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Indefinite Integration - Notes, Concept and All Important Formula

INDEFINITE INTEGRATION

If  f & F are function of \(x\) such that \(F^{\prime}(x)\)\(=f(x)\) then the function \(F\) is called a PRIMITIVE OR ANTIDERIVATIVE OR INTEGRAL of \(\mathrm{f}(\mathrm{x})\) w.r.t. \(\mathrm{x}\) and is written symbolically as \(\displaystyle \int \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\)\(=\mathrm{F}(\mathrm{x})+\mathrm{c} \Leftrightarrow \dfrac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{F}(\mathrm{x})+\mathrm{c}\}\)\(=\mathrm{f}(\mathrm{x})\), where \(\mathrm{c}\) is called the constant of integration.

Note : If \(\displaystyle \int f(x) d x\)\(=F(x)+c\), then \(\displaystyle \int f(a x+b) d x\)\(=\dfrac{F(a x+b)}{a}+c, a \neq 0\)


1. STANDARD RESULTS :

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

(ii) \(\displaystyle \int \dfrac{d x}{a x+b}\)\(=\dfrac{1}{a} \ln|a x+b|+c\)

(iii) \(\displaystyle \int e^{\mathrm{ax}+b} \mathrm{dx}\)\(=\dfrac{1}{\mathrm{a}} e^{\mathrm{ax}+\mathrm{b}}+\mathrm{c}\)

(iv) \(\displaystyle \int a^{p x+q} d x\)\(=\dfrac{1}{p} \dfrac{a^{p x+q}}{\ln a}(a>0)+c\)

(v) \(\displaystyle \int \sin (a x+b) d x\)\(=-\dfrac{1}{a} \cos (a x+b)+c\)

(vi) \(\displaystyle \int \cos (a x+b) d x\)\(=\dfrac{1}{a} \sin (a x+b)+c\)

(vii) \(\displaystyle \int \tan (a x+b) d x\)\(=\dfrac{1}{a} \ln|\sec (a x+b)|+c\)

(viii) \(\displaystyle \int \cot (a x+b) d x\)\(=\dfrac{1}{a} \ln|\sin (a x+b)|+c\)

(ix) \(\displaystyle \int \sec ^{2}(a x+b) d x\)\(=\dfrac{1}{a} \tan (a x+b)+c\)

(x) \(\displaystyle \int \operatorname{cosec}^{2}(a x+b) d x\)\(=-\dfrac{1}{a} \cot (a x+b)+c\)

(xi) \(\displaystyle \int \operatorname{cosec}(a x+b) \cdot \cot (a x+b) d x\)\(=-\dfrac{1}{a} \operatorname{cosec}(a x+b)+c\)

(xii) \(\displaystyle \int \sec (a x+b) \cdot \tan (a x+b) d x\)\(=\dfrac{1}{a} \cdot \sec (a x+b)+c\)

(xiii) \(\displaystyle \int \sec x d x\)\(=\ln |\sec x+\tan x|+c\)

OR \(\displaystyle \int \sec x d x\)\(=\ln \tan \left|\dfrac{\pi}{4}+\dfrac{x}{2}\right|+c\)

(xiv) \(\displaystyle \int \operatorname{cosec} x d x\)\(=\ln|\operatorname{cosec} x-\cot x|+c\)

OR \(\displaystyle \int \operatorname{cosec} x d x\)\(=\ln \left|\tan \dfrac{x}{2}\right|+c\) OR \(-\ln(\operatorname{cosec} x+\cot x)+c\)

(xv) \(\displaystyle \int \dfrac{d x}{\sqrt{a^{2}-x^{2}}}\)\(=\sin ^{-1} \dfrac{x}{a}+c\)

(xvi) \(\displaystyle \int \dfrac{d x}{a^{2}+x^{2}}\)\(=\dfrac{1}{a} \tan ^{-1} \dfrac{x}{a}+c\)

(xvii) \(\displaystyle \int \dfrac{d x}{x \sqrt{x^{2}-a^{2}}}\)\(=\dfrac{1}{a} \sec ^{-1} \dfrac{x}{a}+c\)

(xviii) \(\displaystyle \int \dfrac{d x}{\sqrt{x^{2}+a^{2}}}\)\(=\ln\left[x+\sqrt{x^{2}+a^{2}}\right]+c\)

(xix) \( \displaystyle \int \dfrac{d x}{\sqrt{x^{2}-a^{2}}}\)\(=\ln\left[x+\sqrt{x^{2}-a^{2}}\right]+c\)

