Integration by parts — derivation from product rule, LIATE mnemonic: Worked examples

4.2.7 · D3 · Maths › Calculus II — Integration › Integration by parts — derivation from product rule, LIATE m

Yeh page integration by parts ke liye ek problem gym hai. Parent note ne rule build kiya; yahan hum isse har angle se tackle karte hain. Kuch bhi solve karne se pehle, hum har tarah ke problems ka ek map banate hain jo integration by parts throw kar sakta hai — taaki jab bhi koi naya integral mile, tum already pehchaan lo ki woh kis "cell" mein aata hai.

Recall Woh ek formula jis par hum poori tarah rely karte hain

$\int u d v = uv - \int v d u$ Tum choose karte ho $u$ (woh factor jo differentiate hokar $d u$ banega) aur $d v$ (woh factor jo integrate hokar $v$ banega). Picking rule hai LIATE — Logarithmic, Inverse-trig, Algebraic, Trigonometric, Exponential — pehla letter $u$ banta hai.

Scenario matrix

Integration by parts sirf ek problem nahi hai — yeh ek family hai. Yahan method ke har alag behavior ko list kiya gaya hai. Neeche har example us cell ke saath tagged hai jo woh fill karta hai.

Cell	Case class	Signature / kya hota hai	Filled by
A	Poly × exponential, one pass	$d u$ polynomial ko ek step mein khatam kar deta hai	Ex 1
B	"Hidden product" (single function)	$f (x) = f (x) \cdot 1$ likho taaki $d v = d x$ ho	Ex 2
C	Repeated parts (poly degree $> 1$ )	Parts $n$ baar karo jab tak poly khatam na ho	Ex 3
D	The loop (integral wapas aata hai)	$I$ ke liye algebraically solve karo	Ex 4
E	Inverse-trig akela	$u = arctan x$ , bacha hua $\int v d u$ substitution maangta hai	Ex 5
F	Definite integral	$[uv]_{a}^{b}$ boundary term carry karo; ends par signs dekho	Ex 6
G	Degenerate / limiting case	$\int_{0}^{1} ln x d x$ — ek improper endpoint jahan $ln 0 \to - \infty$	Ex 7
H	Word problem (physics)	Position-dependent, oscillating force se kiya gaya kaam	Ex 8
I	Exam twist (wrong- $u$ trap)	Galat choice dikhao jo explode ho jaati hai, phir LIATE se rescue karo	Ex 9

Signs aur directions jinhe hum explicitly cover karna chahte hain: $uv - \int v d u$ mein leading minus; negative arguments/limits (Ex 6 $0$ se $π$ tak jaata hai jahan $cos$ sign flip karta hai); ek limit jo $- \infty$ tak jaati hai (Ex 7); aur loop constant ke dono signs (Ex 4).

Cell A — Polynomial × exponential (one pass)

Worked example Example 1 —

$\int x e^{2 x} d x$ Forecast: solve karne se pehle guess karo — parts ke ek round ke baad, kya bacha hua integral start se simpler hoga? (Haan, kyunki $x$ ek plain constant mein badal jaata hai.)

LIATE se classify karo. $x$ Algebraic (A) hai, $e^{2 x}$ Exponential (E) hai. A, E se pehle aata hai, isliye $u = x$ . Yeh step kyun? Hume $u$ woh factor banana hai jo differentiate hone par simplify ho jaata hai; $x \to 1$ vanish ho jaata hai, $e^{2 x}$ kabhi simplify nahi hota.
$d u$ aur $v$ compute karo. $u = x \Rightarrow d u = d x$ ; $d v = e^{2 x} d x \Rightarrow v = \frac{1}{2} e^{2 x}$ . Yeh step kyun? $v = \int d v$ ; $e^{2 x}$ integrate karne mein inner factor $2$ se $\frac{1}{2}$ aata hai (chain-rule in reverse — dekho Integration by substitution).
Formula apply karo. $\int x e^{2 x} d x = x \cdot \frac{1}{2} e^{2 x} - \int \frac{1}{2} e^{2 x} d x$ Yeh step kyun? $uv - \int v d u$ mein direct substitution; leading minus ko rakhna mat bhoolo.
Easy integral finish karo. $= \frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x} + C = \frac{1}{4} e^{2 x} (2 x - 1) + C$ Yeh step kyun? $\int \frac{1}{2} e^{2 x} d x = \frac{1}{4} e^{2 x}$ ; $x$ gone hai, toh hum done hain.

