Check the formula sheet of integration

  1. Basic Integration Formulas
  2. Integral of special functions
  3. Integral by Partial Fractions
  4. Integration by Parts
  5. Other Special Integrals
  6. Area as a sum
  7. Properties of definite integration 

  1 Basic Integration  Formula - Chapter 7 Class 12.JPG

2 Basic Integration  Formula Continued - Chapter 7 Class 12.JPG
3 Integration of some special functions - Chapter 7 Class 12.JPG 4 Integration by partial fractions - Chapter 7 Class 12.JPG 5 Integration by parts - Chapter 7 Class 12.JPG 6 Other Special Integrals - Chapter 7 Class 12.JPG 7 Integral of the form px+q root ax2+bx+c - Chapter 7 Class 12.JPG 8 Area as a sum - Chapter 7 Class 12.JPG 9 Definite Integration Properties - Chapter 7 Class 12.JPG

 

 

 

Basic Formula

  1. ∫x  = x n+1 /n+1  + C
  2. ∫cos x    = sin x  + C
  3. ∫sin x    = -cos x  + C
  4. ∫sec 2 x    = tan x  + C
  5. ∫cosec 2 x    = -cot x  + C
  6. ∫sec x tan x    = sec x  + C
  7. ∫cosec  x cot x    = -cosec x  + C
  8. ∫dx/√ 1- x 2  = sin -1  x  + C
  9. ∫dx/√ 1- x 2  = -cos -1  x  + C
  10. ∫dx/√ 1+ x 2  = tan -1  x  + C
  11. ∫dx/√ 1+ x 2  = -cot -1  x  + C
  12. ∫e  = e + C
  13. ∫a  = a x / log a + C
  14. ∫dx/x √ x 2   - 1= sec -1  x  + C
  15. ∫dx/x √ x 2   - 1= cosec -1  x  + C
  16. ∫1/x    = log |x| + c
  17. ∫tan x    = log |sec x| + c
  18. ∫cot x    = log |sin x| + c
  19. ∫sec x    = log |sec x + tan x| + c
  20. ∫cosec x    = log |cosec x - cot x| + c

Practice Basic Formula questions - Part 1 and Basic Formula questions - Part 2.

Integrals of some special function s

  1.  ∫dx/(x 2   - a 2 ) = 1/2a  log⁑ |(x - a) / (x + a)| + c
  2.  ∫dx/(a 2   - x 2 ) = 1/2a  log⁑ |(a + x) / (a - x)| + c
  3. ∫dx / (x 2   + a 2 ) = 1/a  tan (-1) ⁑ x / a + c
  4. ∫dx / √(x 2   - a 2 ) = log |"x" + √(x 2 -a 2 )| + C

  5. 1.∫dx / √(a 2   - x 2 ) = sin-1 x / a + c

  6. ∫dx / √(x 2 + a 2 ) = log |"x" + √(x 2 + a 2 )| + C

Check Practice Questions

Integrals by partial fractions

  1. (px + q) / ((x - a) (x - b)) = A/(x - a) + B / (x - b)

  2. (px + q) / (x - a) 2  = A/(x - a) + B / (x - a) 2   

  3. (px 2   + qx + r) / (x - a) (x - b) (x - c)  = A / (x - a) + B / (x - b) + C / (x - c)
  4. (px 2 + qx + r) / ((x - a) 2 (x - b) ) = A / (x - a) + B / (x - a) 2 + C / (x - b)
  5. (px 2 + qx + r) / (x - a) (x 2 + bx + c)  = A / (x - a) + (Bx + C) / (x 2 + bx + c)

    Where x 2 + bx + c can not be factorised further.

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Integration by parts

  1. ∫𝒇(𝒙) π’ˆβ‘(𝒙)  𝒅𝒙 = 𝒇(𝒙) ∫π’ˆ (𝒙) 𝒅𝒙− ∫(𝒇 ' (𝒙) ∫π’ˆ(𝒙) 𝒅𝒙) 𝒅𝒙

    To decide first function. We use

    I → Inverse (Example sin (-1)  ⁑x)

    L → Log (Example log ⁑x)

    A → Algebra (Example x 2 , x 3 )

    T → Trigonometry (Example sin 2 x)

    E → Exponential (Example e x )

  2. ∫ex [f (x) + f ′(x)] dx = ∫ex f(x) dx + C

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Other Special Integrals

  1.  ∫√ (x - a 2 ) dx = x / 2 √(x - a 2 ) − a / 2 log |x + √(x - a 2 )| + C 
  2. √( x + a 2 ) dx = x / 2 √(x + a 2 ) + a / 2 log |x +√(x + a 2 )| + C 

  3. √( a - x 2 ) dx = x / 2 √(a 2   - x 2 ) + a / 2 sin 1 x / a + C

Check Practice Questions

 

Integral of the form  ∫ (px+q) √( ax + bx + c dx

We solve this using a specific method.

