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Chapter 10

Method of Integration

In this chapter, we first describe how we can reverse the differentiation. Then we explain the following:
  • integrating the fundamental functions,

  • integrating the trigonometric, exponential and logarithmic functions,

  • method of integration: substitution, integration by parts and partial fraction.

    • 10.1 Antiderivatives

    Consider
    f ' (x) = 3x2.
    The natural question aries, what is the function f (x) ? Namely, what is f in terms of x? We know that differentiation decreases the power by 1, so f must contain x3.
    If f (x) =- x3,     then   f ' (x) = 3x2, or
    If f (x) = x2 + 2,   then   f ' (x) = 3x2, or
    If f (x) = x3 - 1⁄2,   then   f ' (x) = 3x2, etc.

    It means that there are many such functions of the form
    f (x) = x3 + C
    shere C is an arbitrary constant.
    We say that
    x3 is the antiderivative of 3x2.
    Next consider
    f ' (x) = x.
    we think the same way as above, the original function f must contain x2. However,
    d⁄dx = x2 = 2x,
    we see the extra factor 2. If we multiply both sides by 1⁄2, then
    d⁄dx ( 1⁄2 x2) = x.

    If   f (x) = 1⁄2 x2 + C where C is an arbitrary constant, then f ' (x) = x, so 1⁄2 x2 is the antiderivative of x.
    If   F (x) is a function where F ' (x) = f (x, then the antiderivative of f (x) is F (x).
    We call the antiderivative as integral and write
    ∫ x dx = 1⁄2 x2 + C,
    where C is the constan of integration.
    We read this as “the integral of x with respect to x.”
    In general,
    if   F ' (x) = f (x)   then   ∫ f (x) dx = F (x + C
    where dx means that the integration is taking place with respect to the variable x.
    Here f (x) is called the integrand. The variable of integration in an integral plays no essential role. It might be x, or t, or u, or anything else:
    ∫ f (x) dx,     ∫ f (t) dt,   ∫ f (u) du,   etc.

    Example 1.
    Find the antiderivative of
    (a) √ x  
    (b) 3x5
    (c) e3x
    (d) 1⁄x3.
    Solution
    Since   d⁄dx x3⁄2 = 3⁄2 x1⁄2   and d⁄dx ( 2⁄3 x3⁄2 ) = x1⁄2 = √ x  ,
    the antiderivative of √ x   is 2⁄3 x3⁄2 .

    (b) Since   d⁄dx x6 = 6x5   and
    d⁄dx ( 1⁄2 x6 ) = 3x5,
    the antiderivative of 3x5 is 1⁄2 x6.

    (c) Since   d⁄dx e3x = 3e3x   and
    d⁄dx ( 1⁄3 e3x ) = e3x ,
    the antiderivative of e3x is 1⁄3 e3x.

    (d) Since d⁄dx x-2 = -2x-3   and
    d⁄dx (- 1⁄2x2 ) = 1⁄x3 ,
    the antiderivative of 1⁄x3 is - 1⁄2x2 .
    We now describe the integration of fundamental functions. We know that
    d⁄dx xn + 1 = (n + 1)xn .
    The reverse of this process is
    ∫ xn dx = 1⁄n + 1 xn + 1 + C     for n ≠ -1.
    When n = 0, it is seen that
    ∫ x0 dx = ∫ 1 dx = ∫ dx = x + C.
    ∫ dx = x + C.
    When n = -1, let us consider the differentiation of ln x.
    Since d⁄dx ln x = 1⁄x, it follows that
    ∫ 1⁄x dx = ln x + C,   for x > 0.
    For x < 0,
    d⁄dx ln |x| = d⁄dx ln (-x)
    = 1⁄x , by Chain Rule.
    ∫ 1⁄x dx = ln |x| + C,   x ≠ 0

    Rules of Integration
    Suppose f (x) and g (x) are continuous functions and k ∈ ℝ
    1. ∫ k dx = kx + C.
    2. ∫ k f (x) dx = k ∫ f (x) dx.
    3. ∫ [ f (x ± g (x) ] dx = ∫ f (x) dx ± ∫ g (x) dx.


    Example 2.
    Evaluate each of the following integrals.
    (a) ∫ (4x5 + 1) dx
    (b) ∫ (2x6 - 1⁄3 x3 + 3⁄x) dx
    (c) ∫ 5x √ x   dx
    (d) ∫ (x - 1)2⁄ √ x   dx.
    Solution
    (a) ∫ (4x5 + 1) dx
    = 4 ∫ x5 dx + ∫ 1 dx
    = (4⁄6 x6 + C1) + (x + C2);   (C1 and C2 are constants of integration)
    = (2⁄3 x6 + x) + C1 + C2
    = (2⁄3 x6 + x) + C.   (C = C1 + C2 is another constant of integration)
    From now on, we shall write constant of integration only in the answer.

