Antiderivative –
 Defination :A function ∅(x) is called the antiderivative (or an integral) of a function f(x) of ∅(x)’ = f(x).
 Example : x^{4}/4 is an antiderivative of x^{3} because (x^{4}/4)’ = x^{3}.
In general, if ∅(x) is antiderivative of a function f(x) and C is a constant.Then, {∅(x)+C}' = ∅(x) = f(x).
Indefinite Integrals –
 Defination :Let f(x) be a function. Then the family of all ist antiderivatives is called the indefinite integral of a function f(x) and it is denoted by ∫f(x)dx.
The symbol ∫f(x)dx is read as the indefinite integral of f(x) with respect to x.
Thus ∫f(x)dx= ∅(x) + C.
Thus, the process of finding the indefinite integral of a function is called integration of the function.
Fundamental Integration Formulas –
 ∫x^{n}dx = (x^{n+1}/(n+1))+C
 ∫(1/x)dx = (log_{e}x)+C
 ∫e^{x}dx = (e^{x})+C
 ∫a^{x}dx = ((e^{x})/(log_{e}a))+C
 ∫sin(x)dx = cos(x)+C
 ∫cos(x)dx = sin(x)+C
 ∫sec^{2}(x)dx = tan(x)+C
 ∫cosec^{2}(x)dx = cot(x)+C
 ∫sec(x)tan(x)dx = sec(x)+C
 ∫cosec(x)cot(x)dx = cosec(x)+C
 ∫cot(x)dx = logsin(x)+C
 ∫tan(x)dx = logsec(x)+C
 ∫sec(x)dx = logsec(x)+tan(x)+C
 ∫cosec(x)dx = logcosec(x)cot(x)+C
Examples –
 Example 1.Evaluate ∫x^{4}dx.
 Solution –
Using the formula, ∫x^{n}dx = (x^{n+1}/(n+1))+C ∫x^{4}dx = (x^{4+1}/(4+1))+C = (x^{5}/(5))+C
 Example 2.Evaluate ∫2/(1+cos2x)dx.
 Solution –
As we know that 1+cos2x = 2cos^{2}x ∫2/(1+cos2x)dx = ∫(2/(2cos^{2}x))dx = ∫sec^{2}x = tan(x)+C
 Example 3.Evaluate ∫((x^{3}x^{2}+x1)/(x1))dx.
 Solution –
∫((x^{3}x^{2}+x1)/(x1))dx = ∫((x^{2}(x1)+(x1))/(x1))dx = ∫(((x^{2}+1)(x1))/(x1))dx = ∫(x^{2}+1)dx = (x^{3}/3)+x+C Using, ∫x^{n}dx = (x^{n+1}/(n+1))+C
 Integration by Substitution :
 Definition –The method of evaluating the integral by reducing it to standard form by proper substitution is called integration by substitution.
If f(x) is a continuously differentiable function, then to evaluate the integral of the form∫g(f(x))f(x)dx
we substitute f(x)=t and f(x)’dx=dt.
This reduces the integral to the form∫g(t)dt
 Examples :
 Example 1.Evaluate the ∫e^{2x3}dx
 Solution
Let 2x3=t => dx=dt/2 ∫e^{2x3}dx = (∫e^{t}dx)/2 = (∫e^{t})/2 = ((e^{2x3})/2)+C
 Example 2.Evaluate the ∫sin(ax+b)cos(ax+b)dx
 Solution
Let ax+b=t => dx=dt/a; ∫sin(ax+b)cos(ax+b)dx = (∫sin(t)cos(t)dt)/a = (∫sin(2t)dt)/2a = (cos(2t))/4a = (cos(2ax+2b)/4a)+C
 Definition –The method of evaluating the integral by reducing it to standard form by proper substitution is called integration by substitution.
 Integration by Parts :
 Theorem :If u and v are two functions of x, then
∫(uv)dx = u(∫vdx)∫(u'∫vdx)dx
where u is a first function of x and v is the second function of x
 Choosing first function :
We can choose first function as the function which comes first in the word ILATE where I – stands for inverse trigonometric functions.
