Important Integration Formulas for JEE Main and Advanced

In this section, students will learn the main indefinite and definite integration formulas as well as some main properties of integration. In general, integration is the reverse operation of differentiation. It is also called antiderivative. The formulas provided here will help students to easily remember them for the exam and score higher marks in the exams. 

Integration Formulas

Indefinite Integration:

  1. If f and g are functions of x such that g’(x) = f(x) then,

∫ f(x)dx = g(x)+c  ⇔ ddx{g(x)+c}=f(x)\frac{d}{dx}\left \{g(x)+c \right \} = f(x) , where c is called the constant of integration.

  1. Standard formula:

(a)  ∫xn dx = xn+1n+1+c\frac{x^{n+1}}{n+1}+c , n ≠ -1

(b) 1xdx=logex+c\int \frac{1}{x}\: dx = \log_{e}\left | x \right |+c

(c) ∫ex dx = ex+c

(d) ∫ax dx = axlogea+c\frac{a^{x}}{\log_{e}a}+c

(e) ∫ sin x dx  = -cos x + c

(f) ∫ cos x dx  = sin x + c

(g) ∫ sec2 x dx  = tan x + c

(h) ∫ cosec2 x dx  = -cot x + c

(h) ∫ sec x tan x dx = sec x + c

(i) ∫ cosec x cot x dx = – cosec x + c

(j) ∫ cot x dx = logsinx+c\log \left | \sin x \right |+c

(k) ∫ tan x dx = logcosx+c-\log \left | \\cos \: x \right |+c

(l) ∫ sec x dx = log secx+tanx+c\left | \\sec \: x + tan \: x\right |+c

(m) ∫ cosec x dx = log cosecxcotx+c\left | \\cosec \: x – \cot \: x\right |+c

(n) 1a2x2  dx=sin1(xa)+c\int \frac{1}{\sqrt{a^{2}-x^{2}}} \; dx = \sin ^{-1}(\frac{x}{a})+c

(o) 1a2x2  dx=cos1(xa)+c\int -\frac{1}{\sqrt{a^{2}-x^{2}}} \; dx = \cos ^{-1}(\frac{x}{a})+c

(p) 1a2+x2  dx=1atan1(xa)+c\int \frac{1}{{a^{2}+x^{2}}} \; dx = \frac{1}{a}\tan ^{-1}(\frac{x}{a})+c

(q) 1a2+x2  dx=1acot1(xa)+c\int -\frac{1}{{a^{2}+x^{2}}} \; dx = \frac{1}{a}\cot ^{-1}(\frac{x}{a})+c

(r) 1xx2a2  dx=1asec1(xa)+c\int \frac{1}{x\sqrt{x^{2}-a^{2}}}\; dx = \frac{1}{a}\sec ^{-1}(\frac{x}{a})+c

(s) 1xx2a2  dx=1acosec1(xa)+c\int -\frac{1}{x\sqrt{x^{2}-a^{2}}}\; dx = \frac{1}{a}\: cosec ^{-1}\left ( \frac{x}{a} \right )+c

If in place of x we have (ax+b), then the same formula is applicable but we must divide by coefficient of x or derivative of (ax+b) i.e. a.

(i) ∫ (ax+b)n dx = (ax+b)n+1a(n+1)+c\frac{(ax+b)^{n+1}}{a(n+1)} \: + c , n ≠ -1

(ii) dxax+b=1a\int \frac{dx}{ax+b} = \frac{1}{a} ln(ax+b) +c

(iii) ∫ eax+b dx = (1/a) eax+b+c

(iv) ∫ apx+q dx = 1papx+qlna+c\frac{1}{p} \frac{a^{px+q}}{\ln a} \: +c , a>0

(v)  ∫ sin (ax+b)dx = -(1/a) cos (ax+b) + c

(vi) ∫ cos (ax+b) dx = (1/a) sin (ax+b) + c

(vii)  ∫ tan (ax+b)dx = (1/a) ln sec (ax+b) + c

(viii) ∫ cot (ax+b)dx = (1/a) ln sin (ax+b) + c

(ix) ∫ sec2 (ax+b)dx = (1/a) tan (ax+b) + c

(x) ∫ cosec2 (ax+b) dx = -(1/a) cot (ax+b) + c

(xi) ∫ sec (ax+b) ⋅ tan (ax+b) dx = (1/a) sec (ax+b) + c

(xii) ∫ cosec (ax+b) ⋅ cot (ax+b) dx = -(1/a) cosec (ax+b) + c

(xiii) ∫ sec x dx = ln (sec x + tan x) + c  or  lntan(π4+x2)\ln tan \left ( \frac{\pi }{4} +\frac{x}{2}\right )

