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Class 12MathsChapter 7NCERTCBSE
Integrals – Class 12 Maths Chapter 7
📌 Quick Overview: Chapter 7 is the longest and most important chapter in Class 12 Maths. It introduces indefinite and definite integration — the reverse process of differentiation. You will learn standard integrals, substitution method, integration by parts, integration using partial fractions, definite integrals, and their properties. This chapter carries the highest marks weightage in CBSE board exams.
Topics Covered:
- > Integration as Anti-Derivative
- > Standard Integral Formulas
- > Integration by Substitution
- > Integrals of Special Forms
- > Integration by Partial Fractions
- > Integration by Parts
- > Integral of sqrt(ax²+bx+c)
- > Definite Integrals
- > Properties of Definite Integrals
- > Important Board Questions
- > MCQ Practice
- > Exam Tips
1. Integration as Anti-Derivative
Definition: Integration is the reverse process of differentiation. If d/dx[F(x)] = f(x), then the indefinite integral of f(x) is:
∫f(x) dx = F(x) + C
where C is the constant of integration (arbitrary constant). F(x) is called the anti-derivative or primitive of f(x).
∫f(x) dx = F(x) + C
where C is the constant of integration (arbitrary constant). F(x) is called the anti-derivative or primitive of f(x).
⭐ Key Point: The constant C is essential — without it, the answer is incomplete. Every indefinite integral has infinitely many anti-derivatives differing only by a constant.
2. Standard Integral Formulas ⭐ (Must Memorise)
| Function f(x) | Integral ∫f(x)dx |
|---|---|
| x^n (n ≠ -1) | x^(n+1)/(n+1) + C |
| 1/x | log|x| + C |
| e^x | e^x + C |
| a^x | a^x / log a + C |
| sin x | -cos x + C |
| cos x | sin x + C |
| tan x | log|sec x| + C = -log|cos x| + C |
| cot x | log|sin x| + C |
| sec x | log|sec x + tan x| + C |
| cosec x | log|cosec x - cot x| + C |
| sec² x | tan x + C |
| cosec² x | -cot x + C |
| sec x tan x | sec x + C |
| cosec x cot x | -cosec x + C |
| 1/sqrt(1-x²) | sin⁻¹ x + C |
| -1/sqrt(1-x²) | cos⁻¹ x + C |
| 1/(1+x²) | tan⁻¹ x + C |
| 1/(x sqrt(x²-1)) | sec⁻¹ x + C |
| 1/sqrt(a²-x²) | sin⁻¹(x/a) + C |
| 1/(a²+x²) | (1/a) tan⁻¹(x/a) + C |
| 1/(x²-a²) (x > a) | (1/2a) log|(x-a)/(x+a)| + C |
| 1/(a²-x²) (x < a) | (1/2a) log|(a+x)/(a-x)| + C |
| 1/sqrt(x²+a²) | log|x + sqrt(x²+a²)| + C |
| 1/sqrt(x²-a²) | log|x + sqrt(x²-a²)| + C |
| sqrt(a²-x²) | (x/2)sqrt(a²-x²) + (a²/2)sin⁻¹(x/a) + C |
| sqrt(x²+a²) | (x/2)sqrt(x²+a²) + (a²/2)log|x+sqrt(x²+a²)| + C |
| sqrt(x²-a²) | (x/2)sqrt(x²-a²) - (a²/2)log|x+sqrt(x²-a²)| + C |
3. Properties of Indefinite Integrals
1. ∫[f(x) + g(x)]dx = ∫f(x)dx + ∫g(x)dx
2. ∫k.f(x)dx = k.∫f(x)dx (k is a constant)
3. d/dx[∫f(x)dx] = f(x)
4. ∫f'(x)dx = f(x) + C
4. Integration by Substitution ⭐
Method: Substitute a suitable variable t = g(x), then dt = g'(x)dx to simplify the integral.