    (b) ∫ (2x6 - 1⁄3 x3 + 3⁄x ) dx
    = 2 ∫ x6 dx - 1⁄3 ∫ x3 dx + 3 ∫ 1⁄x dx
    = 2⁄7 x7 - 1⁄12 x4 + 3 ln |x| + C.

    (c) ∫ 5x √ x   dx
    = 5 ∫ x x1⁄2 dx
    = 5 ∫ x3⁄2 dx
    = 5 (2⁄5 x5⁄2 ) + C.
    = 2x5⁄2 + C.

    (d) ∫ (x - 1)2⁄ √ x   dx = ∫ x2 - 2x + 1⁄2 x1⁄2 dx
    = ∫ (x2⁄ x1⁄2 - 2x⁄x1⁄2 + 1⁄ x1⁄2 ) dx
    = ∫ x3⁄2 dx - 2 ∫ x1⁄2 dx + ∫ x- 1⁄2 dx.
    = 2⁄5 x5⁄2 - 4⁄3 x3⁄2 + 2x1⁄2 + C.

    Integrating Exponential Function

    d⁄dx ax = ax ln a,
    ∫ ax dx = 1⁄ln a ax + C,   where a > 0, a ≠ 1.
    When a = e,
    d⁄dx ex = ex,
    ∫ ex dx = ex + C.

    Example 3.
    Find the following integrals.
    (a) ∫ e-x + 1⁄e-x dx
    (b) ∫ (ex + 2x) dx
    (c) ∫ (ex log 5 + 1⁄x2 ) dx
    (d) ∫ 3x ln 3 dx.
    Solution
    (a) ∫ e-x + 1⁄e-x dx = ∫ (1 + ex) dx
    = ∫ 1 dx + ∫ ex dx = x + ex + C.

    (b) ∫ (ex + 2x) dx = ∫ ex dx + ∫ 2x dx
    = ex + x2 + C.

    (c) ∫ (ex log 5 + 1⁄x2 ) dx = ∫ ex log 5 dx + ∫ x-2 dx
    = ex log 5 - 1⁄x + C.

    (d) ∫ 3x ln 3 dx = 1⁄ln 3 3x Ln 3 + C
    = 3x + C.

    Integrating Trigonometric Functions

    d⁄dx sin x = cos x,     ∫ cos x dx = sin x + C.
    d⁄dx (-cos x) = -(-sin x) = sin x,     ∫ sin x dx = -cos x + C.
    d⁄dx tan x = sec2 x,     ∫ sec2 x dx = tan x + C.

    Example 4.
    Evaluate.
    (a) ∫ (3 cos x - 5 sin x) dx
    (b) ∫ (ex + 2 sin x) dx
    (c) ∫ ex - √ x   ⁄2 dx
    (d) ∫ (1 + tan2 x) dx

    Solution
    (a) ∫ (3 cos x - 5 sin x) dx = 3 ∫ cos x dx - 5 ∫ sin x dx
    = 3 sin x + 5 cos x = C.

    (b) ∫ (ex + 2 sin x) dx = ∫ ex dx + 2 ∫ sin x dx
    = ex - 2 cos x + C.

    (c) ∫ ex - √ x   ⁄2 dx = 1⁄2 ∫ ex dx - 1⁄2 ∫ e1⁄2 dx
    = 1⁄2 ex - 1⁄3 x3⁄2 + C.

    (d) ∫ (1 + tan2 x) dx = ∫ sec2 x dx
    = tan x + C.

    Integrating f (ax + b)

    We have learned the reverse process of differentiation as an integration. Now we consider the integral of function which used the chain rule for differentiation.
    Consider the integral
    ∫ (4x + 1)5 dx.
    By the chain rule,
    d⁄dx (4x + 1)6 = 6(4x + 1)5 d⁄dx (4x + 1)5 ,
    d⁄dx ( 1⁄24 (4x + 1)6 ) = (4x + 1)5.
    Therefore
    ∫ (4x + 1)5 dx = 1⁄24 (4x + 1)6 + C.

    For n ≠ -1,
    d⁄dx ( 1⁄a(n + 1) (ax + b)n + 1 ) = (ax + b)n.