(xiv) ∫ cosec x dx = ln (cosec x – cot x) + c  or  ln tan (x/2) +c  or –ln (cosec x + cot x)+ c

(xv)  dxa2x2=sin1xa+c\int \frac{dx}{\sqrt{a^{2}-x^{2}}} = \sin ^{-1} \frac{x}{a} +c

(xvi) dxa2+x2=12tan1xa+c\int \frac{dx}{a^{2}+x^{2}} = \frac{1}{2} \tan ^{-1} \frac{x}{a} +c

(xvii) dxxx2a2=1asec1xa\int \frac{dx}{\left | x \right |\sqrt{x^{2}-a^{2}}} = \frac{1}{a} \sec ^{-1}\frac{x}{a}

(xviii) dxx2+a2=ln[x+x2+a2]+c\int \frac{dx}{\sqrt{x^{2}+a^{2}}} = \ln [x+\sqrt{x^{2}+a^{2}}]+c

(xix) dxx2a2=ln[x+x2a2]+c\int \frac{dx}{\sqrt{x^{2}-a^{2}}} = \ln [x+\sqrt{x^{2}-a^{2}}]+c

(xx) dxa2x2=12alna+xax+c\int \frac{dx}{{a^{2}-x^{2}}} = \frac{1}{2a}\ln \left | \frac{a+x}{a-x} \right |+c

(xxi) dxx2a2=12alnxax+a+c\int \frac{dx}{{x^{2}-a^{2}}} = \frac{1}{2a}\ln \left | \frac{x-a}{x+a} \right |+c

(xxii) a2x2dx=x2a2x2+a22sin1xa+c\int \sqrt{a^{2}-x^{2}} \: dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin ^{-1}\frac{x}{a}+c

(xxiii) x2+a2dx=x2x2+a2+a22ln(x+x2+a2a)+c\int \sqrt{x^{2}+a^{2}} \: dx = \frac{x}{2}\sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2}\ln \left ( \frac{x+\sqrt{x^{2}+a^{2}}}{a} \right )+c

(xxiv) x2a2dx=x2x2a2a22ln(x+x2a2a)+c\int \sqrt{x^{2}-a^{2}} \: dx = \frac{x}{2}\sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2}\ln \left ( \frac{x+\sqrt{x^{2}-a^{2}}}{a} \right )+c

(xxv) eax.sinbxdx=eaxa2+b2(asinbxbcosbx)+c\int e^{ax}.\sin bx \: dx = \frac{e^{ax}}{a^{2}+b^{2}}\left ( a\, \sin bx-b\cos bx \right )+c

(xxvi) eax.cosbxdx=eaxa2+b2(acosbx+bsinbx)+c\int e^{ax}.\cos bx \: dx = \frac{e^{ax}}{a^{2}+b^{2}}\left ( a\, \cos bx+b\: \sin bx \right )+c

  1. Theorems on integration:

(i) ∫ c f(x) dx = c ∫ f(x) dx

(ii) (f(x)±g(x))dx=f(x)dx±g(x)dx\int (f(x)\pm g(x))\: dx = \int f(x)\: dx\pm g(x)\: dx

(iii) ∫ f(x) dx =  g(x)+c ⇒ ∫ f(ax+b) dx = g(ax+b)a+c\frac{g(ax+b)}{a}+c

  1. Integration by substitutions:

If we substitute f(x) = t, then f’(x) dx = dt

  1. Integration by part:

 ∫( f(x) g(x)) dx = f(x) ∫g(x) dx – (ddx(f(x))(g(x))dx)dx\int\left ( \frac{d}{dx}(f(x))\int (g(x))dx\right ) dx

(i) when you find integral ∫g(x) dx then it will not contain an arbitrary constant.

(ii) ∫g(x) dx should be taken as the same in both terms.

(iii) the choice of f(x) and g(x) is decided by ILATE rule.