Verify karo: $\frac{1}{4} e^{2 x} (2 x - 1)$ differentiate karo: product rule deta hai $\frac{1}{4} (2 e^{2 x} (2 x - 1) + 2 e^{2 x}) = \frac{1}{4} e^{2 x} (4 x) = x e^{2 x}$ . ✓

Cell B — Hidden product

Worked example Example 2 —

$\int arcsin x d x$ Forecast: yahan sirf ek function hai — "by parts" kaise apply hoga? (Trick: $1$ se multiply karo.)

Product banao. $arcsin x = arcsin x \cdot 1$ likho. Phir $u = arcsin x$ (LIATE mein I), $d v = 1 d x$ . Yeh step kyun? Inverse-trig ka koi elementary antiderivative directly nahi dikhta lekin ek clean derivative hai, isliye yeh $u$ hona chahiye.
Differentiate aur integrate karo. $d u = \frac{1}{1 - x ^{2}} d x$ , aur $v = \int 1 d x = x$ . Yeh step kyun? $\frac{d}{d x} arcsin x = \frac{1}{1 - x ^{2}}$ ; bacha hua $1$ integrate hokar $x$ deta hai.
Formula apply karo. $\int arcsin x d x = x arcsin x - \int \frac{x}{1 - x ^{2}} d x$ Yeh step kyun? $uv - \int v d u$ mein substitute karo; naya integral ab ek plain substitution hai.
Substitute karo $w = 1 - x^{2}$ , $d w = - 2 x d x$ : $\int \frac{x}{1 - x ^{2}} d x = - \frac{1}{2} \int w^{- 1/2} d w = - 1 - x^{2}$ Yeh step kyun? Numerator mein $x d x$ exactly (minus half of) $d w$ hai, isliye root untangle ho jaati hai.
Combine karo. $\int arcsin x d x = x arcsin x + 1 - x^{2} + C$

Verify karo: $\frac{d}{d x} (x arcsin x + 1 - x^{2}) = arcsin x + \frac{x}{1 - x ^{2}} - \frac{x}{1 - x ^{2}} = arcsin x$ . ✓

Cell C — Repeated parts (polynomial stages mein khatam hoti hai)

Worked example Example 3 —

$\int x^{2} e^{- x} d x$ Forecast: $x^{2}$ ek pass mein vanish nahi hogi. Kitne rounds mein yeh gone ho jaayegi? (Do — har baar degree ek kam hogi.)

Pehla pass. $u = x^{2}$ , $d v = e^{- x} d x \Rightarrow d u = 2 x d x$ , $v = - e^{- x}$ . Yeh step kyun? Algebraic, Exponential se pehle; $e^{- x}$ integrate karne se $- e^{- x}$ milta hai (minus dhyan rakhna). $\int x^{2} e^{- x} d x = - x^{2} e^{- x} + \int 2 x e^{- x} d x$
Doosra pass $\int 2 x e^{- x} d x$ par: $u = 2 x$ , $d v = e^{- x} d x \Rightarrow d u = 2 d x$ , $v = - e^{- x}$ . Yeh step kyun? Abhi bhi ek poly×exp product hai; tab tak repeat karo jab tak poly constant na ho jaaye. $\int 2 x e^{- x} d x = - 2 x e^{- x} + \int 2 e^{- x} d x = - 2 x e^{- x} - 2 e^{- x}$
Dono passes combine karo. $\int x^{2} e^{- x} d x = - x^{2} e^{- x} - 2 x e^{- x} - 2 e^{- x} + C = - e^{- x} (x^{2} + 2 x + 2) + C$ Yeh step kyun? Bookkeeping: dono results add karo, $- e^{- x}$ factor out karo.

Yeh exactly woh pattern hai jise Tabular integration (DI method) mechanise karta hai — ek table yeh sab likhne se better hai.

Verify karo: $\frac{d}{d x} [- e^{- x} (x^{2} + 2 x + 2)] = e^{- x} (x^{2} + 2 x + 2) - e^{- x} (2 x + 2) = e^{- x} x^{2}$ . ✓

Cell D — Loop trick (integral wapas aata hai)

Worked example Example 4 —

$\int e^{x} cos x d x$ Forecast: do baar parts karo aur tum wahan wapas aa jaoge jahan se shuru kiya tha. Panic karo — ya opportunity lo?

