Class 12 Maths · Chapter 7
Short answer:
Class 12 Maths Chapter 7 Integrals NCERT ka sabse bada chapter hai — 2026-27 rationalised syllabus me isme 22 solved exercise-style questions cover kiye gaye hain: substitution method, trigonometric identities wale integrals, standard forms jaise 1/(x²+a²), partial fractions, integration by parts (ILATE), definite integrals via fundamental theorem, aur definite integrals ki properties (jaise ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a−x)dx). Har question me poora step-by-step working diya hai, indefinite integrals me +C ke saath.
Integration, differentiation ka exact ulta process hai — is chapter me har technique isi ek idea ke around ghoomti hai. Class 11 me hum function ko differentiate karna seekh chuke the; ab yahan reverse problem solve karte hain — agar derivative pata hai to original function (antiderivative) kaise dhoondo. Yeh chapter bada isliye hai kyunki koi ek universal method nahi hai — alag-alag function types ke liye alag technique chahiye: seedha formula, substitution, trig identity, partial fractions, ya by-parts. Board exam me is chapter se sabse zyada weightage aata hai, isliye har technique ko practice ke through pakka karna zaroori hai.
Chapter 7 Summary — 5 Minute Revision
1. Integration as Inverse of Differentiation
Agar d/dx [F(x)] = f(x), to hum kehte hain F(x), f(x) ka antiderivative (indefinite integral) hai, aur likhte hain:
∫f(x)dx = F(x) + C
Jahan C arbitrary constant hai (constant of integration). Ek function ke infinite antiderivatives ho sakte hain — sab ek dusre se ek constant ke farak se differ karte hain. Isi wajah se indefinite integral hamesha "+C" ke saath likhna zaroori hai — geometrically, y = F(x)+C ki saari curves ek dusre ki vertical translate hain, aur kisi bhi point x = a par sabki tangent slope same hoti hai (= f(a)) kyunki dF/dx same rehta hai.
2. Geometrical Interpretation
∫f(x)dx = F(x) + C represents a family of curves. Different C values → parallel curves jo x-axis ke parallel translate hoti hain. Kisi bhi vertical line x = a in sab curves ko jahan-jahan cut karti hai, wahan tangent slopes equal hoti hain (dono ka derivative f(a) hai).
3. Standard Integrals — Direct Formulas
Basic power rule, exponential, trigonometric integrals seedhe formula se solve hote hain (poori list "formulas" table me).
∫xⁿdx = xⁿ⁺¹/(n+1) + C (n ≠ −1), ∫1/x dx = log|x| + C
4. Integration by Substitution
Jab integrand ke andar ek function aur uski derivative dono dikhein, tab substitution use karo: agar t = g(x), to dt = g'(x)dx, aur integral simpler ban jaata hai t ke terms me. Substitution ke baad hamesha wapas original variable me convert karna mat bhoolo.
5. Integration Using Trigonometric Identities
sin²x, cos²x, sin³x jaise powers ko direct integrate nahi kiya ja sakta — pehle double-angle ya product-to-sum identities use karke simplify karo, phir integrate karo.
sin²x = (1 − cos2x)/2, cos²x = (1 + cos2x)/2
6. Integrals of Particular Functions (Standard Forms)
Kuch specific forms itni common hain ki unke ready-made formulas yaad rakhne padte hain — jaise dx/(x²±a²), dx/√(a²−x²), dx/(ax²+bx+c). In forms ko complete the square karke standard shape me convert karke formula lagate hain.
7. Integration by Partial Fractions
Jab integrand ek rational function P(x)/Q(x) ho aur degree(P) ≥ degree(Q) na ho, to use simpler fractions ke sum me todo (denominator ke factors ke basis par), fir har fraction ko alag integrate karo.
8. Integration by Parts
Do functions ke product ko integrate karne ke liye ILATE rule follow karo — Inverse, Logarithmic, Algebraic, Trigonometric, Exponential — jo function is order me pehle aaye use "u" (first function) maano, baaki "v" (second function).
