Evaluate ∫ 0 4 |x − 1| dx .

|x − 1| = −(x − 1) for 0 ≤ x < 1 , and +(x − 1) for 1 ≤ x ≤ 4 . I = ∫ 0 1 (1 − x) dx + ∫ 1 4 (x − 1) dx = [x − x²/2] 0 1 + [x²/2 − x] 1 4 = 1 2 + (4 − (− 1 2 )) = 1 2 + 9 2 = 5 .

Evaluate ∫ 0 π/4 log(1 + tan x) dx .

Let I = ∫ 0 π/4 log(1 + tan x) dx . By King's Property ( x → π/4 − x ): I = ∫ 0 π/4 log[1 + tan( π 4 − x)] dx = ∫ 0 π/4 log[1 + 1 − tan x 1 + tan x ] dx = ∫ 0 π/4 log( 2 1 + tan x ) dx . Add both: 2I = ∫ 0 π/4 log 2 dx = π 4 log 2 ⇒ I = π 8 log 2 .

NCERT Plus Two (Class 12) Mathematics • Chapter 7

Integrals

Q: Evaluate ∫ sec x (sec x + tan x) dx .

∫ (sec² x + sec x tan x) dx = tan x + sec x + C .

Q: Evaluate ∫ eˣ (sin x + cos x) dx .

Here f(x) = sin x and f'(x) = cos x . Using ∫ eˣ[f(x) + f'(x)] dx = eˣ f(x) + C ⇒ eˣ sin x + C .

Q: Evaluate ∫ e^(tan⁻¹ x) 1 + x² dx .

Put tan⁻¹ x = t ⇒ 1 1 + x² dx = dt . Integral becomes ∫ eᵗ dt = eᵗ + C = e^(tan⁻¹ x) + C .

Q: Evaluate ∫ dx x² − 6x + 13 .

Complete the square: x² − 6x + 13 = (x − 3)² + 4 = (x − 3)² + 2² . Using ∫ du u² + a² = 1 a tan⁻¹( u a ) + C : Answer = 1 2 tan⁻¹( x − 3 2 ) + C .

Q: Integrate ∫ x (x − 1)(x − 2) dx using partial fractions.

Write x (x−1)(x−2) = A x−1 + B x−2 ⇒ x = A(x − 2) + B(x − 1) . • Put x = 1 ⇒ A = −1 ; put x = 2 ⇒ B = 2 . • Integrate: 2 log|x − 2| − log|x − 1| + C .

Q: Find ∫ x cos x dx using Integration by Parts.

Take u = x (A), v = cos x (T): ∫ x cos x dx = x ∫ cos x dx − ∫ [1 · (∫ cos x dx)] dx = x sin x − ∫ sin x dx = x sin x + cos x + C .

Q: Evaluate ∫ 0 π/4 log(1 + tan x) dx .

Let I = ∫ 0 π/4 log(1 + tan x) dx . By King's Property ( x → π/4 − x ): I = ∫ 0 π/4 log[1 + tan( π 4 − x)] dx = ∫ 0 π/4 log[1 + 1 − tan x 1 + tan x ] dx = ∫ 0 π/4 log( 2 1 + tan x ) dx . Add both: 2I = ∫ 0 π/4 log 2 dx = π 4 log 2 ⇒ I = π 8 log 2 .

Antiderivatives & +C, substitution & trig identities, 6 special quadratic forms, partial fractions, integration by parts (ILATE), and definite integral properties.

📄 Official NCERT Textbook PDF

Topic 7.2: Integration as Anti-Derivative, Constant +C & Standard Table

If differentiation asks "How fast is the water tank filling right now?", Integration plays the video in reverse: "Given the tap's flow speed, how much water is in the tank?" Wait — if the tank ALREADY had 10 litres (or 50 litres) before you opened the tap, the flow speed is the same! That unknown starting water level is why EVERY indefinite integral carries an arbitrary constant + C!

REVERSE DERIVATIVEIf F'(x) = f(x), then
∫ f(x) dx = F(x) + C

POWER & LOG RULE∫ xⁿ dx = xⁿ⁺¹n+1 + C (n ≠ −1)
∫ 1x dx = log|x| + C

LINEAR CHAIN RULE∫ f(ax+b) dx = F(ax+b)a + C
(DIVIDE by a, don't multiply!)