If ∫f(g(x)) . g'(x) dx, put t = g(x), dt = g'(x)dx → ∫f(t) dt
If ∫f(g(x)) . g'(x) dx, put t = g(x), dt = g'(x)dx → ∫f(t) dt
Key Pattern: If numerator is derivative of denominator (or close to it), use substitution.
∫f'(x)/f(x) dx = log|f(x)| + C
∫f'(x)/f(x) dx = log|f(x)| + C
Example 1: ∫2x/(x²+1) dx
Put t = x²+1, dt = 2x dx
= ∫dt/t = log|t| + C = log|x²+1| + C
Put t = x²+1, dt = 2x dx
= ∫dt/t = log|t| + C = log|x²+1| + C
Example 2: ∫sin³x cos x dx
Put t = sin x, dt = cos x dx
= ∫t³ dt = t⁴/4 + C = sin⁴x/4 + C
Put t = sin x, dt = cos x dx
= ∫t³ dt = t⁴/4 + C = sin⁴x/4 + C
Example 3: ∫e^x / (1 + e^x) dx
Put t = 1 + e^x, dt = e^x dx
= ∫dt/t = log|t| + C = log|1 + e^x| + C
Put t = 1 + e^x, dt = e^x dx
= ∫dt/t = log|t| + C = log|1 + e^x| + C
Example 4: ∫tan x dx
= ∫sin x / cos x dx
Put t = cos x, dt = -sin x dx → -dt = sin x dx
= -∫dt/t = -log|t| + C = -log|cos x| + C = log|sec x| + C
= ∫sin x / cos x dx
Put t = cos x, dt = -sin x dx → -dt = sin x dx
= -∫dt/t = -log|t| + C = -log|cos x| + C = log|sec x| + C
5. Integrals of Special Forms ⭐
TYPE 1: px + q = A(derivative of ax²+bx+c) + B
Used for: ∫(px+q)/(ax²+bx+c) dx
and ∫(px+q)/sqrt(ax²+bx+c) dx
TYPE 2: Complete the square
ax²+bx+c → a[(x + b/2a)² + (c/a - b²/4a²)]
Then use standard forms like 1/(x²+k²) or 1/sqrt(x²-k²)
TYPE 3: ∫(f'(x))/(f(x)) dx = log|f(x)| + C
USEFUL TRIG IDENTITIES:
sin²x = (1 - cos 2x)/2
cos²x = (1 + cos 2x)/2
sin 2x = 2 sin x cos x
cos 2x = 1 - 2sin²x = 2cos²x - 1
sin³x = (3 sin x - sin 3x)/4
cos³x = (3 cos x + cos 3x)/4
Example: ∫1/(x² + 4x + 5) dx
Complete the square: x²+4x+5 = (x+2)² + 1
= ∫1/((x+2)²+1) dx
Put t = x+2, dt = dx
= ∫1/(t²+1) dt = tan⁻¹(x+2) + C
Complete the square: x²+4x+5 = (x+2)² + 1
= ∫1/((x+2)²+1) dx
Put t = x+2, dt = dx
= ∫1/(t²+1) dt = tan⁻¹(x+2) + C
Example 2: ∫(2x+3)/(x²+3x+2) dx
Numerator = derivative of denominator (2x+3)
= log|x²+3x+2| + C
Numerator = derivative of denominator (2x+3)
= log|x²+3x+2| + C
6. Integration by Partial Fractions ⭐
When to use: When integrand is a rational function P(x)/Q(x) where degree of P < degree of Q, and Q(x) can be factored.
