indefinite Integration Formula

Indefinite integrations are the integrations that don’t include any pre-existing value of limits, and due to such, it makes the final value of integral indefinite. Here, it refers to the integration of functions constant. ∫ g'(x) = g(x) + C

Real-Life example of Definite Integration Formula

Let us calculate the distance covered by an object using the integration of functions formulas.

As we know, distance refers to the definite integral of velocity.

Given:  Object velocity = v (t) = -t2 + 5t.

We have to calculate the displacement and total distance covered on (1, 3), which means that the initial and last positions of the object are 1 and 3.

Now, we use the integration formula using the limits 

 = G (b) – G (a)

= v (4) – v (2)

v (t) = -t^2 + 5t

= -t^3/3 + 5t^2 /2

= [(-27/3-1/3) + (45/2 – 5/2)]

= -28/3 +40/3

 =12/3

=4 

Therefore, value of Displacement =4

The distance between the two points refers to the absolute value of its displacement. The complete distance covered between the points (1, 3) and in reverse will be 4+4=8

Therefore, it is calculated that the distance covered by the object is 4 + 4 = 8

Conclusion 

We have explained the interpretation and some of the examples related to Fundamental Standard Integrals involving algebraic trigonometry to provide you with detailed information regarding the implementation of the formula. If you have any doubts or suggestions, let us know by commenting in the box available.