Integral as an Anti-derivative

A function \(\phi(x)\) is called a primitive or anti-derivative of a function \(f(x),\ \text{if}\ \phi'(x)=f(x).\) If \(f_1(x)\ \ and \ \ f_2(x)\) are two anti-derivatives of \(f(x),\) then \(f_1(x)\ \ and \ \ f_2(x)\) differ by a constant. The collection of all its anti-derivatives is called indefinite integral of \(f(x)\) and is denoted by \(\int\ f(x)\ dx.\)

Thus, \(\frac{d}{dx}\ \{\phi\ (x)\ +\ C\}=f(x) \Rightarrow\int\ f(x)\ dx= \phi\ (x)\ +\ C\)

where, \(\phi\ (x)\) is an anti-derivative of \(f (x)\), \(f (x)\) is the integrand and \(C\) is an arbitrary constant known as the constant of integration. Anti-derivative of odd function is always even and of even function is always odd.

Properties of Indefinite Integrals

\(\bullet\ \ \int\ \{f(x)\ \pm\ g(x)\}\ dx=\ \int\ f(x)\ dx\ \pm\ \int\ g(x)\ dx\)

\(\bullet\ \ \int\ k\ \cdot f(x)\ dx=k\ \cdot \int\ f(x)\ dx,\) where \(k\) is any non-zero real number.

\(\bullet\ \ \int\ [k_1\ f_1(x)\ +\ k_2\ f_2(x)\ +... +\ k_n\ f_n(x)]\ dx=\ k_1\ \int\ f_1(x)\ dx\ +\ k_2\ \int\ f_2(x)\ dx\ +...+\ k_n\ \int\ f_n(x)\ dx,\)

     where \(k_1,\ k_2, ...\ k_n\) are non-zero real numbers.

3. Exponential Formulae

\((i)\ \ \int\ e^x\ dx=e^x\ +\ C\\ (ii)\ \ \int\ e^{(ax\ +\ b)}\ dx=\frac{1}{a}\ \cdot e^{(ax\ +\ b)}\ +\ C\\ (iii)\ \ \int\ a^x\ dx=\frac{a^x}{\\log_e\ a}\ +\ C,\ a \gt 0\ and\ a \ne 1\\ (iv)\ \ \int\ a^{(bx\ +\ c)}\ dx=\ \frac{1}{b}\cdot \frac{a^{(bx\ +\ c)}}{\\log_e\ a}\ +\ C,\ a \gt 0\ and\ a \ne 1\)

3. Integration of Different Types of Functions

\(\bullet\)  To evaluate integrals of the form \(\int\ \sin^p\ x\ \cos^q\ x\ dx\)

Where, \(p,\ q \in\ Q\) we use the following rules:

   (i)  If \(p\) is odd, then put \(\cos\ x=t\)

   (ii)  If \(q\) is odd, then put \(\sin\ x=t\)

   (iii)  If both \(p,\ q\) are  odd, then put either \(\sin\ x=t\ \ or\ \ \cos\ x=t\)

   (iv)  If both \(p,\ q\) are even, then use trigonometric identities only.

   (v) If \(p,\ q\) are rational numbers and \(\Big(\frac{p\ +\ q\ -\ 2}{2}\Big)\) is a negative integer, then put \(x=t\ \ or\ \ \tan\ x=t\) as required.

Integration

It is the inverse process of differentiation. If the derivative of \(F(x)\) is \(f(x)\) then we say that anti-derivative or integral of \(f(x)\) is \(F(x)\) and we write \(\int\ f(x)\ dx=\ F(x)\)

\(\frac{d}{dx}\ [F(x)=\ f(x)\Rightarrow\ \int\ f(x)\ dx=\ F(x)\)

Example: Since \(\frac{d}{dx}\ (\sin x)=\ \cos\ x,\) we have \(\int\ \cos\ x\ dx=\ \sin\ x\)

If \(c\) is my constant then \(\frac{d}{dx}\ (\sin\ x\ +\ c)=\ \cos \ x\)

\(\int\ \cos\ x\ dx=\ (\sin\ x\ +\ c)\) different values of \(c\) will give different integrals.

