# CBSE Class 12 Maths Chapter-7 Integrals Formula

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$\textbf{Integration}$ – $\\$ $\\$The process of finding the function $f (x)$ whose differential coeffiicient w.r.t. $‘x’$, denoted by$\\$ $F (x)$ is given, is called the integration of $f (x)$ w.r.t. x and is written as $\ \ \int F(x) \mathrm{d}x =f(x)$ $\\$ Thus, integration is an inverse process of differentiation or integration is anti of differentiation.$\\$ The differential coefficient of a constant is zero. Thus if c is an arbitrary constant independent of x. then$\\$ $\dfrac{\mathrm{d}}{\mathrm{d}x}[f(x)+c] \ \ \$ Thus $\ \ \ \int F(x)\mathrm{d}x=f(x)+c$$\\ The arbitrary constant c is called the constant of integration.\\ \\ \\ \\ \\ 3. \textbf{Trigonometrical transformations} –\\ \\ For the integration of the trigonometrical products such as\\$$\sin ^2 x, \cos ^2 x, \sin ^3 x, \cos ^3 x, \sin ax \cos bx$ etc.$\\$they are expressed as the sum or difference of the sines$\\$ and cosines of multiples of angles.$\\$
$\\ \\ $$4.\textbf{Integration of Some Special Integrals }–\\ \\ (a)For \textbf{\int} \frac{\mathrm d x}{ax^2+bx+c},\int \frac{\mathrm d x}{\sqrt{ax^2+bx+c}} and \\ \int \sqrt{ax^2+bx+c \mathrm d x}$$\\$$ax^2+bx+c=a[x^2+\frac{b}{x}x+\frac{c}{a}]\\a[(x+\frac{b}{2a})^2+\frac{c}{a}-\frac{b^2}{4a^2}] =\\ a[(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a^2}]$$\\$Put $x+\frac{b}{2a}=t,\therefore \mathrm d x=\mathrm d t,\frac{4ac-b^2}{4a^2}=\pm k^2,\\ax^2+bx+c \ \text{changes to }t^2+k^2,t^2-k^2 or k^2-t^2$$\\ (b)For \int \frac{(px+q)\mathrm dx}{ax^2+bx+c},\\ \int\frac{(px+q)\mathrm dx}{ \sqrt{ax^2+bx+c}}\int (px+q) \sqrt{(ax^2+bx+c)}\mathrm dx$$\\$Put$px+q=A\frac{\mathrm d}{\mathrm dx}(ax^2+bx+c)+B$$\\Compare the two sides and find the value of A and B.\\Thus \int \frac{px+q}{ax^2+bx+c}\mathrm dx=\int\frac{A\frac{\mathrm d}{\mathrm dx}(ax^2+bx+c)+B}{(ax^2+bx+c)}\\=A\int \frac{\frac{\mathrm d}{\mathrm dx}(ax^2+bx+c)}{(ax^2+bx+c)}\mathrm dx+B\int \frac{\mathrm dx}{(ax^2+bx+c)}$$\\$Similarly$\int\frac{px+q}{\sqrt{ax^2+bx+c}}\mathrm dx= \\ A\int \frac{\frac{\mathrm d}{\mathrm dx}(ax^2+bx+c)}{(ax^2+bx+c)}\mathrm dx+B\int \frac{\mathrm dx}{(ax^2+bx+c)}$$\\same as do \int (px+q)\sqrt{ax^2+bx+c}\mathrm dx.$$\\$
(c)For $\int \frac{\mathrm dx}{(x+k)\sqrt{ax^2+bx+c}}$put $x+k=\frac{1}{t}$$\\ (d)For \int \frac{\mathrm dx}{\sqrt{(x-\alpha)(x-\beta)}},\int \sqrt{\frac{x-\alpha}{\beta -x}}\mathrm dx$$\\$ $\int \sqrt{(x-\alpha)(x-\beta)}\mathrm dx,$Put $x=\alpha \cos^2 \theta+\beta \sin^2 \theta $$\\ (e)For\int \frac{\mathrm dx}{a+b \cos x},\int \frac{\mathrm dx}{a+b \sin x},\\ \int \frac{\mathrm dx}{a+b \cos x + c \sin x},\\ \sin x=(2 \tan \frac{x}{2})/(1+\tan^2\cfrac{x}{2}),\\ \cos x=(1-\tan \frac{x}{2})/(1+\tan^2 \frac{x}{2}) then put \tan \frac{x}{2}=t$$\\$ (f)For $\int \frac{p \cos x+ q \sin x}{a+b \cos x+ b \sin x}\mathrm dx$$\\ Put p \cos x+q \sin x=A(a+b \cos x+ b \sin x)+B differential of (a+ b\cos x + b \sin x)+C$$\\$$A,B and C can be calculated by equating the coefficients of \cos x. \sin x and the constant terms.