| Form of Q(x) | Partial Fraction Decomposition |
|---|---|
| (x-a)(x-b), a ≠ b | A/(x-a) + B/(x-b) |
| (x-a)² | A/(x-a) + B/(x-a)² |
| (x-a)(x²+bx+c) | A/(x-a) + (Bx+C)/(x²+bx+c) |
| (x-a)²(x-b) | A/(x-a) + B/(x-a)² + C/(x-b) |
Example: ∫1/((x+1)(x+2)) dx
1/((x+1)(x+2)) = A/(x+1) + B/(x+2)
1 = A(x+2) + B(x+1)
Put x = -1: 1 = A(1) → A = 1
Put x = -2: 1 = B(-1) → B = -1
∫[1/(x+1) - 1/(x+2)] dx = log|x+1| - log|x+2| + C = log|(x+1)/(x+2)| + C
1/((x+1)(x+2)) = A/(x+1) + B/(x+2)
1 = A(x+2) + B(x+1)
Put x = -1: 1 = A(1) → A = 1
Put x = -2: 1 = B(-1) → B = -1
∫[1/(x+1) - 1/(x+2)] dx = log|x+1| - log|x+2| + C = log|(x+1)/(x+2)| + C
Example 2: ∫(3x+5)/((x-1)(x²+1)) dx
(3x+5)/((x-1)(x²+1)) = A/(x-1) + (Bx+C)/(x²+1)
3x+5 = A(x²+1) + (Bx+C)(x-1)
Put x = 1: 8 = 2A → A = 4
Compare x²: 0 = A + B → B = -4
Compare constant: 5 = A - C → C = -1
= ∫[4/(x-1) + (-4x-1)/(x²+1)] dx
= 4log|x-1| - 2log|x²+1| - tan⁻¹x + C
(3x+5)/((x-1)(x²+1)) = A/(x-1) + (Bx+C)/(x²+1)
3x+5 = A(x²+1) + (Bx+C)(x-1)
Put x = 1: 8 = 2A → A = 4
Compare x²: 0 = A + B → B = -4
Compare constant: 5 = A - C → C = -1
= ∫[4/(x-1) + (-4x-1)/(x²+1)] dx
= 4log|x-1| - 2log|x²+1| - tan⁻¹x + C
7. Integration by Parts ⭐ (Very Important)
Formula: ∫u.v dx = u.∫v dx - ∫(u' . ∫v dx) dx
Or: ∫u.v dx = u.v₁ - u'.v₂ + u''.v₃ - ...
where v₁ = ∫v dx, v₂ = ∫v₁ dx, etc.
Or: ∫u.v dx = u.v₁ - u'.v₂ + u''.v₃ - ...
where v₁ = ∫v dx, v₂ = ∫v₁ dx, etc.
ILATE Rule — Choose u in this priority order:
I – Inverse trigonometric functions (sin⁻¹x, tan⁻¹x, ...)
L – Logarithmic functions (log x, ln x)
A – Algebraic functions (x, x², x³, ...)
T – Trigonometric functions (sin x, cos x, ...)
E – Exponential functions (e^x, a^x)
Choose u = the function that appears FIRST in ILATE, v = the other function.
I – Inverse trigonometric functions (sin⁻¹x, tan⁻¹x, ...)
L – Logarithmic functions (log x, ln x)
A – Algebraic functions (x, x², x³, ...)
T – Trigonometric functions (sin x, cos x, ...)
E – Exponential functions (e^x, a^x)
Choose u = the function that appears FIRST in ILATE, v = the other function.
