Antiderivative –
· Defination :A function ∅(x) is called the antiderivative (or an integral) of a function f(x) of ∅(x)’ = f(x).
· Example : x4/4 is an antiderivative of x3 because (x4/4)’ = x3.
· 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 –
1. ∫xndx = (xn+1/(n+1))+C
2. ∫(1/x)dx = (loge|x|)+C
3. ∫exdx = (ex)+C
4. ∫axdx = ((ex)/(logea))+C
5. ∫sin(x)dx = -cos(x)+C
6. ∫cos(x)dx = sin(x)+C
7. ∫sec2(x)dx = tan(x)+C
8. ∫cosec2(x)dx = -cot(x)+C
9. ∫sec(x)tan(x)dx = sec(x)+C
10. ∫cosec(x)cot(x)dx = -cosec(x)+C
11. ∫cot(x)dx = log|sin(x)|+C
12. ∫tan(x)dx = log|sec(x)|+C
13. ∫sec(x)dx = log|sec(x)+tan(x)|+C
14. ∫cosec(x)dx = log|cosec(x)-cot(x)|+C
Examples –
· Example 1.Evaluate ∫x4dx.
· Solution –
· Using the formula, ∫xndx = (xn+1/(n+1))+C
· ∫x4dx = (x4+1/(4+1))+C
· = (x5/(5))+C
· Example 2.Evaluate ∫2/(1+cos2x)dx.
· Solution –
· As we know that 1+cos2x = 2cos2x
· ∫2/(1+cos2x)dx = ∫(2/(2cos2x))dx
· = ∫sec2x
· = tan(x)+C
· Example 3.Evaluate ∫((x3-x2+x-1)/(x-1))dx.
· Solution –
· ∫((x3-x2+x-1)/(x-1))dx = ∫((x2(x-1)+(x-1))/(x-1))dx
· = ∫(((x2+1)(x-1))/(x-1))dx
· = ∫(x2+1)dx
· = (x3/3)+x+C
· Using, ∫xndx = (xn+1/(n+1))+C
Methods of Integration –
1. 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 ∫e2x-3dx
· Solution
· Let 2x-3=t => dx=dt/2
· ∫e2x-3dx = (∫etdx)/2
· = (∫et)/2
· = ((e2x-3)/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
2. 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 ∫xsec2xdx
· Solution
· Taking I= x and II = sec2x
· ∫xsin(3x)dx = x(∫sec2xdx)-∫((x)'∫sec2xdx)dx
· = (xtan(x))-∫(1*tan(x))dx
· = xtan(x)+log|cos(x)|+C
3. 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) = (x-a1)(x-a2)(x-a3)….(x-an), then we
assume that
f(x)/g(x) = (A1/(x-a1))+(A2/(x-a2))+(A3/(x-a3))+....(An/(x-an))
·
Case 2.
When g(x) = (x-a)k(x-a1)(x-a2)(x-a3)
….(x-ar),
then we assume that
· f(x)/g(x) = (A1/(x-a)1)+(A2/(x-a)2)+(A3/(x-a)3)
· +....(Ak/(x-a)k)+(B1/(x-a1))+(B2/(x-a2))+(B3/(x-a3))
+....(Br/(x-ar))
· Examples :
· Example 1.∫(x-1)/((x+1)(x-2))dx
· Solution
· Let (x-1)/((x+1)(x-2))= (A/(x+1))+(B/(x-2))
· => x-1 = A(x-2)+B(x+1)
Putting x-2 = 0, we get
B = 1/3
Putting x+1 = 0, we get
A = 2/3
Substituting the values of A and B, we get
(x-1)/((x+1)(x-2))= ((2/3)/(x+1))+((1/3)/(x-2))
∫((2/3)/(x+1))+((1/3)/(x-2))dx
= ((2/3)∫(x+1)dx)+((1/3)∫(x-2)dx)
= ((2/3)log|x+1|)+((1/3)log|x-2|)+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)log|2+t|)+((1/5)log|3+4t|)+C
= ((-1/5)log|2+sin(x)|)+((1/5)log|3+4sin(x)|)+C