\\ 5.\textbf{Integration by parts}$$\\$ $\\[0.5em]$ $\int u.v \mathrm dx=u.\int v \mathrm dx-\mathrm [\frac{\mathrm du}{\mathrm dx}.\int v \mathrm dx]\mathrm dx$$\\i.e., the integral of the product of two functions \\= (first function) × (Integral of the second function - \\Integral of {(dfferential of first function) x (Integral of second function)}\\ This formula is called integration by parts.\\[0.7em] 6.\textbf{Partial Integration}-\\[0.5em]To evaluate\int \frac{P(x)}{Q(x)}\mathrm dx$$\\[0.3em]$The rational functions which we shall consider here for integration purposes will be those whose denominators can be factorised into linear and quadratic factors. If$\ \ \frac{P(x)}{Q(x)} \ \$ is improper fraction, i.e., $\\[0.3em]$degree of numerator is equal or greater than the degree of denominator..$\\$Then first we reduce in proper rational function as$\ \frac{P(x)}{Q(x)}=T(x)+\frac{P_1(x)}{Q(x)}$ where $\ T(x) \$is a polynomial in x$\\[0.3em]$and$\frac{P_1(x)}{Q(x)}\$is a proper rational function.$\\$After this, the integration can be carried out easily using the already known methods.$\\[0.3em]$ The following Table 7.1 indicates the types of simpler partial fractions$\\$ that are to be associated with various kind of rational functions.$\\$
Where $x ^2 + bx + c$ can not be factorised further$\\$ In the above table, $A, B$ and $C$ are real numbers to be determined suitably.$\\[0.7em]$
$\\$$7.\text{Definite Integral}-\\[0.5em] \\The definite integral of f(x) between the limits a to b i.e. in the interval [a,b] is denoted by \int_a^b f(x)\mathrm dx and is defined as follows.\\ \int_a^b f(x)\mathrm dx = [F(x)]_a^b=F(b)-F(a) where \int f(x)\mathrm d x=F(x)$$\\[0.7em]$
$\\$$8.\text{General Properties of Definite Integrals –}$$\\$$\\[0.5em] Prop.I. \int_a^bf(x)\mathrm dx=\int_a^b f(t) \mathrm dt$$\\[0.3em]$ Prop.II. $\int_a^b f(x)\mathrm dx =-\int_b^af(x)\mathrm dx$$\\[0.3em] Prop.III. \int _a^b f(x)\mathrm dx=\int _a^b f(x)\mathrm dx$$\\[0.3em]$ $\ \ \ \ \ \ \ \ \ \ +\int_c^b f(x)\mathrm dx \text{where } \ a < c < b$$\\[0.3em] Prop.IV.\int_a^bf(x)\mathrm dx=\int_a^b f(a+b-x)\mathrm dx$$\\[0.3em]$ In particular $\int_0^af(x) \mathrm dx=\int_0^a f(a-x) \mathrm dx$$\\[0.3em] Prop.V \int_0^2a f(x)\mathrm dx$$\\[0.3em]$ Prop.VI $\int_{-a}^a f(x)\mathrm dx =2\int_0^a f(x)\mathrm dx,$$\\[0.3em] if f(x) is even function\\[0.3em] \int_{-a}^a f(x)\mathrm dx=0, if f(x) is odd function \\[0.3em] Prop.VII. \int _0^{2a} f(x)\mathrm dx =2\int_0^af(x) \mathrm dx+\int_0^a f(2a-x)\mathrm dx$$\\$Prop.VIII.$\int_0^{2a }f(x)\mathrm dx=2\int_0^a f(x)\mathrm dx,$ if $f(2a-x)=f(x)$$\\[0.3em] \int_0^{2a}f(x)\mathrm dx=0, if f(2a-x)=-f(x)$$\\[0.7em]$