Thus, given function may have an indefinite integrals. Because of this property, we call these integrals indefinite integrals \(\frac{d}{dx}\ [F(x)=\ F(x)\ \Rightarrow\ \int\ F(x)\ dx=\ F(x)\ +\ c\) 

where \(c\) is called the constant of integration. Any function to be integrated is known as integral

Formulae

On basis of differentiation and definition of Integration:

1.  \(\frac{d}{dx}\ \Big(\frac{x^{n\ +\ 1}}{n\ +\ 1}\Big)=\ x^n,\ n\ne\ -1\ \Rightarrow\ \int\ x^n\ dx=\ \frac{x^{n\ +\ 1}}{n\ +\ 1}\ +\ c\)

2.  \(\frac{d}{dx}\ (\log\ |\ x\ |)=\ \frac{1}{x}\ \Rightarrow\ \int\ \frac{1}{x}\ dx=\ \log\ |\ x\ |\ +\ c\)

3.  \(\frac{d}{dx}\ (e^x)=\ e^x\ \Rightarrow\ \int\ e^x\ dx=\ e^x\ +\ c\)

4.  \(\frac{d}{dx}\ \Big(\frac{a^x}{\log\ a}\Big)=a^x\ \Rightarrow\ \int\ a^x\ dx=\ \frac{a^x}{\log\ a}\ +\ c\)

5.  \(\frac{d}{dx}\ (\sin\ x)=\ \cos\ x\ \Rightarrow\ \int\ \cos\ x\ dx=\ \sin\ x\ +\ c\)

6.  \(\frac{d}{dx}\ (-\cos\ x)=\ \sin\ x\ \Rightarrow\ \int\ \sin\ x\ dx=\ -\cos\ x\ +\ c\)

7.  \(\frac{d}{dx}\ (\tan\ x)=\ \sec^2\ x\ \Rightarrow\ \int\ \sec^2\ x\ dx=\ \tan\ x\ +\ c\)

\(8.\ \ \frac{d}{dx}\ (-\cot\ x)=\ cosec^2\ x\ \Rightarrow\ \int\ cosec^2\ x\ dx=\ -\cot\ x\ +\ c\)

\(9.\ \ \frac{d}{dx}\ (\sec\ x)=\ \sec\ x\ \tan\ x\ \Rightarrow\ \int\ \sec\ x\ \tan\ x\ dx=\ \sec\ x\ +\ c\)

\(10.\ \ \frac{d}{dx}\ (-\ cosec\ x)=\ cosec\ x\ \cot\ x\ \Rightarrow\ \int\ cosec\ x\ \cot\ x\ dx=\ - cosec\ x\ +\ c\)

\(11.\ \ \frac{d}{dx}\ (\sin^{-1}\ x)=\ \frac{1}{\sqrt{1\ -\ x^2}}\ \Rightarrow\ \int\ \frac{1}{\sqrt{1\ -\ x^2}}\ dx=\ \sin^{-1}\ x\ +\ c\)

\(12.\ \ \frac{d}{dx}\ (\tan^{-1}\ x)=\ \frac{1}{1\ +\ x^2}\ \Rightarrow\ \int\ \frac{1}{1\ +\ x^2}\ dx=\ \tan^{-1}\ x\ +\ c\)

\(13.\ \ \frac{d}{dx}\ (\sec^{-1}\ x)=\ \frac{1}{x\ \sqrt{ x^2\ -\ 1}}\ \Rightarrow\ \int\ \frac{1}{x\ \sqrt{x^2\ -\ 1}}\ dx=\ \sec^{-1}\ x\ +\ c\)

Integration

Theorem 1

      \(\frac{d}{dx}\ \{\int\ f(x)\ dx\}=\ f(x)\)

Theorem 2

     \(\int\ k\ \cdot\ f(x)\ dx= k\ \cdot\ f(x)\ dx\), where \(k\) is a constant

Remark: In general, we have \(\int\ \{\ k_1\ \cdot\ f_1(x)\ \pm\ k_2\ \cdot\ f_2(x)\ \pm\ ...\ \pm\ k_n\ \cdot\ f_n(x)\ \}\ dx\)

\(=k_1\ \cdot\ \int\ f_1(x)\ dx\ \pm\ k_2\ \cdot\ \int\ f_2(x)\ dx\ \pm\ ...\ \pm\ k_n\ \cdot\ \int\ f_n(x)\ dx\)

Integration by Substitution

If we have to evaluate an Integral of the type \(\int \ f\{\ \phi\ (x)\}\ dx\) then we put \(\phi\ (x)=t\) and \(\phi'\ (x)\ dx=dt.\) With this substitution, the integrand becomes easily integrable.

