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Statement I: Integral of an even function is not always an odd function.

Statement II: Integral of an odd function is an even function.

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Important Points to Remember in Chapter -1 - Indefinite Integral from Amit M Agarwal Skills in Mathematics for JEE MAIN & ADVANCED INTEGRAL CALCULUS Solutions

1. Basics of Indefinite Integral

(i) Definition of Indefinite Integration:

If $f & F$ are function of $x$ such that $F^{'} (x) = f (x)$ , then the function $F$ is called a primitive or antiderivative or integral of $f (x)$ w.r.t. $x$ and is written symbolically as

$\int f (x) d x = F (x) + c \Leftrightarrow \frac{d}{d x} \{F (x) + c\} = f (x),$ where $c$ is called the constant of integration.

2. Properties of Indefinite Integrals:

(i) $\int f (x) d x = F (x) + c,$ if $F^{'} (x) = f (x)$ , similarly, ${(\int f (x) d x)}^{'} = f (x)$

(ii) $\int [f (x) \pm g (x)] d x = \int f (x) d x \pm \int g (x) d x$

(iii) $\int k f (x) d x = k \int f (x) d x$ , $k$ is any constant which must be independent of $x$ .

(iv) $\int f (a x + b) d x = \frac{1}{a} F (a x + b) + c$

3. Standard Formulae:

(i) $\int {(a x + b)}^{n} d x = \frac{{(a x + b)}^{n + 1}}{a (n + 1)} + c; n \neq - 1$

(ii) $\int \frac{d x}{a x + b} = \frac{1}{a} l n |a x + b| + c$

(iii) $\int e^{a x + b} d x = \frac{1}{a} e^{a x + b} + c$

(iv) $\int a^{p x + q} d x = \frac{1}{p} \frac{a^{p x + q}}{l n a} + c (a > 0)$

(v) $\int s i n (a x + b) d x = - \frac{1}{a} c o s (a x + b) + c$

(vi) $\int c o s (a x + b) d x = \frac{1}{a} s i n (a x + b) + c$

(vii) $\int t a n (a x + b) d x = \frac{1}{a} l n s e c (a x + b) + c$

(viii) $\int c o t (a x + b) d x = \frac{1}{a} l n s i n (a x + b) + c$

(ix) $\int s e c^{2} (a x + b) d x = \frac{1}{a} t a n (a x + b) + c$

(x) $\int c o s e c^{2} (a x + b) d x = - \frac{1}{a} c o t (a x + b) + c$

(xi) $\int c o s e c (a x + b) \cdot c o t (a x + b) d x = - \frac{1}{a} c o s e c (a x + b) + c$

(xii) $\int s e c (a x + b) \cdot t a n (a x + b) d x = \frac{1}{a} \cdot s e c (a x + b) + c$

(xiii) $\int s e c x d x = l n |s e c x + t a n x| + c$ or $\int s e c x d x = l n t a n |\frac{π}{4} + \frac{x}{2}| + c$

(xiv) $\int c o s e c x d x = l n |c o s e c x - c o t x| + c$ or $- \ln (c o s e c x + c o t x) + c$ or $l n |t a n \frac{x}{2}| + c$

(xv) $\int \frac{d x}{\sqrt{a^{2} - x^{2}}} = {s i n}^{- 1} \frac{x}{a} + c$

(xvi) $\int \frac{d x}{a^{2} + x^{2}} = \frac{1}{a} {t a n}^{- 1} \frac{x}{a} + c$

(xvii) $\int \frac{d x}{x \sqrt{x^{2} - a^{2}}} = \frac{1}{a} {s e c}^{- 1} \frac{x}{a} + c$

(xviii) $\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = l n [x + \sqrt{x^{2} + a^{2}}] + c$

(xix) $\int \frac{d x}{\sqrt{x^{2} - a^{2}}} = l n [x + \sqrt{x^{2} - a^{2}}] + c$

(xx) $\int \frac{d x}{a^{2} - x^{2}} = \frac{1}{2 a} l n |\frac{a + x}{a - x}| + c$

(xxi) $\int \frac{d x}{x^{2} - a^{2}} = \frac{1}{2 a} l n |\frac{x - a}{x + a}| + c$

(xxii) $\int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} {s i n}^{- 1} \frac{x}{a} + c$

(xxiii) $\int \sqrt{x^{2} + a^{2}} d x = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} l n (x + \sqrt{x^{2} + a^{2}}) + c$

(xxiv) $\int \sqrt{x^{2} - a^{2}} d x = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} l n (x + \sqrt{x^{2} - a^{2}}) + c$