∫u·v dx = u∫v dx − ∫[u'·∫v dx]dx
9. Some Special Integrals
√(x²+a²), √(x²−a²), √(a²−x²) wale integrals ke liye derived special formulas hain (in ka use by-parts se hi derive hota hai, par ready formula yaad rakhna faster hai — dekho formulas table).
10. Definite Integrals
Definite integral ∫ₐᵇ f(x)dx ek fixed number deta hai (area under curve, sign ke saath). Fundamental Theorem of Calculus kehta hai: agar F, f ka antiderivative hai, to
∫ₐᵇ f(x)dx = F(b) − F(a)
"As limit of sum" wala definition NCERT me brief cover hota hai — ismein integral ko chhote-chhote rectangles ke area ka limit maante hain jab n → ∞.
11. Properties of Definite Integrals
Kuch properties (P0 se P7 tak) sawaal ko drastically simplify kar deti hain — especially jab function complicated ho par symmetry ho. Sabse zyada use hone wali property: ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a−x)dx. In properties ki poori list formulas table me hai.
Poore Class 12 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q22)
Q1. Find ∫(x² + 2eˣ − 1/x)dx
Term by term standard integrals use karke:
∫x²dx = x³/3, ∫2eˣdx = 2eˣ, ∫(−1/x)dx = −log|x|
Answer: x³/3 + 2eˣ − log|x| + C
Q2. Find ∫(x + 1/x)² dx
Pehle expand karo:
(x + 1/x)² = x² + 2 + 1/x²
Ab term by term integrate:
∫x²dx = x³/3, ∫2dx = 2x, ∫x⁻²dx = −1/x
Answer: x³/3 + 2x − 1/x + C
Q3. Find ∫sin(2x + 5)dx (substitution)
Let t = 2x + 5, so dt = 2dx → dx = dt/2
∫sin t · dt/2 = −(1/2)cos t + C
Answer: −(1/2)cos(2x + 5) + C
Q4. Find ∫2x√(x²+1) dx (substitution)
Let t = x²+1, dt = 2x dx
∫√t dt = (2/3)t^(3/2) + C
Answer: (2/3)(x²+1)^(3/2) + C
Q5. Find ∫tan x dx (substitution using identity)
tan x = sin x / cos x. Let t = cos x, dt = −sin x dx
∫sin x/cos x dx = −∫dt/t = −log|t| + C = −log|cos x| + C
Answer: log|sec x| + C
Q6. Find ∫sin³x dx (trigonometric identity)
sin³x = sin x · sin²x = sin x(1 − cos²x)
∫sin x dx − ∫sin x cos²x dx
Second part: let t = cos x, dt = −sin x dx → ∫sin x cos²x dx = −t³/3 = −cos³x/3
Answer: −cos x + cos³x/3 + C
Q7. Find ∫cos²x dx (trigonometric identity)
cos²x = (1 + cos2x)/2
∫(1 + cos2x)/2 dx = x/2 + (1/2)(sin2x)/2
Answer: x/2 + sin2x/4 + C
Q8. Find ∫sin4x cos3x dx (product to sum)
2 sinA cosB = sin(A+B) + sin(A−B), so sin4x cos3x = (1/2)[sin7x + sinx]
∫(1/2)(sin7x + sinx)dx = (1/2)[−cos7x/7 − cosx]
Answer: −cos7x/14 − cosx/2 + C
Q9. Find ∫dx/(x² + 16) (standard form 1/(x²+a²))
Standard form dx/(x²+a²) = (1/a)tan⁻¹(x/a) + C, here a = 4
Answer: (1/4)tan⁻¹(x/4) + C
Q10. Find ∫dx/√(25 − x²) (standard form 1/√(a²−x²))
Standard form dx/√(a²−x²) = sin⁻¹(x/a) + C, here a = 5
Answer: sin⁻¹(x/5) + C
Q11. Find ∫dx/(x² − 6x + 13) (complete the square)
x² − 6x + 13 = (x−3)² + 4
∫dx/[(x−3)² + 2²] = (1/2)tan⁻¹((x−3)/2) + C
Answer: (1/2)tan⁻¹((x−3)/2) + C