Example 1: ∫x e^x dx
u = x (Algebraic), v = e^x (Exponential)
= x.e^x - ∫1.e^x dx
= x.e^x - e^x + C
= e^x(x-1) + C
u = x (Algebraic), v = e^x (Exponential)
= x.e^x - ∫1.e^x dx
= x.e^x - e^x + C
= e^x(x-1) + C
Example 2: ∫x sin x dx
u = x, v = sin x
= x(-cos x) - ∫1.(-cos x) dx
= -x cos x + sin x + C
= sin x - x cos x + C
u = x, v = sin x
= x(-cos x) - ∫1.(-cos x) dx
= -x cos x + sin x + C
= sin x - x cos x + C
Example 3: ∫log x dx
u = log x, v = 1
= log x . x - ∫(1/x).x dx
= x log x - ∫1 dx
= x log x - x + C = x(log x - 1) + C
u = log x, v = 1
= log x . x - ∫(1/x).x dx
= x log x - ∫1 dx
= x log x - x + C = x(log x - 1) + C
Example 4: ∫e^x sin x dx (Cyclic Method)
Let I = ∫e^x sin x dx
u = sin x, v = e^x
I = e^x sin x - ∫e^x cos x dx
Apply parts again: ∫e^x cos x dx = e^x cos x + ∫e^x sin x dx = e^x cos x + I
I = e^x sin x - (e^x cos x + I)
2I = e^x sin x - e^x cos x
I = e^x(sin x - cos x)/2 + C
Let I = ∫e^x sin x dx
u = sin x, v = e^x
I = e^x sin x - ∫e^x cos x dx
Apply parts again: ∫e^x cos x dx = e^x cos x + ∫e^x sin x dx = e^x cos x + I
I = e^x sin x - (e^x cos x + I)
2I = e^x sin x - e^x cos x
I = e^x(sin x - cos x)/2 + C
⭐ Special Formula: ∫e^x[f(x) + f'(x)] dx = e^x f(x) + C
Example: ∫e^x(sin x + cos x) dx = e^x sin x + C
Example: ∫e^x(1/x + log x) ... wait, ∫e^x(1 + log x) dx → here f(x) = x log x? No.
Standard form: ∫e^x(1/x + (-1/x²)) dx = e^x(1/x) + C [since f=1/x, f'=-1/x²]
Example: ∫e^x(sin x + cos x) dx = e^x sin x + C
Example: ∫e^x(1/x + log x) ... wait, ∫e^x(1 + log x) dx → here f(x) = x log x? No.
Standard form: ∫e^x(1/x + (-1/x²)) dx = e^x(1/x) + C [since f=1/x, f'=-1/x²]
8. Definite Integrals ⭐
Definition: The definite integral of f(x) from a to b is:
∫[a to b] f(x) dx = F(b) - F(a)
where F(x) is the anti-derivative of f(x). This is called the Fundamental Theorem of Calculus.
∫[a to b] f(x) dx = F(b) - F(a)
where F(x) is the anti-derivative of f(x). This is called the Fundamental Theorem of Calculus.
Example: Evaluate ∫[0 to pi/2] sin x dx
= [-cos x] from 0 to pi/2
= -cos(pi/2) + cos(0)
= 0 + 1 = 1
= [-cos x] from 0 to pi/2
= -cos(pi/2) + cos(0)
= 0 + 1 = 1
9. Properties of Definite Integrals ⭐ (Most Important)
Property 1: ∫[a to b] f(x) dx = ∫[a to b] f(t) dt (dummy variable)
Property 2: ∫[a to b] f(x) dx = -∫[b to a] f(x) dx (limits interchange → sign change)
Property 3: ∫[a to b] f(x) dx = ∫[a to c] f(x) dx + ∫[c to b] f(x) dx (additive property)
Property 4: ∫[a to b] f(x) dx = ∫[a to b] f(a+b-x) dx
King's Property — most used in board exams!
King's Property — most used in board exams!