🧠 Memory Trick: Co-answers wear MINUS signs! In integration, answers starting with "c" (−cos x, −cot x, −cosec x) get a negative sign! And NEVER forget + C!

💡 Simple Kid-Check: Integration is literally Un-Differentiating! Ask yourself: "Which function's derivative gives me this expression?" Because shifting ANY graph vertically by + 5 or − 99 doesn't change its slope, all parallel curves share the exact same derivative — so we add + C to represent the whole family!

Integral `∫ f(x) dx`	Result (`+ C`)	Integral `∫ f(x) dx`	Result (`+ C`)
`∫ xⁿ dx` (`n ≠ −1`)	`xⁿ⁺¹n+1 + C`	`∫ eˣ dx` / `∫ aˣ dx`	`eˣ + C` / `aˣlog a + C`
`∫ sin x dx`	`−cos x + C`	`∫ cos x dx`	`sin x + C`
`∫ sec² x dx`	`tan x + C`	`∫ cosec² x dx`	`−cot x + C`
`∫ sec x tan x dx`	`sec x + C`	`∫ cosec x cot x dx`	`−cosec x + C`
`∫ dx√(1−x²)`	`sin⁻¹ x + C`	`∫ dx1+x²`	`tan⁻¹ x + C`

🧒 Super-Easy Simplify-First Example: Evaluate ∫ x³ + 3x + 4√x dx.
• There is NO quotient rule in integration! Split term-by-term first: x^(5/2) + 3x^(1/2) + 4x^(−1/2).
• Add 1 to each power and divide: 27x^(7/2) + 2x^(3/2) + 8√x + C!

Watch THREE classic 1-mark traps! (1) Writing ∫ cos(3x) dx = 3 sin(3x) + C — WRONG! In integration you DIVIDE by 3 → 13 sin(3x) + C! (2) NEVER split ∫ u · v dx as (∫ u dx)(∫ v dx) — algebraically split products/fractions BEFORE integrating! (3) Don't forget + C on indefinite integrals — DHSE deducts ½ mark every time!

Topic 7.3: Integration by Substitution & Trigonometric Identities

Imagine trying to carry a giant 50 kg sack of rice up stairs at once — impossible! Pack it into one neat bag t = g(x) whose derivative g'(x) dx = dt is sitting right beside it, and the giant integral shrinks into a 1-line baby integral! That magic disguise swap is Substitution — and when powers of sin or cos block the road, Trig Identities clear the path!

LOG SHORTCUT∫ f'(x)f(x) dx = log|f(x)| + C
(Derivative on top ⇒ instant log!)

SQUARES RULEsin²x = 1−cos 2x2 | cos²x = 1+cos 2x2
tan²x = sec²x − 1

🧠 Memory Trick: Spot the "Function + Its Derivative Buddy" pair! Whenever you see tan⁻¹ x with 1/(1+x²), or log x with 1/x, substitute t = the boss function!

💡 Simple Kid-Check: In any substitution problem, play "Spot the Pair": ask which inside expression t = g(x) has its derivative g'(x) waiting right inside the numerator to turn into dt!

1
Pick t = inside expression (like tan⁻¹ x, log x, 1 + x², or sin x).
2
Differentiate to swap dx: Replace ALL x and dx cleanly with t and dt.
3
Integrate in t, then substitute back to x! Never leave an indefinite answer in t!

🔄 Classic Substitution + Trig Examples:
1. ∫ e^(tan⁻¹ x)1 + x² dx: Put tan⁻¹ x = t ⇒ dx1+x² = dt ⇒ ∫ eᵗ dt = eᵗ + C = e^(tan⁻¹ x) + C!
2. ∫ sin² x dx = ∫ 1 − cos 2x2 dx = x2 − sin 2x4 + C! (Never write sin³x / 3!)

BIGGEST TRIG TRAP: Writing ∫ sin² x dx = sin³ x3 + C — DEAD WRONG! Where is cos x? You can't use Power Rule without the derivative of sin x! Always use sin² x = 1 − cos 2x2! Also for ∫ 11 + tan x dx, rewrite as cos xcos x + sin x and use 12[(cos+sin) + (cos−sin)]!