Q12. Find ∫(3x+1)/(x²+4x+5) dx (standard particular function)
Denominator derivative = 2x+4. Express numerator: 3x+1 = (3/2)(2x+4) + (1 − 6) = (3/2)(2x+4) − 5
∫(3/2)(2x+4)/(x²+4x+5) dx − 5∫dx/(x²+4x+5)
First part = (3/2)log|x²+4x+5|. Second: x²+4x+5 = (x+2)²+1, so ∫dx/((x+2)²+1) = tan⁻¹(x+2)
Answer: (3/2)log|x²+4x+5| − 5tan⁻¹(x+2) + C
Q13. Find ∫dx/[(x+1)(x+2)] (partial fractions — distinct linear factors)
1/[(x+1)(x+2)] = A/(x+1) + B/(x+2). Solving: 1 = A(x+2) + B(x+1). Put x=−1: A=1. Put x=−2: B=−1
∫[1/(x+1) − 1/(x+2)]dx = log|x+1| − log|x+2|
Answer: log|(x+1)/(x+2)| + C
Q14. Find ∫(3x−2)/[(x−1)(x−2)(x−3)] dx (partial fractions — three linear factors)
(3x−2)/[(x−1)(x−2)(x−3)] = A/(x−1) + B/(x−2) + C/(x−3)
Put x=1: 3(1)−2=1=A(1−2)(1−3)=2A → A=1/2
Put x=2: 3(2)−2=4=B(2−1)(2−3)=−B → B=−4
Put x=3: 3(3)−2=7=C(3−1)(3−2)=2C → C=7/2
∫[(1/2)/(x−1) − 4/(x−2) + (7/2)/(x−3)]dx
Answer: (1/2)log|x−1| − 4log|x−2| + (7/2)log|x−3| + C
Q15. Find ∫x cos x dx (by parts, ILATE: Algebraic before Trig, so u = x)
u = x (algebraic), dv = cos x dx (trig). du = dx, v = sin x
∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x
Answer: x sin x + cos x + C
Q16. Find ∫x² eˣ dx (by parts, ILATE: Algebraic before Exponential, applied twice)
u = x², dv = eˣdx. du = 2x dx, v = eˣ
∫x²eˣdx = x²eˣ − 2∫x eˣdx
Now ∫x eˣdx (by parts again): u=x, dv=eˣdx → = xeˣ − eˣ
= x²eˣ − 2(xeˣ − eˣ) = x²eˣ − 2xeˣ + 2eˣ
Answer: eˣ(x² − 2x + 2) + C
Q17. Find ∫log x dx (by parts, ILATE: Logarithmic first, treat as log x · 1)
u = log x, dv = 1·dx. du = (1/x)dx, v = x
∫log x dx = x log x − ∫x·(1/x)dx = x log x − ∫1 dx
Answer: x log x − x + C
Q18. Find ∫eˣ(sin x + cos x)dx (special ∫eˣ[f(x)+f'(x)]dx form)
Here f(x) = sin x, f'(x) = cos x. Standard result: ∫eˣ[f(x)+f'(x)]dx = eˣf(x) + C
Answer: eˣ sin x + C
Q19. Find ∫₀^(π/2) sin²x dx (definite integral, fundamental theorem)
Antiderivative: ∫sin²x dx = x/2 − sin2x/4
F(x) = x/2 − sin2x/4
Evaluate at π/2 and 0:
F(π/2) = π/4 − 0 = π/4; F(0) = 0
Answer: π/4
Q20. Evaluate ∫₁³ (2x+3)dx (definite integral, direct)
Antiderivative F(x) = x² + 3x
F(3) − F(1) = (9+9) − (1+3) = 18 − 4
Answer: 14
Q21. Evaluate ∫₀^π x sin x dx / [1 + cos²x] (definite integral using property P4: ∫₀ᵃf(x)dx = ∫₀ᵃf(a−x)dx)
Let I = ∫₀^π x sinx/(1+cos²x) dx. Using property, replace x by (π−x):
I = ∫₀^π (π−x) sin(π−x)/(1+cos²(π−x)) dx = ∫₀^π (π−x)sinx/(1+cos²x) dx
Adding both expressions of I:
2I = π∫₀^π sinx/(1+cos²x) dx
Let t = cos x, dt = −sinx dx; limits x=0→t=1, x=π→t=−1
∫₀^π sinx/(1+cos²x)dx = ∫₋₁¹ dt/(1+t²) = [tan⁻¹t]₋₁¹ = π/4 −(−π/4) = π/2
2I = π·(π/2) = π²/2 → I = π²/4
Answer: π²/4
Q22. Evaluate ∫₋₁¹ x|x| dx (definite integral, odd/even function property P7)
f(x) = x|x|. Check: f(−x) = −x|−x| = −x|x| = −f(x), so f is an odd function.