$9. \text{Definite Integral as the limit of a sum}$$\\[0.5em] \int_a^b f(x)\mathrm dx =\lim \limits _{h \to 0 } h[f(a)+f(a+h) +f(a+2h)\\+....+f < a+(n-1)h)]$$\\[0.3em]$or $\int_a^bf(x)\mathrm{d}x =\lim\limits_{h \to 0}h[f(a+h)+f(a+2h)+f(a+3h)+\\.....+f(a+nh)]$ where , $\ \ \ \ \ \ h=\frac{b-a}{n}$$\\[0.3em] \frac{\mathrm d}{\mathrm dx}\int_{u(x)}^{v(x)} f(t) \mathrm dt$$\\[0.3em]$ $= f{v(x)}\frac{\mathrm d}{\mathrm dx}v(x)-f{u(x)}\frac{\mathrm d}{\mathrm dx}u(x)$$\\[0.3em] this rule is called leibnitz’s is Rule.\\[0.7em] \textbf{CONNECTING CONCEPTS}$$\\[0.7em]$ 1. Integration is an operation on function$\\[0.5em]$ 2.$\int [k_1f_1(x)+k_2f_2(x)+.....+k_nf_n(x)]\mathrm dx$$\\[0.5em] =k_1\int f_1(x)\mathrm dx +k_2 \int f_2(x)\mathrm d x +.....+k_n\int f_n(x)\mathrm dx$$\\[0.5em]$ 3. All functions are not integrable and the integral of a function is not unique.$\\[0.5em]$ 4. If a polynomial function of a degree n is integrated we get a polynomial of degree $n + 1$$\\[0.7em] 4.\text{Integration by using standard formulae –}$$\\[0.5em]$
$1.\int k\mathrm dx=k x+c,k \ \text{is constant}$$\\[0.4em] 2.\int k f(x)\mathrm d x=k\int f(x)\mathrm dx +c$$\\[0.4em]$ $3.\int (f_1(x)\pm f_2(x))\mathrm dx =\\ \ \ \ \ \int f_1(x)\mathrm dx \pm \int f_2(x)\mathrm dx+c$$\\[0.4em] 4. \int x^n\mathrm dx = \frac{x^{n+1}}{n+1}+c(n\neq -1)$$\\[0.4em]$ $5. \int \frac{1}{x}\mathrm dx = \log_e|x|+c$$\\[0.4em] 6. \int a^x\mathrm dx=\frac{a^x}{\log_ea}+c,a>0$$\\[0.4em]$ $7. \int e^x \mathrm dx=e^x+c$$\\[0.4em] 8.\int \sin x \mathrm dx=-\cos x+c$$\\[0.4em]$ $9. \int \cos x \mathrm dx =\sin x+c $$\\[0.4em] 10. \int \sec^2 x \mathrm dx =\tan x+c$$\\[0.4em]$ $11. \int \csc^2 x \mathrm dx=-\cot x+c$$\\[0.4em] 12. \int \sec x\tan x\mathrm dx=\sec x+c$$\\[0.4em]$ $13.\int \csc x \cot x \mathrm dx =-\csc x+c$$\\[0.4em] 14.\int \tan x \mathrm dx =\log|\sec x|+c=-\log|\cos x|+c$$\\[0.4em]$ $15. \int \cot x \mathrm dx =\log |\sin x|+c$$\\[0.4em] 16 \int \sec x \mathrm dx =\log|\sec x+\tan x|+c$$\\[0.4em]$ $17.\int \csc x \mathrm dx =\log|\csc x-\cot x|+c$$\\[0.4em] 18. \int \frac{1}{\sqrt{1-x^2}} \mathrm dx =\sin^{-1} +c or -\cos^{-1}x+c$$\\[0.4em]$
$16.\int \frac{1}{1+x^2}\mathrm dx =\tan^{-1}x+c \ or -\cot^{-1}x+c $$\\[0.4em] 17.\int \frac{1}{x\sqrt{x^2-1}} \mathrm dx =\sec^{-1}x+c \ \ or -\csc^{-1}x+c$$\\[0.4em]$ $18.\int \frac{\mathrm dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}(\frac{x}{a})+c$$\\[0.4em] 19. \int \frac{\mathrm dx }{x^2 - a^2}=\frac{1}{2a}\log|\frac{x-a}{x+a}| + c,x>a$$\\[0.4em]$ $20.\int \frac{\mathrm dx}{a^2 -x^2}=\frac{1}{2a} \log | \frac{a+x}{a-x} |+c,x < a $$\\[0.4em] 21.\int \frac{\mathrm dx}{\sqrt{a^2-x^2}}=\sin^{-1}(\frac{x}{a})+c$$\\[0.4em]$ $22.\int \frac{\mathrm dx}{\sqrt{x^2+a^2}} =\log x+\sqrt{a^2+x^2}+c$$\\[0.4em] 23.\int \frac{\mathrm dx}{\sqrt{x^2-a^2}}=\log x+\sqrt{x^2-a^2}+c$$\\[0.4em]$ $24. \int \frac{\mathrm dx}{x\sqrt{x^2-a^2}} =\frac{1}{a}\sec^{-1}(\frac {x}{a})+c$$\\[0.4em] 25.\int \sqrt{a^2-x^2}\mathrm dx =\frac{x}{2}\sqrt{a^2-x^2}+\frac{1}{2}a^2\sin^{-1}(\frac{x}{a})+c$$\\$