 L – stands for logarithmic functions.
 I – stands for algebraic functions.
 I – stands for trigonometric functions.
 I – stands for exponential functions.
 Examples :
 Example 1.Evaluate the ∫xsin(3x)dx
 Solution
Taking I= x and II = sin(3x) ∫xsin(3x)dx = x(∫sin(3x)dx)∫((x)'∫sin(3x)dx)dx = x(cos(3x)/(3))∫(cos(3x)/(3))dx = (xcos(3x)/(3))+(cos(3x)/9)+C
 Example 2.Evaluate the ∫xsec^{2}xdx
 Solution
Taking I= x and II = sec^{2}x ∫xsin(3x)dx = x(∫sec^{2}xdx)∫((x)'∫sec^{2}xdx)dx = (xtan(x))∫(1*tan(x))dx = xtan(x)+logcos(x)+C
 Theorem :If u and v are two functions of x, then
 Integration by Partial Fractions :
 Partial Fractions :
If f(x) and g(x) are two polynomial functions, then f(x)/g(x) defines a rational function of x.
If degree of f(x) < degree of g(x), then f(x)/g(x) is a proper rational function of x.
If degree of f(x) > degree of g(x), then f(x)/g(x) is an improper rational function of x.
If f(x)/g(x) is an improper rational function, we divide f(x) by g(x) so that the rational function can be represented as ∅(x) + (h(x)/g(x)).Now h(x)/g(x) is an proper rational function.
Any proper rational function can be expressed as the sum of rational functions, each having a simple factor of g(x).Each such fraction is called partial fraction .  Cases in Partial Fractions :
 Case 1.
When g(x) = (xa_{1})(xa_{2})(xa_{3})….(xa_{n}), then we assume thatf(x)/g(x) = (A_{1}/(xa_{1}))+(A_{2}/(xa_{2}))+(A_{3}/(xa_{3}))+....(A_{n}/(xa_{n}))

Case 2.
When g(x) = (xa)^{k}(xa_{1})(xa_{2})(xa_{3})
….(xa_{r}),
then we assume thatf(x)/g(x) = (A_{1}/(xa)^{1})+(A_{2}/(xa)^{2})+(A_{3}/(xa)^{3}) +....(A_{k}/(xa)^{k})+(B_{1}/(xa_{1}))+(B_{2}/(xa_{2}))+(B_{3}/(xa_{3})) +....(B_{r}/(xa_{r}))
 Case 1.
 Examples :
 Example 1.∫(x1)/((x+1)(x2))dx
 Solution
Let (x1)/((x+1)(x2))= (A/(x+1))+(B/(x2)) => x1 = A(x2)+B(x+1)
Putting x2 = 0, we get
B = 1/3
Putting x+1 = 0, we get
A = 2/3
Substituting the values of A and B, we get
(x1)/((x+1)(x2))= ((2/3)/(x+1))+((1/3)/(x2)) ∫((2/3)/(x+1))+((1/3)/(x2))dx = ((2/3)∫(x+1)dx)+((1/3)∫(x2)dx) = ((2/3)logx+1)+((1/3)logx2)+C
 Example 2.∫(cos(x))/((2+sin(x))(3+4sin(x)))dx
 Solution
Let I = ∫(cos(x))/((2+sin(x))(3+4sin(x)))dx
Putting sin(x) = t and cos(x)dx = dt, we get
I = ∫dt/((2+t)(3+4t)) Let 1/((2+t)(3+4t))= (A/(2+t))+(B/(3+4t)) => 1 = A(3+4t)+B(2+t)
Putting 3+4t = 0, we get
B = 4/5
Putting 2+t = 0, we get
A = 1/5
Substituting the values of A and B, we get
1/((2+t)(3+4t)) = ((1/5)/(2+t))+((4/5)/(3+4t)) I = (((1/5)/(2+t))dt)+(((4/5)/(3+4t))dt) = ((1/5)log2+t)+((1/5)log3+4t)+C = ((1/5)log2+sin(x))+((1/5)log3+4sin(x))+C
 Partial Fractions :
Methods of Integration –
leave a comment
0 Comments