  1. First we write
         px + q = A (d(√(ax + bx + c))/dx) + B
  2. Then we find A and B
  3. Our equation becomes two seperate identities and then we solve.

 

Some examples are

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Area as a sum

∫a→b f (x)  dx = (b - a)  (lim) (n→∞)  1 / n (f (a) + f (a + h) + f (a + 2h)…+ f (a + (n - 1) h))

Check Practice Questions

Properties of definite integration

  1. P 0 : ∫a→b   f(x) dx = ∫a→b   f(t) dt
  2. P 1 : ∫a→b   f(x) dx = -∫b→a   f(x) dx .In particular, ∫a→a   f(x) dx = 0
  3. P 2 : ∫a→b   f(x) dx = ∫a→c f(x) dx + ∫c→b f(x) dx
  4. P 3 : ∫a→b f(x) dx= ∫a→b   f(a + b - x) dx.
  5. P 4 : ∫0→a f(x)dx = ∫0→a   f(a - x) dx
  6. P 5 : ∫0→2a   f(x) dx = ∫0→a   f(x) dx + ∫0→a f(2a - x) dx
  7. P 6 :  ∫0→2a f(x) = {(2∫0→a   f(x) dx,  if f (2a - x) = f (x) , if f (2a - x) = -f(x))
  8. P 7 :  ∫(-a)→a f(x) = {(2∫0→a f(x) dx,  if f(-x) = f(x), if f ( -x) = -f(x)


Check Practice Questions

 

You can also download the pdf here

 