    ∫ (ax + b)n dx = 1⁄a (ax + b)n + 1 ⁄(n + 1) + C,   n ≠ -1.
    When n = -1,
    d⁄dx ( 1⁄a ln (ax + b)) = 1⁄a ( a⁄ax + b )
    = 1⁄ax + b ,     a ≠ 0.

    ∫ 1⁄ax + b dx = 1⁄a ln |ax + b| + C.
    In general, if f is a differentiable functiion of x, then
    d⁄dx f (ax + b) = f ' (ax + b) d⁄dx (ax + b)
    = af ' (ax + b)
    and reversing this we get
    ∫ f ' (ax + b) dx = 1⁄a f(ax + b) + C.
    For the trigonometric functions of the form f (ax) + b),
    ∫ cos(ax + b) dx = 1⁄a sin (ax + b) + C,
    ∫ sin (ax + b) dx = - 1⁄a cos (ax + b) + C.,
    ∫ sec2 (ax + b) dx = 1⁄a tan (ax + b) + C.

    For the exponential functions of the form f (px + q),
    ∫ apx + q dx = 1⁄p 1⁄ln a apx + q + C
    where a > 0,   a ≠ 1.
    When a = e,
    ∫ epx + q dx = 1⁄p epx + q + C.
    Example 5.
    Evaluate the following integrals.
    (a) ∫ 1⁄3 - 4x dx
    (b) ∫ √ 8x - 7   dx
    (c) ∫ e2x + 1 dx
    (d) ∫ 23x + 1 dx
    (e) ∫ cosx dx
    (f) ∫ sin 4x cos 3x dx

    Solution
    (a) ∫ 1⁄3 - 4x dx = 1⁄4 |3 - 4x| + C.

    (b) ∫ √ 8x - 7   dx = 1⁄8 [ 2⁄3 (8x - 7)1⁄2 ] + C
    = 1⁄12 (8x - 7)1⁄ 2 + C.

    (c) ∫ e2x + 1 dx = 1⁄2 e2x + 1 + C.

    (d) ∫ 23x + 1 dx = 1⁄2 1⁄ln 2 23x + 1 + C.

    (e) We use   cos2 x= 1⁄2 (1 + cos 2x), then
    ∫ cos2 x dx = 1⁄2 ∫ (1 + cos 2x) dx
    = 1⁄2 (x + 1⁄2 sin 2x) + C
    = 1⁄2 x + 1⁄4 sin 2x + C.

    (f) We use sin 4x cos 3x = 1⁄2 (sin 7x + sin x), then
    ∫ sin 4x cos 3x dx = 1⁄2 ∫ (sin 7x + sin x) dx
    = 1⁄2 (-1⁄7 cos 7x - cos x) + C
    = - 1⁄14 cos 7x - 1⁄2 cos x + C.


    Answer to 1.(h)
    ပံ့ပိုးကူညီသူ


    Exercise 10.1
    1. Evaluate the following integrals:
    (a) ∫ 4xx dx
    Solution
    ∫ 4xx dx = 4 ∫ x9 dx
    = 4 ⋅ x8 + 1 ⁄ 8 + 1 + C
    = 4 ⁄ 9 x9 + C.

    (b) ∫ 3 ⁄ 2 x2 ∛ x   dx
    Solution
    ∫ 3 ⁄ 2 x2 ∛ x   dx = ∫ 3 ⁄ 2 x2 ⋅ x 1 ⁄ 3 dx
    = 3 ⁄ 2 ∫ x 7 ⁄ 3 dx
    = 3 ⁄ 2 ⋅ x 7 ⁄ 3 + 1 ⁄ 7 ⁄ 3 + 1 + C
    = 3 ⁄ 2 ⋅ x 10 ⁄ 3 ⁄ 10 ⁄ 3 + C
    = 3 ⁄ 2 ⋅ 3 ⁄ 10 x 10 ⁄ 3 + C
    = 9 ⁄ 20 x 10 ⁄ 3 + C.

    (c) ∫ (5x + 2) dx
    ∫ (5x + 2) dx = ( 1 ⁄ ln 5 5x + 2 ⋅ x0 + 1 ⁄ 0 + 1 ) + C
    = 5x ⁄ ln 5 + 2x + C

    for sin2 α = 1 ⁄ 2 (1 - cos 2α),
    cos 2α = 1 - 2 sin2 α
    2 sin2 α + cos 2α = 1
    2 sin2 α = 1 - cos 2α
    sin2α = 1 ⁄ 2 (1 - cos 2α)

    for ∫ cos 2x dx = sin 2x ⁄ 2,
    Let 2x = u
    Then 2dx = du
    dx = 1 ⁄ 2 du
    ∫ cos 2x dx = ∫ cos u 1 ⁄ 2 du
    = 1 ⁄ 2 ∫ cos u du
    = 1 ⁄ 2 sin u + C
    = 1 ⁄ 2 sin 2x + C
    = sin 2x ⁄ 2 + C.