Topic 7.4: Six Special Quadratic Forms & Completing the Square

Look closely at three tiny sign swaps inside a quadratic denominator: 1x² + 4, 1x² − 4, and 14 − x²! Only one + or − flips, yet the answer transforms completely from tan⁻¹ into log! Master the 6 Special Quadratic Formulas plus Completing the Square and every quadratic integral falls like dominoes!

NO ROOT (1/2a LOG vs TAN⁻¹)∫ dxx²−a² = 12a log|x−ax+a|
∫ dxa²−x² = 12a log|a+xa−x|
∫ dxx²+a² = 1a tan⁻¹(xa)

WITH √ ROOT∫ dx√(a²−x²) = sin⁻¹(xa)
∫ dx√(x²±a²) = log|x + √(x²±a²)|

LINEAR ON TOP∫ px + qax²+bx+c dx:
Write px+q = A(2ax+b) + B!

🧠 Memory Trick: Watch the signs without √: x² + a² is always 1/a tan⁻¹; x² − a² starts with (x−a) on top; a² − x² starts with (a+x) on top! With √: a² − x² is sin⁻¹ (NO 1/a outside!), others are log|x + √...|!

💡 Simple Kid-Check: How do you turn a messy quadratic like x² − 6x + 13 into one of the 6 clean forms? Complete the Square! Take half of the x-coefficient, square it, add & subtract it: (x − 3)² + 2² → instant 12 tan⁻¹(x−32) + C!

1
Factor out coefficient of x² FIRST (especially if it's 4 or −1, like in √(9 − 4x²) = 2√((3/2)² − x²)!).
2
Complete the Square: Rewrite x² + Bx + C as (x + B/2)² + (C − B²/4).
3
Match to one of the 6 Formulas (plus divide by k if kx appeared!).

🎯 Linear-on-Top Example: Evaluate ∫ x + 22x² + 6x + 5 dx.
• Set x + 2 = A · ddx(2x² + 6x + 5) + B = A(4x + 6) + B.
• Compare coefficients: 4A = 1 ⇒ A = 14, and 6A + B = 2 ⇒ B = 12.
• First part gives 14 log|2x² + 6x + 5|; second part completes the square into tan⁻¹!

Watch TWO sneaky traps! (1) In ∫ dx√(9 − 4x²), students forget the 4 on x² and write sin⁻¹(2x/3) WITHOUT dividing by 2! Correct answer: 12 sin⁻¹(2x3) + C! (2) Notice sin⁻¹(x/a) has NO 1/a out front, whereas tan⁻¹(x/a) HAS 1/a out front!

Topic 7.5: Integration by Partial Fractions

Adding 2x−1 + 3x+2 into one single fraction 5x+1(x−1)(x+2) is super easy — Partial Fractions simply runs that process BACKWARDS! Breaking a bulky rational polynomial into bite-sized fractions turns scary integrals into instant log and tan⁻¹ sums!

RULE #0: PROPER CHECK!Need deg(Top) < deg(Bottom).
If deg(Top) ≥ deg(Bottom), LONG DIVIDE FIRST!

REPEATED LINEAR...(x−a)² → Ax−a + B(x−a)²

IRREDUCIBLE QUADRATIC...(x−a)(x²+c) → Ax−a + Bx+Cx²+c

🧠 Memory Trick: Linear factor gets 1 letter (A); Irreducible quadratic gets 2 letters (Bx + C)! And ALWAYS check degrees before splitting!

💡 Simple Kid-Check: Think of Partial Fractions like making change for a ₹500 note into smaller ₹100 and ₹50 notes! And for distinct linear factors like (x−1)(x−2), just plug in x = 1 and x = 2 after clearing denominators to find A and B in 5 seconds flat!

Denominator Type	Exact Partial Fraction Template
`(x − a)(x − b)`	`Ax−a + Bx−b`
`(x − a)²` (Repeated!)	`Ax−a + B(x−a)²`
`(x − a)(x² + bx + c)`	`Ax−a + Bx + Cx² + bx + c`

✂️ 3-Mark Partial Fraction Example: Evaluate ∫ x(x − 1)(x − 2) dx.
• Set x(x−1)(x−2) = Ax−1 + Bx−2 ⇒ x = A(x − 2) + B(x − 1).
• Put x = 1 ⇒ 1 = −A ⇒ A = −1. Put x = 2 ⇒ 2 = B(1) ⇒ B = 2.
• Integrate: ∫ (−1x−1 + 2x−2) dx = 2 log|x − 2| − log|x − 1| + C!