  1. Integration of type

dxax2+bx+c\int \frac{dx}{ax^{2}+bx+c}  , dxax2+bx+c\int \frac{dx}{\sqrt{ax^{2}+bx+c}}  , ax2+bx+c  dx\int \sqrt{ax^{2}+bx+c}\; dx

Make the substitution x+b2a=tx+\frac{b}{2a} = t

  1. Integration of type

(1) px+qax2+bx+cdx\int \frac{px+q}{ax^{2}+bx+c}\: dxpx+qax2+bx+cdx\int \frac{px+q}{\sqrt{ax^{2}+bx+c}}\: dx , (px+q)ax2+bx+c  dx\int (px+q)\sqrt{ax^{2}+bx+c}\; dx

Make the substitution x+b2a=tx+\frac{b}{2a} = t , then split the integral as sum of two integrals one containing the linear term and the other containing constant term.

8. Integration of trigonometric functions

(1) dxa+bsin2x\int \frac{dx}{a+b\: \sin ^{2}x}  or dxa+bcos2x\int \frac{dx}{a+b\: \cos ^{2}x}  or dxasin2x+bsinxcosx+ccos2x\int \frac{dx}{a \sin ^{2}x+b \sin x\cos x+c\cos ^{2}x}  put tan x = t.

(2) dxa+bsinx\int \frac{dx}{a+b\sin x}  or dxa+bcosx\int \frac{dx}{a+b\cos x}  or 

dxa+bsinx+ccosx\int \frac{dx}{a+b\sin x+c \cos x}  put tan (x/2) = t.

(3) acosx+bsinx+clcosx+msinx+ndx\int \frac{a\cos x+ b\sin x+c}{l\cos x+ m\sin x+n}dx

Express Nr = A(Dr)+Bddx(Dr)+cA(Dr)+B\frac{d}{dx}(Dr)+c  and proceed.

  1. Integration of type ∫sinmx. cosnx dx 

Case 1. If m and n are even natural numbers then express sinmx cosnx in the terms of sines and cosines of multiples of x by using trigonometric results or De’ Moivere’s theorem.

Case 2. If m is an odd natural number then put cos x = t.

If n is an odd natural number then put sin x = t.

If both m and n are odd natural numbers then put either sin x = t or cos x = t.

Case 3. When m+n is a negative even integer then put tan x = t.

  1. Integration of type  x2±1x4+Kx2+1dx\int \frac{x^{2}\pm 1}{x^{4}+Kx^{2}+1}\: dx where K is any constant.

Divide numerator and denominator by x2 and put x±1x=tx\pm \frac{1}{x} = t

  1. Integration of type dx(ax+b)px+q\int \frac{dx}{(ax+b)\sqrt{px+q}} or dx(ax2+bx+c)px+q\int \frac{dx}{(ax^{2}+bx+c)\sqrt{px+q}} :

Put px+q = t2

  1. Integration of type

(i) dx(ax+b)px2+qx+r\int \frac{dx}{(ax+b)\sqrt{px^{2}+qx+r}} : put ax+b = 1t\frac{1}{t}

(ii) dx(ax2+b)px2+q\int \frac{dx}{(ax^{2}+b)\sqrt{px^{2}+q}} : put x = 1t\frac{1}{t}

  1. Integration of type

(i) xαβxdx\int \sqrt{\frac{x-\alpha }{\beta -x}} \: dx or (xα)(βx)\int \sqrt{(x-\alpha )(\beta -x)} : put x = αcos2θ+βsin2θ\alpha \cos ^{2}\theta +\beta \sin ^{2}\theta

(ii) xαxβdx\int \sqrt{\frac{x-\alpha }{x -\beta }} \: dx or (xα)(xβ)\int \sqrt{(x-\alpha )(x-\beta )} : put x = αsec2θβtan2θ\alpha \sec ^{2}\theta -\beta \tan ^{2}\theta

(iii) dx(xα)(xβ)\int \frac{dx}{\sqrt{(x-\alpha )(x-\beta )}} : put x-α = t2 or x-β = t2

  1. Reduction formula of ∫ tann x dx, ∫ cotn x dx, ∫ secn x dx, ∫ cosecn x dx:

(i) If In = ∫ tann x dx, then In = tann1xn1In2\frac{tan^{n-1}x}{n-1}-I_{n-2}

(ii) If In = ∫ cotn x dx, then  In = cotn1xn1In2-\frac{\cot ^{n-1}x}{n-1}-I_{n-2}