    (d) ∫ sin2 x dx
    Solution
    ∫ sin2 x dx = ∫ 1 ⁄ 2 (1 - cos 2x) dx
    = 1 ⁄ 2 ∫ (1 - cos 2x) dx
    = 1 ⁄ 2 ( 1 ⋅ x0 + 1 ⁄ 0 + 1 - sin 2x ⁄ 2) + C
    = 1 ⁄ 2 (x - sin 2x ⁄ 2 ) + C.

    (e) ∫ x + 3 ⁄ √ x   dx
    Solution
    ∫ x + 3 ⁄ √ x   dx = ∫ ( x ⋅ x 1 ⁄ 2 + 3x - 1 ⁄ 2 ) dx
    = ∫ (x 1 ⁄ 2 + 3x- 1 ⁄ 2 ) dx
    = x 1 ⁄ 2 + 1 ⁄ 1 ⁄ 2 + 1 + 3 ⋅ x- 1 ⁄ 2 + 1 ⁄ - 1 ⁄ 2 + 1 + C
    = x 3 ⁄ 2 ⁄ 3 ⁄ 2 + 3 ⋅ x 1 ⁄ 2 ⁄ 1 ⁄ 2 + C
    = 2 ⁄ 3 x 3 ⁄ 2 + 6 x 1 ⁄ 2 + C.

    (f) ∫ ( 1 ⁄ 2x + 5) dx
    Solution
    ∫ ( 1 ⁄ 2x + 5) dx = ∫ ( 1 ⁄ 2 ⋅ 1 ⁄ x + 5 ) dx
    = 1 ⁄ 2 ln |x| + 5 ⋅ x0 + 1 ⁄ 0 + 1 + C
    = 1 ⁄ 2 ln |x| + 5x + C.

    (g) ∫ ( ex + 2 ⁄ x ) dx
    Solution
    ∫ ( ex + 2 ⁄ x ) dx = ex + 2 ln |x| + C.

    (h) ∫ ( 1 ⁄ x5 + 4ex ) dx
    Solution
    ∫ ( 1 ⁄ x5 + 4ex ) dx = x-5 + 1 ⁄ -5 + 1 + 4 ex + C
    = x-4 ⁄ -4 + 4 ex + C.

    (i) ∫ ( 3 ⁄ x + ex + 10 )dx
    Solution
    ∫ ( 3 ⁄ x + ex + 10 )dx = 3 ln |x| + ex + 10 x0 + 1 ⁄ 0 + 1 + C
    = 3 ln |x| + ex + 10 x + C.

    (j) ∫ sin2 3x dx
    Solution
    Let u = 3x
    Then du = 3 dx
    1 ⁄ 3 du = dx

    ∫ sin2 3x dx = ∫ sin2 u 1 ⁄ 3 du
    = 1 ⁄ 3 ∫ sin2 u du
    = 1 ⁄ 3 [ 1 ⁄ 2 (u - sin 2u ⁄ 2 )] + C
    Substituting u = 3x,
    = 1 ⁄ 3 [ 1 ⁄ 2 (3x - sin 2⋅ 3x ⁄ 2 )] + C
    = 1 ⁄ 2 x - 1 ⁄ 12 sin 6x + C

    sin α sin β = 1 ⁄ 2 [cos (α - β) - cos (α + β)]
    (k) ∫ sin 5x sin 2x dx
    Solution
    ∫ sin 5x sin 2x dx = 1 ⁄ 2 ∫ (cos 3x - cos 7x) dx
    For cos 3x dx,
    let u = 3x
    Then du = 3 dx
    1 ⁄ 3 du = dx
    ∫ cos 3x dx = ∫ cos u 1 ⁄ 3 du
    = 1 ⁄ 3 ∫ cos u du
    = 1 ⁄ 3 sin u + C
    = 1 ⁄ 3 sin 3x + C
    = sin 3x ⁄ 3 + C

    For cos 7x dx,
    Let v = 7x
    Then dv = 7 dx
    1 ⁄ 7 dv = dx
    ∫ cos 7x dx = ∫ cos v 1 ⁄ 7 dv
    = 1 ⁄ 7 ∫ cos v dv
    = 1 ⁄ 7 sin v + C
    = 1 ⁄ 7 sin 7x + C
    = sin 7x ⁄ 7 + C