Watch THREE huge traps! (1) Splitting an IMPROPER fraction like x² + 1x² − 5x + 6 without dividing first — degree of top (2) equals bottom (2), so DIVIDE first to get 1 + 5x − 5(x−2)(x−3)! (2) On (x − 1)², missing the linear term A/(x − 1)! (3) In ∫ x² dx(x²+1)(x²+4), you can set x² = y ONLY for algebraic splitting (don't substitute dx!), then restore x² before integrating to tan⁻¹!

Topic 7.6: Integration by Parts (ILATE) & Special eˣ [f(x) + f'(x)]

When two completely unrelated functions multiply — like x · eˣ or x · sin x — substitution doesn't work! So we flip Differentiation's Product Rule upside down to get Integration by Parts! Pick the bossy First function u using the ILATE priority ladder so the integral gets simpler, NOT uglier — and watch for the 1-line Kerala board miracle eˣ[f(x) + f'(x)]!

BY PARTS FORMULA∫ u·v dx = u ∫ v dx − ∫ [u' · (∫ v dx)] dx

ILATE PRIORITY (Pick u!)Inverse trig → Log → Algebraic → Trig → Exp

1-LINE MAGIC FORMULA∫ eˣ [f(x) + f'(x)] dx = eˣ f(x) + C

🧠 Memory Trick: ILATE puts I and L first because we don't know their integrals directly — so when integrating ∫ log x dx or ∫ tan⁻¹ x dx alone, multiply by invisible 1 as Second function v!

💡 Simple Kid-Check: "1st function × (Integral of 2nd) MINUS Integral of [(Derivative of 1st) × (Integral of 2nd)]"! Notice how the 1st function u gets differentiated (so x vanishes to 1, and log x vanishes to 1/x!), while the 2nd function v gets integrated!

3 Bonus Square-Root Formulas (from By Parts with · 1):
• ∫ √(a² − x²) dx = x2√(a²−x²) + a²2 sin⁻¹(xa) + C
• ∫ √(x² ± a²) dx = x2√(x²±a²) ± a²2 log|x + √(x²±a²)| + C

⚡ Two Instant Winners:
1. ∫ x eˣ dx: Take u = x (A), v = eˣ (E) ⇒ x eˣ − ∫ 1·eˣ dx = eˣ(x − 1) + C!
2. ∫ eˣ (1x − 1x²) dx: Spot f(x) = 1x & f'(x) = −1x² ⇒ Instant Answer eˣx + C!

Watch TWO common traps! (1) In ∫ x log x dx, students automatically pick u = x because x is written first! WRONG! By ILATE, L beats A, so pick u = log x and v = x! (2) In ∫ eˣ sin x dx (two cyclic functions T and E), apply By Parts TWICE until the original integral I reappears on the right, then move −I to the left to solve 2I = ...!

Topic 7.7 – 7.10: Definite Integrals, Fundamental Theorem, King's & Even/Odd Rules

When you integrate between two exact fences a and b — like finding the area under a wave from x = 0 to x = π — the constant + C cancels out cleanly (C − C = 0) leaving ONE exact number! Better still, mirror-symmetry properties like King's Property and Even/Odd Cancellation solve monstrous 4-mark definite integrals in three lines without even integrating!

FUNDAMENTAL THEOREM∫ₐᵇ f(x) dx = F(b) − F(a)
(NO + C! Upper minus Lower!)

KING'S PROPERTY (P4)∫₀ᵃ f(x) dx = ∫₀ᵃ f(a − x) dx
General: ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a+b−x) dx

EVEN / ODD ON [−a, a]Odd (f(−x)=−f(x)): ∫_−aᵃ f(x) dx = 0
Even (f(−x)=f(x)): 2 ∫₀ᵃ f(x) dx

🧠 Memory Trick: Add I + I = 2I with King's Property! Replace x with a − x, add the new integral to original I, and ugly fractions magically collapse to 1!

💡 Simple Kid-Check: Picture an odd curve like y = x³ or sin x from −a to +a: the area hanging below the x-axis on the left is the EXACT negative twin of the area rising above on the right — add them together and they wipe out to ZERO!