(xx) \(\displaystyle \int \dfrac{d x}{a^{2}-x^{2}}\)\(=\dfrac{1}{2 a} \ln \left|\dfrac{a+x}{a-x}\right|+c\)

(xxi) \(\displaystyle \int \dfrac{d x}{x^{2}-a^{2}}\)\(=\dfrac{1}{2 a} \ln \left|\dfrac{x-a}{x+a}\right|+c\)

(xxii) \(\displaystyle \int \sqrt{a^{2}-x^{2}} d x\)\(=\dfrac{x}{2} \sqrt{a^{2}-x^{2}}+\dfrac{a^{2}}{2} \sin ^{-1} \dfrac{x}{a}+c\)

(xxiii) \(\displaystyle \int \sqrt{x^{2}+a^{2}} d x\)\(=\dfrac{x}{2} \sqrt{x^{2}+a^{2}}+\dfrac{a^{2}}{2} \ln \left(x+\sqrt{x^{2}+a^{2}}\right)+c\)

(xxiv) \(\displaystyle \int \sqrt{x^{2}-a^{2}} d x\)\(=\dfrac{x}{2} \sqrt{x^{2}-a^{2}}-\dfrac{a^{2}}{2} \ln \left(x+\sqrt{x^{2}-a^{2}}\right)+c\)

(xxv) \(\displaystyle \int e^{\mathrm{ax}} \cdot \sin b x d x\)\(=\dfrac{e^{\operatorname{ax}}}{a^{2}+b^{2}}(a \sin b x-b \cos b x)+c\)

(xxvi) \( \displaystyle \int e^{a x} \cdot \cos b x d x\)\(=\dfrac{e^{\operatorname{ax}}}{a^{2}+b^{2}}(a \cos b x+b \sin b x)+c\)


2. TECHNIQUES OF INTEGRATION :

(a) Substitution or change of independent variable:

Integral \(\mathrm{I}\)\(=\displaystyle \int \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\) is changed to \(\displaystyle \int \mathrm{f}(\phi(\mathrm{t})) \phi^{\prime}(\mathrm{t})\,\,  \mathrm{dt}\), by a suitable substitution \(\mathrm{x}\)\(=\phi\) (t) provided the later integral is easier to integrate.

Some standard substitution:

(1) \(\displaystyle \int[f(x)]^{n} f^{\prime}(x) d x \quad O R \displaystyle \int \dfrac{f^{\prime}(x)}{[f(x)]^{n}} d x\) put \(f(x)=t \) & proceed.

(2) \(\displaystyle \int \dfrac{d x}{a x^{2}+b x+c},\) \( \displaystyle \int \dfrac{d x}{\sqrt{a x^{2}+b x+c}},\) \( \displaystyle \int \sqrt{a x^{2}+b x+c}\,\,  d x\)

Express \(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\) in the form of perfect square & then apply the standard results.

(3) \(\displaystyle \int \dfrac{\mathrm{px}+\mathrm{q}}{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}} \mathrm{dx},\) \( \displaystyle \int \dfrac{\mathrm{px}+\mathrm{q}}{\sqrt{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}}} \mathrm{dx}\)

Express \(\mathrm{px}+\mathrm{q}\)\(=\mathrm{A}\) (differential coefficient of quadratic term of denominator ) \(+\mathrm{B}\).

(4) \(\displaystyle \int e^{x}\left[f(x)+f^{\prime}(x)\right] d x\)\(=e^{x} \cdot f(x)+c\)

(5) \(\displaystyle \int\left[f(x)+x f^{\prime}(x)\right] d x\)\(=x f(x)+c\)

(6) \(\displaystyle \int \dfrac{d x}{x\left(x^{n}+1\right)} \,\, n \in N\), take \(x^{n}\) common & put \(1+x^{-n}\)\(=t\).

(7) \(\displaystyle \int \dfrac{d x}{x^{2}\left(x^{n}+1\right)^{(n-1) / n}}\,\, n \in N\), take \(x^{n}\) common & put \(1+x^{-n}\)\(=t^{n}\)

(8) \(\displaystyle \int \dfrac{d x}{x^{n}\left(1+x^{n}\right)^{1 / n}}\), take \(x^{n}\) common and put \(1+x^{-n}\)\(=t\).