Case 1:

When the integrand is of the form \(f(ax\ +\ b),\) we put \((ax\ +\ b)=t\) and \(dx=\frac{1}{a}\ dt\)

Case 2:

When the integrand is of the form \(x^{n\ -\ 1}.\ F(x^n)\) we put \(x^n=t\) and \(nx^{n\ -\ 1}\ dx=dt\)

Case 3:

When the integrand is of the form \(\{f(x)\}^n\ \cdot\ f'(x),\) we put \(f(x)=t\) and \(f'(x)\ dx=\ dt\)

Case 4:

When the integrand is of the form \(\frac{f'(x)}{f(x)},\) we put \(f(x)=t\) and \(f'(x)\ dx=\ dt\)

Theorem 1:

\(\int\ (ax\ +\ b)^n\ dx=\ \frac{(ax\ +\ b)^{n\ +\ 1}}{a(n\ +\ 1)}\ +\ c,\ n\ne-1\)

Theorem 1:

\((i)\ \ \int\ \cos\ (ax\ +\ b)\ dx=\ \frac{1}{a}\ \sin\ (ax\ +\ b)\ +c\\ (ii)\ \ \int\ cosec^2\ (ax\ +\ b)\ dx=\ -\frac{1}{a}\ \cot\ (ax\ +\ b)\ +c\)

Integration Using Trigonometric Identities

When the integrand consists of trigonometric functions, we use known identities to convert it into a form which can easily be integrated. Some of the identities useful for this purpose are given below:

\(1.\ \ 2\ \sin^2\ \Big(\frac{x}{2}\Big)=\ (1\ -\ cos\ x)\\ 2.\ \ 2\ \cos^2\ \Big(\frac{x}{2}\Big)=\ (1\ +\ cos\ x)\\ 3.\ \ 2\ \sin\ A\ \cos\ B=\ \sin\ (A\ +\ B)\ +\ \sin\ (A\ -\ B)\\ 4.\ \ 2\ \cos\ A\ \sin\ B=\ \sin\ (A\ +\ B)\ -\ \sin\ (A\ -\ B)\\ 5.\ \ 2\ \cos\ A\ \cos\ B=\ \cos\ (A\ +\ B)\ +\ \cos\ (A\ -\ B)\\ 6.\ \ 2\ \sin\ A\ \sin\ B=\ \cos\ (A\ -\ B)\ -\ \cos\ (A\ -\ B)\)

Integration by Parts

Theorem

If \(\mu\ \ and\ \ v\) are two functions of \(x\) then \(\int \ (\mu\ v)\ dx=\ [\ \mu\ \cdot\int\ v\ dx\ ]\ -\ \int\ \Big\{\frac{du}{dx}\ \cdot\ \int\ v\ dx\ \Big\}\ dx\) 

We can express this result as given below:

Integral of product of two functions \(=\ (1st\ function)\ \times\ (Integral\ of\ 2nd)\ -\ \int\ \{(derivative\ of\ 1st)\ \times\ (Integral\ of\ 2nd)\}\ dx\)

Remarks:

(i)  If the integrand is of the form \(f(x)\ x^n,\) we consider \(x^n\) as the first function and \(f(x)\) as the second function.

(ii)  If the integrand contains a logarithmic function, we take it as first function. In all such cases, if the second function is not given, we take it as \(1\).

Integration by Parts Based Theorems

Theorem 1:

\(\int\ e^x\ \{\ f(x)\ +\ f'(x)\ dx=e^x\ \cdot\ f(x)\ +\ c\)

Theorem 2:

\(\int\ e^{kx}\ \{\ k\ \cdot f(x)\ +\ f'(x)\}\ dx=e^{kx}\ \cdot\ f(x)\ +\ c\)

Theorem 3:

\(\int\ e^{ax}\ \cos\ (bx\ +\ c)\ dx=\ e^{ax}\ \cdot\ \frac{\cos\ [\ bx\ +\ c\ -\ \tan^{-1}\ (b/a)\ ]}{\sqrt{a^2\ +\ b^2}}\)

\(\int\ e^{ax}\ \sin\ (bx\ +\ c)\ dx=\ e^{ax}\ \cdot\ \frac{\sin\ [\ bx\ +\ c\ -\ \tan^{-1}\ (b/a)\ ]}{\sqrt{a^2\ +\ b^2}}\)

Some Special Integrals

Theorem 1:

\((i)\ \ \int\ \frac{dx}{a^2\ -\ x^2}=\frac{1}{2a}\ \log\ \Big|\frac{a\ +\ x}{a\ -\ x}\Big|\ +c\\ (ii)\ \ \int\ \frac{dx}{x^2\ -\ a^2}=\frac{1}{2a}\ \log\ \Big|\frac{x\ -\ a}{x\ +\ a}\Big|\ +c\\ (iii)\ \ \int\ \frac{dx}{x^2\ +\ a^2}=\frac{1}{a}\ \tan^{-1}\ \frac{x}{a}\ +c\)

Remark:

If we have an integral of the form \(\int\ \frac{dx}{(ax^2\ +\ bx\ +\ c)}\) then we put the denominator in the form \([\ (x\ +\ \propto)^2\ \pm\ \beta^2\ ]\) and then Integrate.