    Therefore
    ∫ sin 5x sin 2x dx = 1 ⁄ 2 ∫ (cos 3x - cos 7x) dx
    = 1 ⁄ 2 [ sin 3x ⁄ 3 - sin 7x ⁄ 7 ] + C.

    cos A cos B = 1 ⁄ 2 [ cos (A + B) + cos (A - B)]
    (l) ∫ cos 7x cos 4x dx
    Solution
    ∫ cos 7x cos 4x dx = 1 ⁄ 2 ∫ (cos 11x + cos 3x dx
    For ∫ cos 11x dx ,
    Let u = 11x
    Then du = 11dx
    1 ⁄ 11 du = dx
    ∫ cos 11 x dx = ∫ cos u 1 ⁄ 1 du
    = 1 ⁄ 11 ∫ cos u du
    = 1 ⁄ 11 sin u + C

    For cos 3x dx,
    Let v = 3x ,
    Then dv = 3 dx
    1 ⁄ 3 dv = dx
    ∫ cos 3x dx = ∫p v 1 ⁄ 3 dv
    = 1 ⁄ 3 sin v + C.

    Therefore
    ∫ cos 7x cos 4x dx = 1 ⁄ 2 ∫(cos 11x + cos 3x) dx
    = 1 ⁄ 2 [ sin 11x ⁄ 11 + sin 3x ⁄ 3 ] + C.

    2. Evaluate the following integrals:
    (a) ∫ (1 - 2x)3 dx
    Solution
    Let u = 1 - 2x ,
    Then du = -2dx
    - 1 ⁄ 2 du = dx
    ∫ (1 - 2x)3 dx = ∫ u3 - 1 ⁄ 2 du
    = - 1 ⁄ 2 ∫ u3 du
    = - 1 ⁄ 2 ⋅ u3 + 1 ⁄ 3 + 1 + C
    = - 1 ⁄ 2 ⋅ u4 ⁄ 4 + C
    = - 1 ⁄ 2 (1 - 2x)4 ⁄ 4 + C
    = - 1 ⁄ 8 (1 - 2x)4 + C

    (b) ∫ sin (2πx + 7) dx
    Solution
    Let u = 2πx + 7
    Then du = 2π dx
    = 1 ⁄ 2π du = dx
    ∫ sin (2πx + 7) dx = ∫ sin u 1 ⁄ 2 u 1 ⁄ 2π du
    = 1 ⁄ 2π ∫ sin u du
    = 1 ⁄ 2π (- cos u) + C
    = - 1 ⁄ 2π cos (2πx + 7) + C

    (c) ∫ cos (3x - 7) dx
    Solution
    Let u = 3x - 7 ,
    du = 3dx
    1 ⁄ 3 du = dx
    ∫ cos (3x - 7) dx = ∫ cos u 1 ⁄ 3 du
    = 1 ⁄ 3 ∫ cos u du
    = 1 ⁄ 3 sin u + C
    1 ⁄ 3 sin (3x - 7) + C

    (d) ∫ 35x - 2 dx
    Solution
    ∫ 35x - 2 dx = 1 ⁄ 5 1 ⁄ ln 3 35x - 2 + C

    (e) ∫ 1 ⁄ 7x - 6 dx
    Soution
    ∫ 1 ⁄ 7x - 6 dx = 1 ⁄ 7 ln |7x - 6| + C

    (f) ∫ sin 2x ⁄ sin x dx
    Solution
    ∫ sin 2x ⁄ sin x dx = ∫ 2 sin x cos x ⁄ sin x dx
    = ∫ 2 cos x dx
    = 2 sin x + C   (sin x ≠ 0)

    (g) ∫ sec2(2x + 3) dx
    Solution
    ∫ sec2(2x + 3) dx = 1 ⁄ 2 tan (2x + 3) + C

    (h) ∫ e7x - 3 dx
    Solution
    ∫ e7x - 3 dx = 1 ⁄ 7 e7x - 3 + C

    (i) ∫ (1 + tan2 2πx) dx
    Solution
    Let u = 2πx ,
    Then du = 2π dx
    1 ⁄ 2π du = dx
    ∫ (1 + tan2 2πx) dx = ∫ sec2 2π x dx
    = ∫ sec2 u 1 ⁄ 2π du
    = 1 ⁄ 2π ∫ sec2 u du
    = 1 ⁄ 2π tan u + C
    = 1 ⁄ 2π tan 2πx + C