Property: ∫₋ₐᵃ f(x)dx = 0 if f is odd
Answer: 0
Important Equations — Ek Nazar Me
| Type | Formula |
|---|---|
| Power rule | ∫xⁿdx = xⁿ⁺¹/(n+1) + C, n ≠ −1 |
| Reciprocal | ∫1/x dx = log|x| + C |
| Exponential | ∫eˣdx = eˣ + C, ∫aˣdx = aˣ/log a + C |
| Trigonometric | ∫sinx dx = −cosx + C, ∫cosx dx = sinx + C, ∫sec²x dx = tanx + C, ∫cosec²x dx = −cotx + C, ∫secx·tanx dx = secx + C, ∫cosecx·cotx dx = −cosecx + C |
| Other trig integrals | ∫tanx dx = log|secx| + C, ∫cotx dx = log|sinx| + C, ∫secx dx = log|secx+tanx| + C, ∫cosecx dx = log|cosecx−cotx| + C |
| Special form 1 | ∫dx/(x²+a²) = (1/a)tan⁻¹(x/a) + C |
| Special form 2 | ∫dx/(x²−a²) = (1/2a)log|(x−a)/(x+a)| + C |
| Special form 3 | ∫dx/(a²−x²) = (1/2a)log|(a+x)/(a−x)| + C |
| Special form 4 | ∫dx/√(a²−x²) = sin⁻¹(x/a) + C |
| Special form 5 | ∫dx/√(x²+a²) = log|x+√(x²+a²)| + C |
| Special form 6 | ∫dx/√(x²−a²) = log|x+√(x²−a²)| + C |
| √(x²+a²) integral | ∫√(x²+a²)dx = (x/2)√(x²+a²) + (a²/2)log|x+√(x²+a²)| + C |
| √(x²−a²) integral | ∫√(x²−a²)dx = (x/2)√(x²−a²) − (a²/2)log|x+√(x²−a²)| + C |
| √(a²−x²) integral | ∫√(a²−x²)dx = (x/2)√(a²−x²) + (a²/2)sin⁻¹(x/a) + C |
| Integration by parts | ∫u·v dx = u∫v dx − ∫[u'·∫v dx]dx (u chosen by ILATE order) |
| Special by-parts result | ∫eˣ[f(x)+f'(x)]dx = eˣf(x) + C |
| Fundamental Theorem | ∫ₐᵇ f(x)dx = F(b) − F(a), where F'(x) = f(x) |
| Property P1 | ∫ₐᵇ f(x)dx = −∫ᵇₐ f(x)dx |
| Property P2 | ∫ₐᵇ f(x)dx = ∫ₐᶜ f(x)dx + ∫ᶜᵇ f(x)dx |
| Property P3 | ∫ₐᵇ f(x)dx = ∫ₐᵇ f(a+b−x)dx |
| Property P4 | ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a−x)dx |
| Property P5 | ∫₀²ᵃ f(x)dx = ∫₀ᵃ f(x)dx + ∫₀ᵃ f(2a−x)dx |
| Property P6 | ∫₀²ᵃ f(x)dx = 2∫₀ᵃ f(x)dx if f(2a−x)=f(x), else 0 |
| Property P7 | ∫₋ₐᵃ f(x)dx = 2∫₀ᵃ f(x)dx if f is even; = 0 if f is odd |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Indefinite integral me +C bhool jaana. Har indefinite integral ka answer +C ke bina incomplete maana jaata hai, chahe working sahi ho — board marking me ismein marks katte hain.
