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Integration by parts and integration by partial fractions have been discussed in the later sections. In this section, our main emphasis would be on integration of functions using the substitution method.
While solving integrals by the substitution method, some integrals can be computed using the direct substitutions while some need indirect substitutions. There are certain types of functions in which some standard substitutions are to be applied. Hence, in the substitution method, there are three major types of substitutions:
We shall discuss all these methods in detail one by one. First of all we begin with the direct substitution method:
If the integral is of the form ∫ f(g(x)) g'(x) dx, then put g(x) = t, provided ∫ f(t) exists.
The given integral is ∫ f'(x)/f(x) dx
Put f(x) = t ⇒ f'(x) dx = dt
⇒ ∫ dt/t = ln |t| + c = ln |f (x)| + c.
The given integral is ∫ f'(x)/√f(x) dx
Put f (x) = t
Then the given integral becomes
= ∫ dt/√t = 2 √t + c = 2 √f(x) + c.
We now discuss certain examples based on direct substitution:
In certain standard integrands, some standard substitutions prove helpful:
For terms of the form (x2 + a2) or √(x2 + a2), put x = a tan θ or a cot θ
For terms of the form (x2 - a2) or √(x2 – a2), put x = a sec θ or a cosec θ
For terms of the form (a2 - x2) or √(x2 + a2), put x = a sin θ or a cos θ
If both √(a+x), √(a–x) are present, then put x = a cos 2θ.
For terms of the type 2x/(a2 - x2), 2x/(x2 + a2), (x2 - a2)/(x2 + a2),the substitution x = a tan θ proves useful.
For the type √(x–a)(b–x), put x = a cos2θ + b sin2θ
For the type (√(x2 + a2) ± x)n or (x ± √(x2 - a2)n, put the expression within the bracket = t.
For the type (n ∈ N, n > 1), put (x + b)/(x + a) = t
For 1/(x + a)n1 (x + b)n2, n1, n2 ∈ N (and > 1), again put (x + a) = t(x + b)
We now discuss some illustrations based on application of these substitutions:
Example 1:
Evaluate ∫ dx/(x + 1)6/5 (x – 3)4/5
Solution:
The given integral is I = ∫ dx/(x + 1)6/5 (x – 3)4/5
= ∫ dx / [(x + 1)2{(x – 3)/(x +1)}4/5]
This gives dt = 4/(x+1)2 dx
Hence, I = ¼ ∫ dt / t4/5 = 5/4 t1/5 + c
= 5/4 {(x-3)/(x+1)}1/5 + c.
If the integrand f(x) can be written as the product of two functions of the form f(x)g(x), where g(x) is a function of the integral of f(x), then put integral of f(x) = t.
If ∫ f(x) dx = g(x), then ∫ f(ax + b) dx = 1/a g(ax + b)
∫ f’(x)/ f(x) dx = log {f(x)} + c
∫ f’(x)/√f(x) dx = 2√f(x) + c
∫ [f(x)]n f’(x) dx = [f(x)]n+1/(n+1) + C, n ≠ 1
In order to evaluate integrals of the form ∫ sinmx dx, ∫ cosmx dx where m ≤ 4, we express sinmx, cosmx in terms of sines and cosines of multiples of x by using trigonometric identities for sin2x, cos2x, sin 3x and cos 3x.
While evaluating integrals of the form sinmx cosnx dx, m, n ∈ N:
First check the exponents of sin x and cos x.
If the exponent of sin x is an odd positive integer put cos x = t
If the exponent of cos x is an odd positive integer put sin x = t
If the exponents of sin x and cos x both are odd positive integers put either sin x or cos x = t
If the exponents of sin x and cos x both are even positive integers then express sinmx cosnx in terms of sines and cosines of multiples of x by using trigonometric results of De’Moivre’s theorem.
Example 1:
Evaluate
Solution:
We first compute the integral of the numerator
The Integral of the numerator = x3/2 /(3/2)
Put x3/2 = t.
We get I = 2/3 ∫dt/√(t2 + a3)
= 2/3 In |x3/2 + √x3 + a3| + c.
Remark:
Sometimes it is useful to write the integral as a sum of two related integrals which can be evaluated by making suitable substitutions.
Examples of such integrals are:
A. Algebraic Twins
B. Trigonometric twins
Method of evaluating these integrals is illustrated by means of the following examples:
Example 1:
Evaluate ∫ 5/(1+x4) dx.
For I2, we write (x + 1/x) = t ⇒ (1 – 1/x2)dx = dt
Combining the two integrals, we get
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Example 2:
Evaluate ∫ √tan x dx.
Solution:
Put tan x = t2 ⇒ sec2x dx = 2t dt
⇒ dx = 2t dt/(1 + t4)
This integral can be computed on similar lines as discussed in the previous example.
Q1. In integrals of the form √(x–a)(b–x), the substitution that proves useful is
(a) x = a cos2θ + b sin2θ
(b) put x = a cos θ and solve
(c put x = sec θ + cosec θ
(d) none of the above
Q2. In case of integrals of the form (a2 - x2) or √(x2 + a2),
(a) use x = a sin θ only
(b) can use x = a cos θ only
(c) may use x = a sin θ or a cos θ
(d) x = a sin θ + a cos θ
Q3. For terms of the type 2x/(a2 - x2), 2x/(x2 + a2), (x2 - a2)/(x2 + a2)
(a) use x = a sin θ only
(b) can use x = a cos θ only
(c) may use x = a sin θ or a cos θ
(d) x = a tan θ
Q4. While evaluating integrals of the form sinmx cosnx dx, where the exponent of cos x is an odd positive integer one must
(a) sin x = t
(b) sin x – cos x = t
(c) sin x + cos x = t
(d) cos x = t
Q5. For 1/(x + a)n1 (x + b)n2, n1, n2 ∈ N (and > 1),
(a) put (x + b) = t(x + a)
(b) put (x + a) = t(x + b)
(c) put (x + a)2 = t(x + b)
Q1. |
Q2. |
Q3. |
Q4. |
Q5. |
(a) |
(c) |
(d) |
(a) |
(b) |
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