Property 5: ∫[0 to a] f(x) dx = ∫[0 to a] f(a-x) dx
Special case of King's Property with b=a, a=0
Special case of King's Property with b=a, a=0
Property 6: ∫[0 to 2a] f(x) dx = ∫[0 to a] f(x) dx + ∫[0 to a] f(2a-x) dx
Special case:
> If f(2a-x) = f(x): ∫[0 to 2a] f(x) dx = 2∫[0 to a] f(x) dx
> If f(2a-x) = -f(x): ∫[0 to 2a] f(x) dx = 0
Special case:
> If f(2a-x) = f(x): ∫[0 to 2a] f(x) dx = 2∫[0 to a] f(x) dx
> If f(2a-x) = -f(x): ∫[0 to 2a] f(x) dx = 0
Property 7 (Even/Odd Functions):
> If f(-x) = f(x) [even function]: ∫[-a to a] f(x) dx = 2∫[0 to a] f(x) dx
> If f(-x) = -f(x) [odd function]: ∫[-a to a] f(x) dx = 0
> If f(-x) = f(x) [even function]: ∫[-a to a] f(x) dx = 2∫[0 to a] f(x) dx
> If f(-x) = -f(x) [odd function]: ∫[-a to a] f(x) dx = 0
⭐ King's Property in Action (Classic Board Example):
Prove: ∫[0 to pi/2] log(tan x) dx = 0
Let I = ∫[0 to pi/2] log(tan x) dx
Using Property 5 (replace x with pi/2 - x):
I = ∫[0 to pi/2] log(tan(pi/2 - x)) dx = ∫[0 to pi/2] log(cot x) dx
Adding: 2I = ∫[0 to pi/2] [log(tan x) + log(cot x)] dx
= ∫[0 to pi/2] log(tan x . cot x) dx
= ∫[0 to pi/2] log(1) dx = 0
∴ I = 0 ✓
Prove: ∫[0 to pi/2] log(tan x) dx = 0
Let I = ∫[0 to pi/2] log(tan x) dx
Using Property 5 (replace x with pi/2 - x):
I = ∫[0 to pi/2] log(tan(pi/2 - x)) dx = ∫[0 to pi/2] log(cot x) dx
Adding: 2I = ∫[0 to pi/2] [log(tan x) + log(cot x)] dx
= ∫[0 to pi/2] log(tan x . cot x) dx
= ∫[0 to pi/2] log(1) dx = 0
∴ I = 0 ✓
Example (King's Property): Evaluate ∫[0 to pi] x sin x / (1 + cos²x) dx
Let I = ∫[0 to pi] x sin x/(1+cos²x) dx ...(1)
Using property: replace x with pi-x:
I = ∫[0 to pi] (pi-x) sin(pi-x)/(1+cos²(pi-x)) dx
= ∫[0 to pi] (pi-x) sin x/(1+cos²x) dx ...(2)
Adding (1) and (2):
2I = ∫[0 to pi] pi sin x/(1+cos²x) dx
2I = pi [-tan⁻¹(cos x)] from 0 to pi
2I = pi [(-tan⁻¹(-1)) - (-tan⁻¹(1))]
2I = pi [pi/4 + pi/4] = pi²/2
I = pi²/4
Let I = ∫[0 to pi] x sin x/(1+cos²x) dx ...(1)
Using property: replace x with pi-x:
I = ∫[0 to pi] (pi-x) sin(pi-x)/(1+cos²(pi-x)) dx
= ∫[0 to pi] (pi-x) sin x/(1+cos²x) dx ...(2)
Adding (1) and (2):
2I = ∫[0 to pi] pi sin x/(1+cos²x) dx
2I = pi [-tan⁻¹(cos x)] from 0 to pi
2I = pi [(-tan⁻¹(-1)) - (-tan⁻¹(1))]
2I = pi [pi/4 + pi/4] = pi²/2
I = pi²/4
10. Definite Integral as Limit of Sum
Formula:
∫[a to b] f(x) dx = lim(n→inf) h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)]
where h = (b-a)/n
∫[a to b] f(x) dx = lim(n→inf) h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)]
where h = (b-a)/n
USEFUL SUMMATION FORMULAS:
Sum of 1 to n = n(n+1)/2