Theorem 2:

Integrals of the form \(\int\ \frac{(px\ +\ q)}{(ax^2\ +\ bx\ +c)}\ dx\)

METHOD:

Let \((px\ +\ q)=A\ \cdot\ \frac{d}{dx}\ (ax^2\ +\ bc\ +\ c)\ +\ B\)

Find \(A \ \ and\ \ B\).

Now, the integrand so obtained can be integrated easily.

Theorem 3:

\((i)\ \ \int\ \frac{dx}{\sqrt{a^2\ -\ x^2}}=\ \sin^{-1}\ \frac{x}{a}\ +\ c\\ (ii)\ \ \int\ \frac{dx}{\sqrt{x^2\ -\ a^2}}=\ \log\ |x\ +\ \sqrt{x^2\ -\ a^2}\ |\ +c\\ (iii)\ \ \int\ \frac{dx}{\sqrt{x^2\ +\ a^2}}=\ \log\ |x\ +\ \sqrt{x^2\ +\ a^2}\ |\ +c\)

Theorem 4:

Integrals of the form \(\int\ \frac{(px\ +\ q)}{\sqrt{ax^2\ +\ bx\ +\ c}}\ dx\)

METHOD:

Let \((px\ +\ q)=\ A\ \cdot\ \frac{d}{dx}\ (ax^2\ +\ bx\ +\ c)\ +B\)

Now, the value of the integral can be obtained easily.

Theorem 5:

\((i)\ \ \int\ \sqrt{a^2\ -\ x^2}\ dx=\frac{x}{\alpha}\ \sqrt{a^2\ -\ x^2}\ +\ \frac{a^2}{\alpha}\ \sin^{-1}\ \frac{x}{a}\ +c\)

\((ii)\ \ \int\ \sqrt{x^2\ -\ a^2}\ dx=\frac{x}{\alpha}\ \sqrt{x^2\ -\ a^2}\ -\ \frac{a^2}{\alpha}\ \log\ |\ x\ +\ \sqrt{x^2\ -\ a^2}\ |\ +c\)

\((iii)\ \ \int\ \sqrt{x^2\ +\ a^2}\ dx=\frac{x}{\alpha}\ \sqrt{x^2\ +\ a^2}\ +\ \frac{a^2}{\alpha}\ \log\ |\ x\ +\ \sqrt{x^2\ +\ a^2}\ |\ +c\)

Theorem 6:

Integrals of the form \(\int\ \sqrt{ax^2\ +\ bx\ +\ c}\ dx\)

METHOD:

Express \((ax^2\ +\ bx\ +\ c)\) as \(a\ [\ (x\ +\ \alpha)^2\ \pm\ \beta^2\ ]\) and obtain an integral which can be evaluated easily.

Theorem 7:

Integrals of the form \(\int\ (px\ +\ q)=\ \sqrt{ax^2\ +\ bx\ +\ c}\ \ dx\)

METHOD:

Let \((px\ +\ q)=\ A\ \cdot\ \frac{d}{dx}\ (ax^2\ +\ bx\ +\ c)\ +B\)

Find \(A\ \ and\ \ B.\)

Then, we get the integrant which is easily integrable

Integration using partial fractions

Rational Fractions:

If \(f(x)\ \ and\ \ g(x)\) are polynomial fractions such that \(g(x)\ \ne 0\). Then, \(\frac{f(x)}{g(x)}\) is called rational fraction.

If the degree \(f(x)\ \lt\ degree\ \ g(x)\) then \(\frac{f(x)}{g(x)}\) is called proper rational fraction.

If the degree \(f(x)\ \ge\ degree\ \ g(x)\) then \(\frac{f(x)}{g(x)}\) is called improper rational fraction.

If \(\frac{f(x)}{g(x)}\) is an improper rational fraction then by dividing \(f(x)\ \ by\ \ g(x),\) we can express \(\frac{f(x)}{g(x)}\) as sum of a polynomial and proper rational fraction.