  1. Chapter 7 Class 12 Integrals
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Transcript

Chapter 7 Class 12 Integration Formula Sheet by teachoo.com Basic Formulae βˆ«β–’γ€–π‘₯𝑛 𝑑π‘₯=γ€— π‘₯^(𝑛+1)/(𝑛+1)+𝐢,π‘›β‰ βˆ’1. π‘ƒπ‘Žπ‘Ÿπ‘‘π‘–π‘π‘’π‘™π‘Žπ‘Ÿπ‘™π‘¦ , βˆ«β–’γ€–π‘‘π‘₯=π‘₯+𝑐〗 βˆ«β–’γ€–π‘π‘œπ‘ β‘π‘₯ 𝑑π‘₯=γ€— sin x + C βˆ«β–’γ€–π‘ π‘–π‘›β‘π‘₯ 𝑑π‘₯=γ€— βˆ’cos x + C βˆ«β–’γ€–π‘ π‘’π‘2⁑π‘₯ 𝑑π‘₯=γ€— tan x + c βˆ«β–’γ€–π‘π‘œπ‘ π‘’π‘2⁑π‘₯ 𝑑π‘₯= γ€—βˆ’cot x + c βˆ«β–’γ€–π‘ π‘’π‘β‘π‘₯ π‘‘π‘Žπ‘›β‘π‘₯ 𝑑π‘₯= γ€—sec x + c βˆ«β–’γ€–π‘π‘œπ‘ π‘’π‘β‘π‘₯ π‘π‘œπ‘‘β‘π‘₯ 𝑑π‘₯= γ€—βˆ’cosec x + c βˆ«β–’γ€–π‘‘π‘₯/√(1 βˆ’ π‘₯^2 )= γ€—sin-1 x + c βˆ«β–’γ€–π‘‘π‘₯/√(1 βˆ’ π‘₯^2 )= 〗– cos-1 x + c βˆ«β–’γ€–π‘‘π‘₯/(1 + π‘₯^2 )= γ€—tan-1 x + c Questions in Ex 7.2 and Ex 7.3 βˆ«β–’γ€–π‘‘π‘₯/(1 + π‘₯^2 )= 〗– cot-1 x + c βˆ«β–’γ€–π‘’^π‘₯⁑𝑑π‘₯=𝑒^π‘₯ γ€— + c βˆ«β–’γ€–π‘Ž^π‘₯ 𝑑π‘₯=π‘Ž^π‘₯/logβ‘π‘Ž γ€— + c βˆ«β–’γ€–π‘‘π‘₯/(π‘₯√(π‘₯^2 βˆ’ 1))= γ€—sec-1 x + c βˆ«β–’γ€–π‘‘π‘₯/(π‘₯√(π‘₯^2 βˆ’ 1))= γ€—βˆ’cosec -1 x + c βˆ«β–’γ€–1/π‘₯ 𝑑π‘₯= γ€—log |π‘₯| + c βˆ«β–’γ€–π‘‘π‘Žπ‘›β‘π‘₯ 𝑑π‘₯=γ€— log |"sec x" | + c βˆ«β–’γ€–π‘π‘œπ‘‘β‘π‘₯ 𝑑π‘₯=γ€— log |"sin x" | + c βˆ«β–’γ€–π‘ π‘’π‘β‘π‘₯ 𝑑π‘₯=γ€— log |"sec x" +tan⁑π‘₯ | + c βˆ«β–’γ€–π‘π‘œπ‘ π‘’π‘ π‘₯ 𝑑π‘₯=γ€— log |"cosec x" βˆ’γ€–cot 〗⁑π‘₯ | + c Integrals of some special functions βˆ«β–’π‘‘π‘₯/(π‘₯^2 βˆ’ π‘Ž^2 ) = 1/2π‘Ž log⁑〖|(π‘₯ βˆ’ π‘Ž)/(π‘₯ + π‘Ž)|+𝑐〗 βˆ«β–’π‘‘π‘₯/(π‘Ž^2 βˆ’ π‘₯^2 ) = 1/2π‘Ž log⁑〖|(π‘Ž + π‘₯)/(π‘Ž βˆ’ π‘₯)|+𝑐〗 βˆ«β–’π‘‘π‘₯/(π‘₯^2 + π‘Ž^2 ) = 1/π‘Ž tan^(βˆ’1)⁑〖π‘₯/π‘Ž+𝑐〗 βˆ«β–’π‘‘π‘₯/√(π‘₯^2 βˆ’ π‘Ž^2 ) = log |"x" + √(π‘₯^2βˆ’π‘Ž^2 )|+ C βˆ«β–’π‘‘π‘₯/√(π‘Ž^2 βˆ’ π‘₯^2 ) = sin-1 π‘₯/π‘Ž+𝑐 βˆ«β–’π‘‘π‘₯/√(π‘₯^2 + π‘Ž^2 ) = log |"x" + √(π‘₯^2+π‘Ž^2 )|+ C Integrals by partial fractions 1. (𝑝π‘₯ + π‘ž)/((π‘₯ βˆ’ π‘Ž)(π‘₯ βˆ’ 𝑏)) = 𝐴/(π‘₯ βˆ’ π‘Ž) + 𝐡/(π‘₯ βˆ’ 𝑏), π‘Ž β‰  b 2. (𝑝π‘₯ + π‘ž)/(π‘₯ βˆ’ π‘Ž)^2 = 𝐴/(π‘₯ βˆ’ π‘Ž) + 𝐡/(π‘₯ βˆ’ π‘Ž)^2 3. (𝑝π‘₯^2 + π‘žπ‘₯ + π‘Ÿ)/(π‘₯ βˆ’ π‘Ž)(π‘₯ βˆ’ 𝑏)(π‘₯ βˆ’ 𝑐) = 𝐴/(π‘₯ βˆ’ π‘Ž) + 𝐡/(π‘₯ βˆ’ 𝑏) + 𝐢/(π‘₯ βˆ’ 𝑐) 4. (𝑝π‘₯^2 + π‘žπ‘₯ + π‘Ÿ)/((π‘₯ βˆ’ π‘Ž)^2 (π‘₯ βˆ’ 𝑏) ) = 