(iii) If In = ∫ secn x dx, then In = tanxsecn2xn1+n2n1In2\frac{\\tan x \: \sec ^{n-2}x}{n-1}+\frac{n-2}{n-1}\: I_{n-2}

(iv) If In = ∫ cosecn x, then  In = cotx  cosecn2xn1+n2n1In2\frac{\cot x \; cosec^{n-2}x}{n-1}+\frac{n-2}{n-1}I_{n-2}

Definite Integration:

Properties of definite integral:

(1) abf(x)dx=abf(t)dt\int_{a}^{b} f(x)dx = \int_{a}^{b}f(t)dt

(2) abf(x)dx=baf(x)dx\int_{a}^{b} f(x)dx = -\int_{b}^{a}f(x)dx

(3) abf(x)dx=acf(x)dx+cbf(x)dx\int_{a}^{b} f(x)dx = \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx

(4) aaf(x)dx=0a(f(x)+f(x))dx\int_{-a}^{a} f(x)dx = \int_{0}^{a} (f(x)+f(-x))dx = {20af(x)dx,f(x)=f(x)0,f(x)=f(x)\left\{\begin{matrix} 2\int_{0}^{a}f(x)dx\: ,&f(-x) = f(x) \\ 0\: ,& f(-x) = -f(x) \end{matrix}\right.

(5) abf(x)dx=abf(a+bx)dx\int_{a}^{b} f(x)dx = \int_{a}^{b}f(a+b-x)dx

(6) 0af(x)dx=0af(ax)dx\int_{0}^{a} f(x)dx = \int_{0}^{a}f(a-x)dx

(7) 02af(x)dx=0a(f(x)+f(2ax))dx\int_{0}^{2a} f(x)dx = \int_{0}^{a}(f(x)+f(2a-x))dx = {20af(x)dx,f(2ax)=f(x)0,f(2ax)=f(x)\left\{\begin{matrix} 2\int_{0}^{a}f(x)dx\: ,&f(2a-x) = f(x) \\ 0\: ,& f(2a-x) = -f(x) \end{matrix}\right.

(8) If f(x) is a periodic function with period T, then 

(i) 0nTf(x)dx=n0Tf(x)dx,\int_{0}^{nT}f(x)dx = n\int_{0}^{T}f(x)dx ,, n∈Z .

(ii) 0a+nTf(x)dx=n0Tf(x)dx,\int_{0}^{a+nT}f(x)dx = n\int_{0}^{T}f(x)dx ,, n∈Z, a∈R .

(iii) mtnTf(x)dx=(nm)0Tf(x)dx,\int_{mt}^{nT}f(x)dx = (n-m)\int_{0}^{T}f(x)dx ,, m, n∈Z.

(iv) nta+nTf(x)dx=0af(x)dx,\int_{nt}^{a+nT}f(x)dx = \int_{0}^{a}f(x)dx ,, n∈Z, a∈R .

(9) If ψ(x)f(x)ϕ(x)foraxb thenabψ(x)dxabf(x)dxabϕ(x)\psi (x)\leq f(x)\leq \phi (x) for a\leq x\leq b\ then \int_{a}^{b}\psi (x)dx \leq \int_{a}^{b}f(x)dx\leq \int_{a}^{b}\phi (x)

(10) If m≤ f(x) ≤ M for a≤x≤b, then m(b-a) ≤ abf(x)dxM(ba)\int_{a}^{b}f(x) dx\leq M(b-a)

(11) abf(x)dxabf(x)dx\left | \int_{a}^{b}f(x)dx \right |\leq \int_{a}^{b}\left | f(x) \right |dx

(12) If f(x) ≥0 on [a,b] then abf(x)dx0\int_{a}^{b}f(x)dx\geq 0

Leibnitz Theorem:

If F(x) = g(x)h(x)f(t)dt\int_{g(x)}^{h(x)}f(t)dt, then dF(x)dx=\frac{dF(x)}{dx} = h’(x)f(h(x))-g’(x)f(g(x))

Definite integrals as a limit of sum:

(i) abf(x)dx=limnr=0n1hf(a+rh)=limnr=0n1(ban)f(a+(ba)rn)\int_{a}^{b}f(x)dx=\lim_{n\to\infty }\sum_{r=0}^{n-1}h\: f(a+rh) = \lim_{n\to\infty }\sum_{r=0}^{n-1}\left ( \frac{b-a}{n} \right )f\left ( a+\frac{(b-a)r}{n} \right )

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