(9) \(\displaystyle \int \dfrac{d x}{a+b \sin ^{2} x} \quad\) OR \(\quad \displaystyle \int \dfrac{d x}{a+b \cos ^{2} x}\)

OR \(\displaystyle \int \dfrac{d x}{a \sin ^{2} x+b \sin x \cos x+c \cos ^{2} x}\)

Multiply \(\mathrm{N}^{\mathrm{r}} \) & \( \mathrm{D}^{\mathrm{r}}\) by \(\sec ^{2} \mathrm{x} \) & put \(\tan \mathrm{x}\)\(=\mathrm{t}\)

(10) \(\displaystyle \int \dfrac{d x}{a+b \sin x}\) OR \(\displaystyle \int \dfrac{d x}{a+b \cos x}\) OR \(\displaystyle \int \dfrac{d x}{a+b \sin x+c \cos x}\)

Convert sines & cosines into their respective tangents of half the angles, put tan \(\dfrac{x}{2}\)\(=t\)

(11) \(\displaystyle \int \dfrac{a \cdot \cos x+b \cdot \sin x+c}{p \cdot \cos x+q \cdot \sin x+r} \,\, d x\)

Express Numerator \(\left(\mathrm{N}^{r}\right) \equiv \ell\left(\mathrm{D}^{r}\right)+\mathrm{m} \dfrac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{D}^{r}\right)+\mathrm{n} \) & proceed.

(12) \(\displaystyle \int \dfrac{x^{2}+1}{x^{4}+K x^{2}+1} d x\) OR \(\displaystyle \int \dfrac{x^{2}-1}{x^{4}+K x^{2}+1} d x\),

where \(\mathrm{K}\) is any constant. Divide \(N r \) & Dr by \(x^{2}\), then put \(x-\dfrac{1}{x}\)\(=t\) OR \(x+\dfrac{1}{x}\)\(=t\) respectively & proceed

(13) \(\displaystyle \int \dfrac{d x}{(a x+b) \sqrt{p x+q}} \) & \(\displaystyle \int \dfrac{d x}{\left(a x^{2}+b x+c\right) \sqrt{p x+q}} ;\) put \(p x+q\)\(=t^{2}\)

(14) \(\displaystyle \int \dfrac{d x}{(a x+b) \sqrt{p x^{2}+q x+r}}\), put \(a x+b\)\(=\dfrac{1}{t}\);

\(\displaystyle \int \dfrac{d x}{\left(a x^{2}+b x+c\right) \sqrt{p x^{2}+q x+r}}\), put \(x\)\(=\dfrac{1}{t}\)

(15) \(\displaystyle \int \sqrt{\dfrac{x-\alpha}{\beta-x}}\, d x \) OR \(\displaystyle \int \sqrt{(x-\alpha)(\beta-x)} ;\) put \(x=\alpha \cos ^{2} \theta+\beta \sin ^{2} \theta\)

\(\displaystyle \int \sqrt{\dfrac{x-\alpha}{x-\beta}} \, d x\) OR \(\displaystyle \int \sqrt{(x-\alpha)(x-\beta)} ;\) put \(x=\alpha \sec ^{2} \theta-\beta \tan ^{2} \theta\)

\(\displaystyle \int \dfrac{\mathrm{dx}}{\sqrt{(x-\alpha)(x-\beta)}} ;\) put \(x-\alpha=t^{2}\) or \(x-\beta=t^{2}\)

(16) To integrate \(\displaystyle \int \sin ^{m} x \cos ^{n} x d x\).

(i) If \(m\) is odd positive integer put \(\cos x\)\(=t\).

(ii) If \(n\) is odd positive integer put \(\sin x\)\(=t\)

(iii) If \(m+n\) is negative even integer then put \(\tan x=t.\)

(iv) If \(\mathrm{m}\) and \(\mathrm{n}\) both even positive integer then use

\(\sin ^{2} x\)\(=\dfrac{1-\cos 2 x}{2}, \cos ^{2} x\)\(=\dfrac{1+\cos 2 x}{2}\)


(b) Integration by parts:

\(\displaystyle \int \mathrm{u} . \mathrm{v}\, \mathrm{dx}\)\(=\mathrm{u} \displaystyle \int \mathrm{v}\, \mathrm{dx}-\displaystyle \int\left[\dfrac{d \mathrm{u}}{\mathrm{dx} } \cdot \int \mathrm{v}\, \mathrm{dx}\right] \mathrm{dx}\) where \(\mathrm{u} \) & \(\mathrm{v}\) are differentiable functions.