(xxv) $\int e^{a x} \cdot s i n b x d x = \frac{e^{a x}}{a^{2} + b^{2}} (a s i n b x - b c o s b x) + c$

(xxvi) $\int e^{a x} \cdot c o s b x d x = \frac{e^{a x}}{a^{2} + b^{2}} (a c o s b x + b s i n b x) + c$

4. Methods of Integration

(i) Substitution or Change of Independent Variable Method:

The integral $I = \int f (x) d x$ can be changed into $\int f (ϕ (t)) ϕ^{'} (t) d t,$ by a suitable substitution i.e., $x = ϕ (t)$ provided $\int f (ϕ (t)) ϕ^{'} (t) d t,$ is easier to integrate.

Some Basic Substitution:

(a) $\int {[f (x)]}^{n} f^{'} (x) d x$ or $\int \frac{f^{'} (x)}{{[f (x)]}^{n}} d x$ put $f (x) = t$ & proceed.

(b) $\int \frac{d x}{x (x^{n} + 1)} n \in N,$ take $x^{n}$ common & put $1 + x^{- n} = t$ .

(d) $\int \frac{d x}{x^{n} {(1 + x^{n})}^{\frac{1}{n}}},$ take $x^{n}$ common and put $1 + x^{- n} = t$ .

(ii) Integration by Parts Method:

$\int u \cdot v d x = u \int v d x - \int (\frac{d u}{d x} \cdot \int v d x) d x$ where $u & v$ are differentiable functions.

Note: While using integration by parts, choose $u$ & $v$ such that $\int v d x & \int (\frac{d u}{d x} \cdot \int v d x) d x$ is simple to integrate. This is generally obtained by keeping the order of $u & v$ as per the order of the letters in ILATE, where I-Inverse function, L-Logarithmic function, A-Algebraic function, T-Trigonometric function & E-Exponential function.

$\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} \cdot f (x) + c$ .

This is an important formula and should be always remembered.

(iii) Partial Fraction Method:

Rational function is defined as the ratio of two polynomials in the form $\frac{P (x)}{Q (x)},$ where $P (x)$ and $Q (x)$ are polynomials in $x$ and $Q (x) \neq 0 .$ If the degree of $P (x)$ is less than the degree of $Q (x)$ , then the rational function is called proper, otherwise, it is called improper. The improper rational function can be reduced to the proper rational functions by a long division process. Thus, if $\frac{P (x)}{Q (x)}$ is improper, then $\frac{P (x)}{Q (x)} = T (x) + \frac{P_{1} (x)}{Q (x)}$ where $T (x)$ is a polynomial in $x$ and $\frac{P_{1} (x)}{Q (x)}$ is a proper rational function. It is always possible to write the integrand as a sum of simpler rational functions by a method called partial fraction decomposition. After this, the integration can be carried out easily using the already known methods.

S. No.	Form of the rational function	Form of the partial fraction
1	$\frac{p x^{2} + q x + r}{(x - a) (x - b) (x - c)}$	$\frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c}$
2	$\frac{p x^{2} + q x + r}{{(x - a)}^{2} (x - b)}$	$\frac{A}{x - a} + \frac{B}{{(x - a)}^{2}} + \frac{C}{x - b}$
3	$\frac{p x^{2} + q x + r}{(x - a) (x^{2} + b x + c)}$ , where $x^{2} + b x + c$ cannot be factorized further i.e. $D < 0$	$\frac{A}{x - a} + \frac{B x + C}{x^{2} + b x + c}$

Note:

In competitive exams, partial fractions are generally found by inspection by noting the following fact: $\frac{1}{(x - α) (x - β)} = \frac{1}{(α - β)} (\frac{1}{x - α} - \frac{1}{x - β})$

It can also be applied to the case when $x^{2}$ or any other function is there in place of x.

5. Integration of Rational Functions

(i) For $\int \frac{d x}{a x^{2} + b x + c}$ or $\int \frac{d x}{\sqrt{a x^{2} + b x + c}}$ or $\int \sqrt{a x^{2} + b x + c} d x$ , express $a x^{2} + b x + c$ in the form of perfect square & then apply the standard results.

(ii) For $\int \frac{p x + q}{a x^{2} + b x + c} d x$ or $\int \frac{p x + q}{\sqrt{a x^{2} + b x + c}} d x$ , express $p x + q = A$ (differential coefficient of denominator) $+ B$ .