Sum of 1² to n² = n(n+1)(2n+1)/6
Sum of 1³ to n³ = [n(n+1)/2]²
Sum of GP: a(r^n - 1)/(r-1) if r > 1
11. Important Board Exam Questions
Q1. Evaluate ∫ x/(x+1)(x+2) dx
Partial fractions: x/((x+1)(x+2)) = A/(x+1) + B/(x+2)
x = A(x+2) + B(x+1)
x = -1: -1 = A(1) → A = -1
x = -2: -2 = B(-1) → B = 2
= ∫[-1/(x+1) + 2/(x+2)] dx
= -log|x+1| + 2log|x+2| + C = log|(x+2)²/(x+1)| + C
x = A(x+2) + B(x+1)
x = -1: -1 = A(1) → A = -1
x = -2: -2 = B(-1) → B = 2
= ∫[-1/(x+1) + 2/(x+2)] dx
= -log|x+1| + 2log|x+2| + C = log|(x+2)²/(x+1)| + C
Q2. Evaluate ∫ x² e^x dx
u = x², v = e^x (Algebraic before Exponential)
= x² e^x - ∫2x e^x dx
= x² e^x - 2[x e^x - e^x] + C
= x² e^x - 2x e^x + 2e^x + C
= e^x(x² - 2x + 2) + C
= x² e^x - ∫2x e^x dx
= x² e^x - 2[x e^x - e^x] + C
= x² e^x - 2x e^x + 2e^x + C
= e^x(x² - 2x + 2) + C
Q3. Evaluate ∫[0 to pi/2] sin²x dx
∫[0 to pi/2] (1-cos2x)/2 dx
= (1/2)[x - sin2x/2] from 0 to pi/2
= (1/2)[(pi/2 - 0) - (0 - 0)]
= pi/4
= (1/2)[x - sin2x/2] from 0 to pi/2
= (1/2)[(pi/2 - 0) - (0 - 0)]
= pi/4
Q4. Evaluate ∫[0 to pi/2] (sin x - cos x)/(1 + sin x cos x) dx
Let I = ∫[0 to pi/2] (sin x - cos x)/(1+sin x cos x) dx
Using property (replace x with pi/2 - x):
I = ∫[0 to pi/2] (cos x - sin x)/(1+cos x sin x) dx = -I
2I = 0 → I = 0
Using property (replace x with pi/2 - x):
I = ∫[0 to pi/2] (cos x - sin x)/(1+cos x sin x) dx = -I
2I = 0 → I = 0
Q5. Evaluate ∫[1 to 3] (x² + 5x) dx as limit of sum.
a = 1, b = 3, h = 2/n, f(x) = x² + 5x
= lim(n→inf) (2/n) sum[r=0 to n-1] f(1 + 2r/n)
= lim(n→inf) (2/n) sum[(1+2r/n)² + 5(1+2r/n)]
= lim(n→inf) (2/n)[n + 5n + (2/n)(2)(n(n-1)/2) + 5(2/n)(n(n-1)/2) + (4/n²)(n(n-1)(2n-1)/6)]
After simplification: = 56/3
Verify: [x³/3 + 5x²/2] from 1 to 3 = (9+45/2) - (1/3+5/2) = 54+45/2-1/3-5/2 = 56/3 ✓
= lim(n→inf) (2/n) sum[r=0 to n-1] f(1 + 2r/n)
= lim(n→inf) (2/n) sum[(1+2r/n)² + 5(1+2r/n)]
= lim(n→inf) (2/n)[n + 5n + (2/n)(2)(n(n-1)/2) + 5(2/n)(n(n-1)/2) + (4/n²)(n(n-1)(2n-1)/6)]
After simplification: = 56/3
Verify: [x³/3 + 5x²/2] from 1 to 3 = (9+45/2) - (1/3+5/2) = 54+45/2-1/3-5/2 = 56/3 ✓
Q6. Find ∫ sqrt(x²+2x+5) dx
x²+2x+5 = (x+1)² + 4 = (x+1)² + 2²
Put t = x+1:
= ∫sqrt(t²+4) dt
= (t/2)sqrt(t²+4) + (4/2)log|t+sqrt(t²+4)| + C
= ((x+1)/2)sqrt(x²+2x+5) + 2 log|x+1+sqrt(x²+2x+5)| + C
Put t = x+1:
= ∫sqrt(t²+4) dt
= (t/2)sqrt(t²+4) + (4/2)log|t+sqrt(t²+4)| + C
= ((x+1)/2)sqrt(x²+2x+5) + 2 log|x+1+sqrt(x²+2x+5)| + C
12. Key Formulas at a Glance
FUNDAMENTAL THEOREM:
∫[a to b] f(x) dx = F(b) - F(a) where F'(x) = f(x)
SUBSTITUTION:
∫f(g(x)).g'(x) dx → put t = g(x), dt = g'(x)dx
INTEGRATION BY PARTS (ILATE):