𝐴/(π‘₯ βˆ’ π‘Ž) + 𝐡/(π‘₯ βˆ’ π‘Ž)^2 + 𝐢/(π‘₯ βˆ’ 𝑏) 5. (𝑝π‘₯^2 + π‘žπ‘₯ + π‘Ÿ)/(π‘₯ βˆ’ π‘Ž)(π‘₯^2 + 𝑏π‘₯ + 𝑐) = 𝐴/(π‘₯ βˆ’ π‘Ž) + (𝐡π‘₯ + 𝐢)/(π‘₯^2 + 𝑏π‘₯ + 𝑐) Where π‘₯^2+ bx + c can not be factorised further. Integration by parts 1. ∫1▒〖𝒇(𝒙) π’ˆβ‘(𝒙) γ€— 𝒅𝒙=𝒇(𝒙) ∫1β–’π’ˆ(𝒙) π’…π’™βˆ’βˆ«1β–’(𝒇^β€² (𝒙) ∫1β–’π’ˆ(𝒙) 𝒅𝒙) 𝒅𝒙 To decide first function. We use I β†’ Inverse (Example 〖𝑠𝑖𝑛〗^(βˆ’1)⁑π‘₯) L β†’ Log (Example log⁑π‘₯) A β†’ Algebra (Example x2, x3) T β†’ Trignometry (Example sin2 x) E β†’ Exponential (Example ex) 2. βˆ«β–’γ€–π‘’π‘₯ γ€— [𝑓(π‘₯)+𝑓′(π‘₯)] dx = βˆ«β–’γ€–π‘’π‘₯ γ€—f(x) dx + C Other Special Integrals βˆ«β–’βˆš(𝒙^πŸβˆ’π’‚^𝟐 ) 𝒅𝒙 = π‘₯/2 √(π‘₯^2βˆ’π‘Ž^2 ) βˆ’ π‘Ž^2/2 log |π‘₯+√(π‘₯^2βˆ’π‘Ž^2 )| + C βˆ«β–’βˆš(𝒙^𝟐+𝒂^𝟐 ) 𝒅𝒙 = π‘₯/2 √(π‘₯^2+π‘Ž^2 ) + π‘Ž^2/2 log |π‘₯+√(π‘₯^2+π‘Ž^2 )| + C βˆ«β–’βˆš(𝒂^πŸβˆ’π’™^𝟐 ) 𝒅𝒙 = π‘₯/2 √(π‘Ž^2βˆ’π‘₯^2 ) + π‘Ž^2/2 sin^1 π‘₯/π‘Ž + C Limit as a sum ∫1▒𝑓(π‘₯) 𝑑π‘₯ =(π‘βˆ’π‘Ž) (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑓(π‘Ž)+𝑓(π‘Ž+β„Ž)+𝑓(π‘Ž+2β„Ž)…+ 𝑓(π‘Ž+(π‘›βˆ’1)β„Ž)) Properties of definite integration P0 : ∫_π‘Ž^𝑏▒〖𝑓(π‘₯)𝑑π‘₯=∫_π‘Ž^𝑏▒〖𝑓(𝑑)𝑑𝑑=γ€—γ€— P1 : ∫_π‘Ž^𝑏▒〖𝑓(π‘₯)𝑑π‘₯=βˆ’βˆ«_𝑏^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯ γ€—γ€—.In particular,∫_π‘Ž^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯=0 γ€— P2 : ∫_π‘Ž^𝑏▒〖𝑓(π‘₯)𝑑π‘₯=∫_π‘Ž^𝑐▒〖𝑓(π‘₯)𝑑π‘₯ γ€—γ€—+∫_𝑐^𝑏▒𝑓(π‘₯)𝑑π‘₯ P3 : ∫_π‘Ž^𝑏▒〖𝑓(π‘₯)𝑑π‘₯=∫_π‘Ž^𝑏▒〖𝑓(π‘Ž+π‘βˆ’π‘₯)𝑑π‘₯. γ€—γ€— P4 : ∫_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯=∫_0^π‘Žβ–’γ€–π‘“(π‘Žβˆ’π‘₯)𝑑π‘₯ γ€—γ€— P5 : ∫_0^2π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯=∫_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯+ γ€—γ€— ∫_0^π‘Žβ–’π‘“(2π‘Žβˆ’π‘₯)𝑑π‘₯ P6 : ∫24_0^2π‘Žβ–’π‘“(π‘₯) ={β–ˆ(2∫24_0^π‘Žβ–’π‘“(π‘₯)𝑑π‘₯, 𝑖𝑓 𝑓(2π‘Žβˆ’π‘₯)=𝑓(π‘₯) @&0, 𝑖𝑓 𝑓(2π‘Žβˆ’π‘₯)=βˆ’π‘“(π‘₯))─ P6 : ∫_(βˆ’π‘Ž)^π‘Žβ–’π‘“(π‘₯) ={β–ˆ(2∫_0^π‘Žβ–’π‘“(π‘₯)𝑑π‘₯, 𝑖𝑓 𝑓(βˆ’π‘₯)=𝑓(π‘₯) @&0, 𝑖𝑓 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯))─

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
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