$26. \int \sqrt{x^2+a^2}\mathrm dx= \\ \ \ \ \ \ \ \ \ \ \frac{x}{2}\sqrt{x^2+a^2}+\frac{1}{2}a^2 \log|x+\sqrt{x^2+a^2}|+c$$\\ 27. \int \sqrt{x^2-a^2}\mathrm dx=\\ \ \ \ \ \ \ \frac{x}{2}\sqrt{x^2-a^2}-\frac{1}{2}a^2 \log x+\sqrt{x^2-a^2}+c$$\\$ $28. \int e ^x[f(x)+f'(x)]\mathrm dx =e^x f(x)+c$$\\ 29. \textbf{ Use of Trigonometric Identities in Integration.}$$\\[0.5em]$ $(i)\sin^2 x=\frac{1-\cos 2x}{2},\cos^2 x=\frac{1+\cos 2x}{2}$$\\ (ii)\sin^3 x=\frac{3 \sin x-\sin 3x}{4},\cos^3 x=\frac{3 \cos x+\cos 3 x}{4}$$\\$ $(iii) 2\sin A \cos B=\sin(A+B)+\sin(A-B)\\ 2 \cos A \sin B= \sin(A+B)-\sin(A-B)\\ 2\cos A \cos B=\cos(A+B)+\cos(A-B)\\ 2\sin A \sin B=\cos(A-B)+\cos(A+B)$$\\ (iv) \sin x=2\sin(\frac{x}{2}).\cos(\frac{x}{2})$$\\$
$30.(i) 1+2+3+.....+n=\frac{n(n+1)}{2}\\ (ii)1^2+2^2+3^2+....+n^2=\frac{n(n+1)(2n-1)}{6}\\ (iii)1^3+2^3+3^3+....n^3=[\frac{n(n+1)}{2}]^2$$\\ (iv ) a+(a+d)+(a+2d)+....+[a+(n-1)d]\\ =\frac{n}{2}[2a+(n-1)d]$$\\$ $(v) a+ar+ar^2+....+ar^{n+1}=\frac{a(r^n-1)}{r-1}$$\\ 2.\textbf{Integration by Substitution} \\ \\ \\ (a) \textbf{To evaluate the integral \int f(ax+b)\mathrm{d}x}$$\\$ Put $\ ax+b=t,$ so that $a\mathrm{d}x =\mathrm{d}t \ \$ i.e.,$\ \ \mathrm{d}x=\frac{1}{a}\mathrm{d}t$$\\ \int f(ax+b)\mathrm{d}x=\int f(t).\frac{1}{a}\mathrm{d}t=\frac{1}{a}F(t),$$\\$ where $\int f(t)\mathrm{d}t=F(t)=F(ax+b)$$\\ If a function is not in some suitable form to find the integration, then we transform it into some suitable form by changing the independent variable x to t by substituting x = g (t).$$\\$ Consider $\ \ \ \ I=\int f(x)\mathrm{d}x$$\\ Put \ \ \ \ x=g(t), so that \frac{\mathrm{d}x}{\mathrm{d}t}=g'(t)$$\\$ We write $\ \ \ \ \mathrm{d}t=g'(t)\mathrm{d}t$$\\ Thus \ \ \ \ I=\int f(x).\mathrm{d}x=\int f(g(t))g'(t)\mathrm{d}t$$\\$ But it is very important to guess, what will be the useful substitution.$\\$ (b) $\int \frac{f'(x)}{f(x)}\mathrm{d}x=\log f(x)+c$$\\ (c) \int [f(x)]^n f'(x) \mathrm d x=f(x)^{n+1}/(n+1)+c$$\\$ (d) Some important substitutions$\\$ function $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \$Substitutions$\\$ $\sqrt{a^2-x^2}$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \$$x=a \sin \theta or x= a \cos \theta $$\\ \sqrt{a^2+x^2}$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$x=a \tan \theta \\ \sqrt{x^2-a^2}$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$$x=a \sec \theta$
$\begin{array}{|c|c|c|c|} \hline \textbf{S.No}& \text{Form of the rational function} & \text{Form of the partial fraction} \\ \hline 1. &\frac{px-q}{(x-a)(x-b)},a\neq b&\frac{A}{x-a}+\frac{B}{x-b}\\ \hline 2. &\frac{px+q}{(x-a)^2} &\frac{A}{x-a}+\frac{B}{(x-b)^2} \\ \hline 3.&\frac{px^2+qx+r}{(x-a)(x-b)(x-c)} &\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}\\ \hline 4.& \frac{px^2+qx+r}{(x-a)^2(x-b)}& \frac{A}{x-a}+\frac{B}{(x-b)^2}+\frac{C}{x-b}\\ \hline 5.& \frac{px^2+qx+r}{(x-a)(x^2+bx+c)} & \frac{A}{x-a}+\frac{Bx+c}{x^2+bx+c}\\ \hline \end{array}$