∫u.v dx = u.(∫v dx) - ∫[u'.(∫v dx)] dx
SPECIAL FORMULA:
∫e^x[f(x) + f'(x)] dx = e^x.f(x) + C
PARTIAL FRACTIONS:
Always: degree of numerator < degree of denominator
KEY PROPERTIES:
King's: ∫[a to b]f(x)dx = ∫[a to b]f(a+b-x)dx
Even fn: ∫[-a to a]f(x)dx = 2∫[0 to a]f(x)dx
Odd fn: ∫[-a to a]f(x)dx = 0
TRIG REDUCTION:
∫sin²x dx = x/2 - sin(2x)/4 + C
∫cos²x dx = x/2 + sin(2x)/4 + C
13. MCQ Practice (1 Mark)
1. ∫ e^x(1+x)/cos²(xe^x) dx equals:
(a) -cot(xe^x) + C (b) tan(xe^x) + C (c) tan(e^x) + C (d) cot(xe^x) + C
Answer: (b) tan(xe^x) + C — put t = xe^x, dt = e^x(1+x)dx; ∫sec²t dt = tan t + C
2. ∫[-pi to pi] sin³x dx equals:
(a) pi (b) 2 (c) 0 (d) pi/2
Answer: (c) 0 — sin³x is an odd function, so integral on [-a,a] = 0
3. ∫ 1/(sin²x cos²x) dx equals:
(a) tan x + cot x + C (b) tan x - cot x + C (c) -tan x + cot x + C (d) None
Answer: (b) tan x - cot x + C — use sin²x+cos²x=1 in numerator, split into sec²x + cosec²x
4. ∫[0 to 1] tan⁻¹x dx equals:
(a) pi/4 - log2/2 (b) pi/4 + log2/2 (c) pi/4 - log2 (d) pi/4
Answer: (a) pi/4 - log2/2 — use integration by parts with u=tan⁻¹x, v=1
5. The value of ∫[0 to pi/2] log(sin x) dx is:
(a) pi log 2 (b) -(pi/2)log 2 (c) (pi/2)log 2 (d) 0
Answer: (b) -(pi/2)log 2 — classic result using King's property
14. Exam Tips
- Always add + C in indefinite integrals — marks are deducted if forgotten.
- For substitution: look for a function and its derivative together in the integrand.
- ILATE is for choosing u in integration by parts — Inverse trig comes first, Exponential last.
- For ∫e^x[f(x)+f'(x)]dx — identify f(x) such that f'(x) is also present; answer is simply e^x.f(x)+C.
- King's Property is the most powerful tool for definite integral "prove" questions — practice at least 10 examples.
- For even/odd function integrals on symmetric limits [-a,a], always check symmetry first — it can save all the work.
- Complete the square for quadratic expressions under square root or in denominator before integrating.
- Partial fractions: If degree of numerator ≥ degree of denominator, perform long division first.
✅ Summary: Integrals is the highest-weightage chapter in Class 12 Maths. Expect 3-4 questions from this chapter in board exams carrying 15-20 marks total. The key areas are: substitution, integration by parts, partial fractions, and properties of definite integrals (especially King's Property). Memorise all standard formulas and practice daily — this chapter is purely based on practice and pattern recognition.
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