ICAI Notes: Differential and Integral Calculus- 3 | Mathematics for GRE Paper II PDF Download

9 The Fundamental Theorem in Terms of Differentials

Fundamental Theorem of Calculus: If F (x) is one antiderivative of the function f (x), i.e., F'(x) = f (x), then

Thus, the integral of the diffierential of a function F is equal to the function itself plus an arbitrary constant. This is simply saying that diffierential and integral are inverse math operations of each other. If we rst dierentiate a function F(x) and then integrate the derivative F'(x) = f (x), we obtain F (x) itself plus an arbitrary constant. The opposite also is true. If we first integrate a function f (x) and then dierentiate the resulting integral F (x) + C , we obtain F'(x) = f (x) itself.

Example:

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10 Integration by Subsitution

Substitution is a necessity when integrating a composite function since we cannot write down the antiderivative of a composite function in a straightforward manner.
Many students nd it difficult to gure out the substitution since for dierent functions the subsitutions are also different. However, there is a general rule in substitution, namely, to change the composite function into a simple, elementary function.

Example: 

Solution: Note that sin(√x) is not an elementary sine function but a composite function. The rst goal in solving this integral is to change sin(√x) into an elementary sine function through substitution. Once you realize this, u = √x is an obvious subsitution. Thus, du = u'dx = 1/2√x dx,  or dx = 2√x du = 2udu. Substitute into the integral, we obtain

Once you become more experienced with subsitutions and diffierentials, you do not need to do the actual substitution but only symbolically. Note that x = (√x)²,

Thus, as soon as you realize that √x is the substitution, your goal is to change the diffierential in the integral dx into the diffierential of px which is d√x.

If you feel that you cannot do it without the actual substitution, that is ne. You can always do the actual substitution. I here simply want to teach you a way that actual subsitution is not a necessity!

Solution: Note that cos(x⁵ ) is a composite function that becomes a simple cosine function only if the subsitution u = x⁵ is made. Since du = u'dx = 5x⁴ dx, x⁴ dx = 1/5du. Thus,

Or alternatively,

Example: 

Solution: Note that    is a composite function. We realize that u = x² + 1 is a substitution. du = u'dx = 2xdx implies xdx = 1/2du. Thus,

Or alternatively,

In many cases, substitution is required even no obvious composite function is involved.

Solution: The integrand ln x/x is not a composite function. Nevertheless, its antiderivative is not obvious to calculate. We need to gure out that (1/x)dx = d ln x, thus by introducing the substitution u = ln x, we obtained a diffierential of the function ln x which also appears in the integrand. Therefore,

It is more natural to consider this substitution is an attempt to change the diffierential dx into something that is identical to a function that appears in the integrand, namely d ln x. Thus,

Solution: tan x is not a composite function. Nevertheless, it is not obvious to gure out tan x is the derivative of what function. However, if we write tan x = sin x/cos x, we can regard 1/cos x as a composite function. We see that u = cosx is a candidate for substitution and du = u'dx = sin(x)dx. Thus,

Or alternatively,

Some substitutions are standard in solving speci c types of integrals.

Example: Integrands of the type 

In this case both x = asinu and x = acosu will be good. x = a tanh u also works (1 tanh² u = sech² u). Let's pick x = asinu in this example. If you ask how can we nd out that x = asinu is the substitution, the answer is a² - x² = a²(1 - sin²u) = a² cos²u. This will help us eliminate the half power in the integrand. Note that with this substitution, u = sin^-1(x/a), sinu = x/a, and cosu = √1 - x²/a²

 , we can make a simple substitution x = au, thus

Similarly,

x = asinh(u) is a good substitution since a² + x² = a² + a²sinh² (u) = a² (1 + sinh² (u)) = a² cosh²(u), where the hyperbolic identity 1 + sinh² (u) = cosh²(u) was used. (x = a tan u is also good since 1 + tan² u = sec²u!). Thus,

where the following hyperbolic identities were used: sinh(2u) = 2sinh(u)cosh(u), sinh^-1(x/a) = 

Example: Integrands of the type 

x = acosh(u) is prefered since x² a² = a²scos² (u) a² = a²[cosh² (u) 1] = a² sinh² (u), where the hyperbolic identity cosh² (u) 1 = sinh² (u) was used. (x = a sec u is also good since sec² u 1 = tan² u!). Thus,

Similarly

where cosh u = x/a, sinhu   

1. Substitution aimed at eliminating a composite function

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2. Substitution to achieve a function in diffierential that appears in the integrand

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3. Special Trigonometric Substitutions 

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11 Integration by Parts

Integration by Parts is the integral version of the Product Rule in dierentiation. The Product Rule in terms of diffierentials reads,

d(uv) = vdu + udv

Integrating both sides, we obtain

now that   , the above equation can be expressed in the following form,

Generally speaking, we need to use Integration by Parts to solve many integrals that involve the product between two functions. In many cases, Integration by Parts is most efficient in solving integrals of the product between a polynomial and an exponential, a logarithmic, or a trigonometric function. It also applies to the product between exponential and trigonometric functions.

Solutions: In order to eliminate the power function x, we note that (x)' = 1. Thus,

  = xe^x - e^x + C

Solution: In order to eliminate the power function x² , we note that (x² )" = 2. Thus, we need to use Integration by Patrs twice.

= x²sinx + 2xcosx - 2sinx + C

Solution: In order to eliminate the power function x² , we note that (x² )"= 2. Thus, we need to use Integration by Patrs twice. However, the number (-2) can prove extremely annoying and easily cause errors. Here is how we use substitution to avoid this problem.

When integrating the product between a polynomial and a logarithmic function, the main goal is to eliminate the logarithmic function by dierentiating it. This is because (ln x)' = 1/x.

Solutions:  

Solutions:  

More Exercises on Integration by Parts: 

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12 Integration by Partial Fractions

Rational functions are defined as the quotient between two polynomials:

where P_n(x) and Q_m (x) are polynomials of degree n and m respectively. The method of partial fractions is an algebraic technique that decomposes R(x) into a sum of terms:

where p(x) is a polynomial and F_i(x); (i = 1; 2; ..... ; k) are fractions that can be integrated easily.
The method of partial fractions is an area many students nd very difficult to learn. It is related to algebraic techniques that many students have not been trained to use. The most typical claim is that there is no fixed formula to use. It is not our goal in this course to cover this topics in great details (Read Edwards/Penney for more details). Here, we only study two simple cases.

Case I: Q_m(x) is a power function, i.e., Q_m(x) = (x - a)^m (Q_m (x) = x^m if a = 0!).

Example:

This integral involves the simplest partial fractions:

Some may feel that it is easier to write the fractions in the following form: D^-1(A + B + C ) = D^-1A + D^-1B + D^-1C . Thus,

Solutions: 

Case II: P_n(x) = A is a constant and Q_m (x) can be factorized into the form Q₂ (x) = (x - a)(x - b), Q₃ (x) = (x- a)(x - b)(x - c), or Q_m (x) = (x - a₁ )(x - a₂) ..... (x - a_m )

Example:

Since,   implies that A(x + 1) + B (x - 3) = 1. Setting x = 3 in this equation, we obtain A = 1/4. Setting x = -1, we obtain B = -1/4. Thus,

Example:

Since   implies that A(x - 2)(x 3) + B (x -1)(x - 3) + C (x - 1)(x - 2) = 1. Setting x = 1 in this equation, we obtain A = 1/2. Setting x = 2, we obtain B = -1. Setting x = 3, we get C = 1/2. Thus,

More Exercises on Integration by Partial Fractions: 

Examples:

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