- ILATE order ulta lagana in by-parts. Wrong function ko u (first function) banana calculation ko bahut lamba ya impossible bana deta hai — e.g. ∫x eˣdx me eˣ ko u banaoge to answer nahi milega, x ko u banana zaroori hai.
- Definite integral me limits substitution ke baad change karna bhool jaana. Jab t = g(x) substitution use karte ho definite integral me, to naye limits (t ke terms me) use karo — original x-limits use karke t ka answer nikalna galat hai.
- Indefinite formula seedha definite integral pe apply kar dena bina evaluate kiye. ∫ₐᵇ f(x)dx sirf ek number hota hai — antiderivative F(x) likhne ke baad F(b)−F(a) evaluate karna zaroori hai, sirf F(x)+C likh dena galat hai.
- Trigonometric substitution me sign error. √(a²−x²) type integrals me x = a sinθ substitute karte waqt cos θ ka sign range (usually 0 se π/2) ke hisaab se positive lena bhool jaana common mistake hai.
- Partial fractions me degree(numerator) ≥ degree(denominator) check na karna. Agar numerator ka degree denominator ke barabar ya usse zyada hai, to pehle long division karke proper fraction banana zaroori hai — seedha partial fractions apply karna galat result dega.
Board-Style Important Questions
- 1 mark: Evaluate ∫sec²(7−4x)dx.
- 2 marks: Find ∫x/(x+1) dx.
- 2 marks: Evaluate ∫₀^(π/4) tan²x dx.
- 4 marks: Find ∫(2x)/[(x²+1)(x²+3)] dx using partial fractions.
- 4 marks: Evaluate ∫x² log x dx using integration by parts.
- 6 marks: Evaluate ∫₀^π x/(1+sinx) dx using properties of definite integrals.
Aksar Poochhe Jaane Wale Sawaal
Integration aur differentiation me kya farak hai?
Differentiation ek function ka rate of change (derivative) nikalta hai, jabki integration uska ulta hai — derivative se wapas original function (antiderivative) dhoondna. Isi liye integration ko 'anti-differentiation' bhi kehte hain.
Indefinite integral me +C kyun likhte hain?
Kyunki kisi bhi function ka derivative lete waqt constant term zero ho jaata hai, isliye antiderivative unique nahi hota — infinite functions same derivative de sakte hain, jo sirf ek constant ke farak se alag hote hain. +C is poore family ko represent karta hai.
ILATE rule kya hai aur kab use karte hain?
Integration by parts me do function ka product integrate karte waqt decide karna padta hai kaunsa function 'u' (pehla) banega. ILATE order — Inverse, Logarithmic, Algebraic, Trigonometric, Exponential — batata hai ki jo function is list me pehle aaye use u maano, taaki calculation simplify ho.
Definite aur indefinite integral me kya farak hai?
Indefinite integral ek function (family of functions, +C ke saath) deta hai, jabki definite integral ek fixed number deta hai jo do limits ke beech curve ke under ka signed area represent karta hai.
Definite integrals ki properties itni important kyun hain?
Kai definite integrals directly integrate karna mushkil ya impossible hota hai, par properties (jaise ∫₀ᵃf(x)dx = ∫₀ᵃf(a−x)dx) use karke complex expressions simplify ho jaate hain aur answer bina full antiderivative nikale mil jaata hai.
Partial fractions kab use karte hain?
Jab integrand ek rational function P(x)/Q(x) ho jisme denominator factorize ho sakta ho, aur seedha koi standard formula fit na ho, tab fraction ko simpler parts me todkar (denominator ke factors ke basis par) har part ko alag-alag integrate karte hain.
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