(iii) For $\int \frac{x^{2} + 1}{x^{4} + K x^{2} + 1} d x$ or $\int \frac{x^{2} - 1}{x^{4} + K x^{2} + 1} d x$ , where $K$ is any constant. Divide $N^{r} & D^{r}$ by $x^{2},$ then put $x - \frac{1}{x} = t$ or $x + \frac{1}{x} = t$ respectively & proceed.

6. Integration of Irrational Functions

(i) $\int \frac{d x}{(a x + b) \sqrt{p x + q}}$ or $\int \frac{d x}{(a x^{2} + b x + c) \sqrt{p x + q}}$ , put $p x + q = t^{2}$

(ii) $\int \frac{d x}{(a x + b) \sqrt{p x^{2} + q x + r}},$ put $a x + b = \frac{1}{t}, \int \frac{d x}{(a x^{2} + b x + c) \sqrt{p x^{2} + q x + r}}$ , put $x = \frac{1}{t}$

(iii) $\int \sqrt{\frac{x - α}{β - x}} d x$ or $\int \sqrt{(x - α) (β - x)};$ put $x = α {c o s}^{2} θ + β {s i n}^{2} θ$

(iv) $\int \sqrt{\frac{x - α}{x - β}} d x$ or $\int \sqrt{(x - α) (x - β)};$ put $x = α {s e c}^{2} θ - β {t a n}^{2} θ$

(v) $\int \frac{d x}{\sqrt{(x - α) (x - β)}};$ put $x = α - t^{2}$ or $x - β = t^{2}$

Integration Using Trigonometric Identities:

(a) For $\int \frac{d x}{a + b {s i n}^{2} x} o r \int \frac{d x}{a + b {c o s}^{2} x}$ or $\int \frac{d x}{a {s i n}^{2} x + b s i n x c o s x + c {c o s}^{2} x}$ , multiply $N^{r} {& D}^{r}$ by ${s e c}^{2} \times &$ put $t a n x = t$ .

(b) For $\int \frac{d x}{a + b s i n x}$ or $\int \frac{d x}{a + b c o s x}$ or $\int \frac{d x}{a + b s i n x + c c o s x}$ , convert sines & cosines into their respective tangents of half angles and then put tan $\frac{x}{2} = t$ .

(c) For $\int \frac{a c o s x + b s i n x + c}{p c o s x + q s i n x + r} d x$ , express Numerator $(N^{r}) = l (D^{r}) + m \frac{d}{d x} (D^{r}) + n$ & proceed.

(d) To integrate $\int s i n^{m} x c o s^{n} x d x$

If $m$ is odd positive integer, put $c o s x = t$ .

If $n$ is odd positive integer, put $s i n x = t$ .

If $m + n$ is negative even integer, then put $t a n x = t$ .

If $m$ and $n$ are both even positive integer, then use $s i n^{2} x = \frac{1 - c o s 2 x}{2}, c o s^{2} x = \frac{1 + c o s 2 x}{2}$ .

7. Reduction Formula:

A reduction formula is one that enables us to solve an integral problem by reducing it to a problem of solving an easier integral problem, and then reducing that to the problem of solving an easier problem, and so on. Reduction formulae are generated using Integration by Part method.

Some basic reduction formulae are given below. These are not necessarily to be remembered but process of finding these formulae is expected.

(i) $I_{n} = \int x^{n} e^{a x} d x = \frac{x^{n} e^{a x}}{a} - \frac{n}{a} I_{n - 1}$

(ii) $I_{n} = \int s i n^{n} x d x = - \frac{(\sin^{n - 1} x) (\cos x)}{n} - \frac{n - 1}{n} I_{n - 2}$

(iii) $I_{n} = \int x^{a} {(\ln x)}^{n} ⅆ x = \frac{x^{a + 1} {(\ln x)}^{n}}{a + 1} - \frac{n}{a + 1} I_{n - 1}$

(iv) $I_{n} = \int \frac{ⅆ x}{{(x^{2} + a^{2})}^{n}} = \frac{x}{2 (n - 1) a^{2} {(x^{2} - a^{2})}^{n - 1}} - \frac{2 n - 3}{2 (n - 1) a^{2}} I_{n - 1}$

Statement I: Integral of an even function is not always an odd function. Statement II: Integral of an odd function is an even function.

Important Points to Remember in Chapter -1 - Indefinite Integral from Amit M Agarwal Skills in Mathematics for JEE MAIN & ADVANCED INTEGRAL CALCULUS Solutions

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Statement I: Integral of an even function is not always an odd function.

Statement II: Integral of an odd function is an even function.