Property	Rule	When to Use
Swap Limits	`∫ₐᵇ f(x) dx = −∫_bᵃ f(x) dx`	Flipping upside-down limits negates sign
Split Point	`∫ₐᵇ f = ∫ₐᶜ f + ∫_cᵇ f`	Modulus `\|x − c\|` or piecewise functions!
King's (P4)	`∫₀ᵃ f(x) dx = ∫₀ᵃ f(a−x) dx`	`sin/cos` fractions, `x · f(sin x)`, `log(1+tan x)`
Even/Odd (P7)	`∫_−aᵃ f = 0` (Odd) / `2∫₀ᵃ f` (Even)	Symmetric limits `[−a, a]`!

👑 Iconic King's Property Proof: Evaluate I = ∫₀^(π/2) √sin x√sin x + √cos x dx.
• Replace x → π2 − x (sin ↔ cos): I = ∫₀^(π/2) √cos x√cos x + √sin x dx.
• Add both equations: 2I = ∫₀^(π/2) √sin x + √cos x√sin x + √cos x dx = ∫₀^(π/2) 1 dx = π2 ⇒ I = π4!

DEADLY DEFINITE SUBSTITUTION TRAP: When you put t = g(x) in a definite integral, students plug the OLD x-limits into the t-antiderivative! ALWAYS convert limits to t immediately when substituting (or switch back to x before plugging limits)! Also remember for modulus ∫₀⁴ |x − 1| dx, ALWAYS split at x = 1 where the inside hits zero!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Evaluate ∫ sec x (sec x + tan x) dx.

∫ (sec² x + sec x tan x) dx = tan x + sec x + C.

Q2 (Kerala HSE Say • 1 Marks): Evaluate ∫ eˣ (sin x + cos x) dx.

Here f(x) = sin x and f'(x) = cos x.
Using ∫ eˣ[f(x) + f'(x)] dx = eˣ f(x) + C ⇒ eˣ sin x + C.

Q3 (Kerala HSE March • 2 Marks): Evaluate ∫ e^(tan⁻¹ x)1 + x² dx.

Put tan⁻¹ x = t ⇒ 11 + x² dx = dt.
Integral becomes ∫ eᵗ dt = eᵗ + C = e^(tan⁻¹ x) + C.

Q4 (Kerala HSE Model • 2 Marks): Evaluate ∫ dxx² − 6x + 13.

Complete the square: x² − 6x + 13 = (x − 3)² + 4 = (x − 3)² + 2².
Using ∫ duu² + a² = 1a tan⁻¹(ua) + C:
Answer = 12 tan⁻¹(x − 32) + C.

Q5 (Kerala HSE March • 3 Marks): Integrate ∫ x(x − 1)(x − 2) dx using partial fractions.

Write x(x−1)(x−2) = Ax−1 + Bx−2 ⇒ x = A(x − 2) + B(x − 1).
• Put x = 1 ⇒ A = −1; put x = 2 ⇒ B = 2.
• Integrate: 2 log|x − 2| − log|x − 1| + C.

Q6 (Kerala HSE Say • 3 Marks): Find ∫ x cos x dx using Integration by Parts.

Take u = x (A), v = cos x (T):
∫ x cos x dx = x ∫ cos x dx − ∫ [1 · (∫ cos x dx)] dx
= x sin x − ∫ sin x dx = x sin x + cos x + C.

Q7 (NCERT / DHSE March • 3 Marks): Evaluate ∫₀⁴ |x − 1| dx.

|x − 1| = −(x − 1) for 0 ≤ x < 1, and +(x − 1) for 1 ≤ x ≤ 4.
I = ∫₀¹ (1 − x) dx + ∫₁⁴ (x − 1) dx
= [x − x²/2]₀¹ + [x²/2 − x]₁⁴ = 12 + (4 − (−12)) = 12 + 92 = 5.

Q8 (Kerala HSE March • 4 Marks): Evaluate ∫₀^π/4 log(1 + tan x) dx.

Let I = ∫₀^π/4 log(1 + tan x) dx.
By King's Property (x → π/4 − x):

I = ∫₀^π/4 log[1 + tan(π4 − x)] dx = ∫₀^π/4 log[1 + 1 − tan x1 + tan x] dx = ∫₀^π/4 log(21 + tan x) dx

.
Add both: 2I = ∫₀^π/4 log 2 dx = π4 log 2 ⇒ I = π8 log 2.