Note : While using integration by parts, choose u & v such that

(i) \(\displaystyle \int \mathrm{v \, dx} \)   &   (ii) \(\displaystyle \int\left[\dfrac{\mathrm{du}}{\mathrm{dx}} \cdot \displaystyle \int \mathrm{v} \, \mathrm{dx}\right] \mathrm{dx}\) is simple to integrate.

This is generally obtained, by keeping the order of \(\mathrm{u}\) & 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.


(c) Partial fraction :

Rational function is defined as the ratio of two polynomials in the form \(\dfrac{\mathrm{P}(\mathrm{x})}{\mathrm{Q}(\mathrm{x})}\), where \(\mathrm{P}(\mathrm{x})\) and \(\mathrm{Q}(\mathrm{x})\) are polynomials in \(x\) and \(Q(x) \neq 0\). If the degree of \(P(x)\) is less than the degree of \(\mathrm{Q}(\mathrm{x})\), 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 long division process. Thus, if \(\dfrac{\mathrm{P}(\mathrm{x})}{\mathrm{Q}(\mathrm{x})}\) is improper, then \(\dfrac{\mathrm{P}(\mathrm{x})}{\mathrm{Q}(\mathrm{x})}\)\(=\mathrm{T}(\mathrm{x})+\dfrac{\mathrm{P}_{1}(\mathrm{x})}{\mathrm{Q}(\mathrm{x})}\), where \(\mathrm{T}(\mathrm{x})\) is a polynomial in \(\mathrm{x}\) and \(\dfrac{\mathrm{P}_{1}(\mathrm{x})}{\mathrm{Q}(\mathrm{x})}\) is 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.

\(\begin{array}{|l|l|l|} \hline \text { S. No. } & \begin{array}{l} \text { Form of the } \\ \text { rational function } \end{array} & \begin{array}{l} \text { Form of the } \\ \text { partial fraction } \end{array} \\ \hline \text { 1. } & \frac{\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}}{(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})(\mathrm{x}-\mathrm{c})} & \frac{\mathrm{A}}{\mathrm{x}-\mathrm{a}}+\frac{\mathrm{B}}{\mathrm{x}-\mathrm{b}}+\frac{\mathrm{C}}{\mathrm{x}-\mathrm{c}} \\ \hline \text { 2. } & \frac{\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}}{(\mathrm{x}-\mathrm{a})^{2}(\mathrm{x}-\mathrm{b})} & \frac{\mathrm{A}}{\mathrm{x}-\mathrm{a}}+\frac{\mathrm{B}}{(\mathrm{x}-\mathrm{a})^{2}}+\frac{\mathrm{C}}{\mathrm{x}-\mathrm{b}} \\ \hline \text { 3. } & \frac{\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}}{(\mathrm{x}-\mathrm{a})\left(\mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}\right)} & \frac{\mathrm{A}}{\mathrm{x}-\mathrm{a}}+\frac{\mathrm{Bx}+\mathrm{C}}{\mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}} \\ \hline \end{array}\)

where \(x^2 + bx + c\) cannot be factorised further

Note: 

In competitive exams, partial fraction are generally found by inspection by noting following fact :

\(\dfrac{1}{(\mathrm{x}-\alpha)(\mathrm{x}-\beta)}\)\(=\dfrac{1}{(\alpha-\beta)}\left(\dfrac{1}{\mathrm{x}-\alpha}-\dfrac{1}{\mathrm{x}-\beta}\right)\)

It can be applied to the case when \(\mathrm{x}^{2}\) or any other function is there in all places of \(x\). 

Example : 

(1) \(\dfrac{1}{\left(x^{2}+1\right)\left(x^{2}+3\right)}\)\(=\dfrac{1}{2}\left(\dfrac{1}{t+1}-\dfrac{1}{t+3}\right) \quad\) take \(x^{2}=t\) 

(2) \(\dfrac{1}{x^{4}\left(x^{2}+1\right)}\)\(=\dfrac{1}{x^{2}}\left(\dfrac{1}{x^{2}}-\dfrac{1}{x^{2}+1}\right)\)\(=\dfrac{1}{x^{4}}-\left(\dfrac{1}{x^{2}}-\dfrac{1}{x^{2}+1}\right)\) 

(3) \(\dfrac{1}{x^{3}\left(x^{2}+1\right)}\)\(=\dfrac{1}{x}\left(\dfrac{1}{x^{2}}-\dfrac{1}{x^{2}+1}\right)\)\(=\dfrac{1}{x^{3}}-\dfrac{1}{x\left(x^{2}+1\right)}\)

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