Partial Fractions

Any proper rational fraction \(\frac{p(x)}{q(x)}\) can be expressed as the sum of rational fractions, each having a simplest factor \(q(x).\) Each such fraction is known as partial fraction and the process of obtaining them is called the decomposition or resolving of the given fraction into partial fraction.

      Fraction in the denominator     Corresponding partial fraction

(i)      \((x\ -\ a)\)                                      \(\frac{A}{x\ -\ a}\)

(ii)     \((x\ -\ b)^2\)                                    \(\frac{A}{x\ -\ b}\ +\ \frac{B}{(x\ -\ b)^2}\)

(iii)    \((x\ -\ c)^3\)                                    \(\frac{A}{x\ -\ c}\ +\ \frac{B}{(x\ -\ c)^2}\ +\ \frac{C}{(x\ -\ c)^3}\)

(iv)    \((ax^2\ +\ bx\ +\ c)\)                        \(\frac{Ax\ +\ B}{(ax^2\ +\ bx\ +\ c)}\)

Properties of Definite Integrals

\(1.\ \ \int\limits_a^b\ f(x)\ dx=\int\limits_a^b\ f(t)\ dt\)

\(2.\ \ \int\limits_a^b\ f(x)\ dx=\ -\ \int\limits_a^b\ f(x)\ dx\)

\(3.\ \ \int\limits_a^b\ f(x)\ dx=\int\limits_a^c\ f(x)\ dx\ +\ \int\limits_c^b\ f(x)\ dx,\) where \(a\ \lt\ \lt\ \lt\ b\)

Remark:

If \(a\ \lt\ c_1\lt\ c_2\lt\ . . .\ \lt\ c_n\ \lt\ b\) then \(\int\limits_a^b\ f(x)\ dx=\int\limits_a^{c_1}\ f(x)\ dx\ +\ \int\limits_{c_1}^{c_2}\ f(x)\ dx\ +\ \int\limits_{c_n}^b\ f(x)\ dx\)

\(4.\ \ \int\limits^a_0\ f(x)\ dx=\ \int\limits^a_0\ f(a\ -\ x)\ dx\)

\(5.\ \ \int\limits^b_a\ f(a\ +\ b\ -\ x)\ dx=\ \int\limits^b_a\ f(x)\ dx\)

\(6.\ \ \int\limits^b_a\ \{\ f(x)\ +\ g(x)\ \}\ dx=\ \int\limits^b_a\ f(x)\ dx\ +\ \int\limits^b_a\ g(x)\ dx\)

\(7.\ \ \int\limits^a_{-a}\ f(x)\ dx=\ \begin{cases}0,\ &\ \text{when f(x)is an odd function}\\ 2\ \int\limits^a_0\ f(x)\ dx,\ &\ \text{when f(x) is even function}\end{cases}\)

\(8.\ \ \int\limits^{2a}_0\ f(x)\ dx=\ \begin{cases}0,\ &\ \text{if f(2a - x) = - F(x)}\\ 2\ \int\limits^a_0\ f(x)\ dx,\ &\ \text{if f(2a - x) = f(x)}\end{cases}\)

To evaluate a definite Integral by substitution

In \(\int\limits^b_a\ f(x)\ dx,\) when the variable \(x\) is converted into a new variable \(t\) by some relation then we put \(x=a\) and \(x=b\) in that relation to obtain the corresponding values of \(t\), giving the lower limit and upper limit respectively of the new integrand in \(t.\)

Fundamental Theorem of Integral Calculus

Let \(F(x)\) be continuous function defined on an integral \([\ a,\ b\ ]\) and let the anti-derivative of \(f(x)\) be \(F(x)\). Then, the definite integral of \(f(x) \) over \([\ a,\ b\ ],\) denoted by \(\int\limits^b_a\ f(x)\ dx\) is given by \(\int\limits^b_a\ f(x)\ dx = \ [\ F(x)\ ]^b_a\ =\ F(b)\ -\ F(a)\)

\(\overline{\underline{\text{NOTE}}}\):

Here \(a\ \ and\ \ b\) are respectively known as the lower limit and the upper limit of the integral.

If \(\int\ f(x)\ dx\ =\ F(x)\ +\ c\) then 

\(\int\limits^b_a\ f(x)\ dx\ =\ [\ F(x)\ +\ c\ ]^b_a\ =\ \{F(b)\ +\ c\}\ -\ \{F(a)\ +\ c\}=\ F(b)\ -\ F(a).\)