Integral ko naam do. $I = \int e^{x} cos x d x$ lao. $u = cos x$ , $d v = e^{x} d x \Rightarrow d u = - sin x d x$ , $v = e^{x}$ lo. Yeh step kyun? Koi bhi choice loop karti hai; $I$ ko naam dene se hum baad mein algebraically solve kar sakte hain. $I = e^{x} cos x + \int e^{x} sin x d x$
Doosra pass $\int e^{x} sin x d x$ par: $u = sin x$ , $d v = e^{x} d x \Rightarrow d u = cos x d x$ , $v = e^{x}$ . Yeh step kyun? $u$ ke liye same type (trig) dono baar rakho, warna pass 1 unwind ho jaayega. $\int e^{x} sin x d x = e^{x} sin x - \int e^{x} cos x d x = e^{x} sin x - I$
Substitute back karo aur solve karo. $I = e^{x} cos x + e^{x} sin x - I \Rightarrow 2 I = e^{x} (cos x + sin x)$ $I = \frac{1}{2} e^{x} (sin x + cos x) + C$ Yeh step kyun? $I$ dono sides par appear karta hai — ise unknown treat karo aur isolate karo. Koi infinite regress nahi.

Verify karo: $\frac{d}{d x} [\frac{1}{2} e^{x} (sin x + cos x)] = \frac{1}{2} e^{x} (sin x + cos x) + \frac{1}{2} e^{x} (cos x - sin x) = e^{x} cos x$ . ✓

Cell E — Inverse-trig, bacha hua integral substitution maangta hai

Cell F — Ek definite integral (boundary term carry karo)

Definite integrals ke liye rule ban jaata hai $\int_{a}^{b} u d v = [uv]_{a}^{b} - \int_{a}^{b} v d u,$ jahan $[uv]_{a}^{b} = u (b) v (b) - u (a) v (a)$ (yeh $uv$ term par Fundamental Theorem of Calculus apply karna hai). Boundary bookkeeping ke liye dekho Definite integrals.

Worked example Example 6 —

$\int_{0}^{π} x cos x d x$ Forecast: $[0, π]$ par cosine pehle positive phir negative hota hai — kya tum positive ya negative answer expect karte ho? (Guess karo, phir check karo.)

Choose karo. $u = x$ (A), $d v = cos x d x \Rightarrow d u = d x$ , $v = sin x$ .
Limits ke saath apply karo. $\int_{0}^{π} x cos x d x = [x sin x]_{0}^{π} - \int_{0}^{π} sin x d x$
Boundary term evaluate karo. $[x sin x]_{0}^{π} = π sin π - 0 \cdot sin 0 = π \cdot 0 - 0 = 0$ . Yeh step kyun? $sin π = 0$ aur lower end $0$ hai — poora boundary term collapse ho jaata hai. Figure dekho: $x sin x$ ki height dono ends par zero hai.
Remaining integral evaluate karo. $\int_{0}^{π} sin x d x = [- cos x]_{0}^{π} = - cos π + cos 0 = 1 + 1 = 2$ .
Combine karo. $\int_{0}^{π} x cos x d x = 0 - 2 = - 2$ Negative kyun? $[\frac{π}{2}, π]$ par cosine negative hai aur $x$ bada hai, isliye negative area pehle wale positive area se zyada hai — figure mein red region dominate karta hai.

Verify karo: antiderivative hai $x sin x + cos x$ ; evaluate karo $[\cdot]_{0}^{π} = (π \cdot 0 - 1) - (0 + 1) = - 2$ . ✓

Cell G — Degenerate / limiting endpoint (improper integral)

Worked example Example 7 —

$\int_{0}^{1} ln x d x$ Forecast: $ln x \to - \infty$ jab $x \to 0^{+}$ . Kya area blow up ho jaata hai, ya finite rehta hai? (Finite rehta hai — spike patli hai.)

Parts se antiderivative nikalo. Parent note se, $\int ln x d x = x ln x - x$ (lo $u = ln x$ , $d v = d x$ ).
Bad end ko limit treat karo. Kyunki $ln x$ $x = 0$ par undefined hai, define karo $\int_{0}^{1} ln x d x = lim_{ε \to 0^{+}} [x ln x - x]_{ε}^{1} .$ Yeh step kyun? Hum $ln 0$ directly kabhi evaluate nahi kar sakte; hum $ε$ se uske paas jaate hain (figure mein shrinking sliver dekho).
Ends plug in karo. $x = 1$ par: $1 \cdot 0 - 1 = - 1$ . $x = ε$ par: $ε ln ε - ε$ .
Limit lo. $ε ln ε \to 0$ jab $ε \to 0^{+}$ ( $ε$ , $ln$ ke slow $- \infty$ ko crush kar deta hai), aur $ε \to 0$ . Toh lower end $0$ contribute karta hai. Yeh step kyun? Yeh crucial degenerate check hai: $x ln x$ mein $x$ factor singularity ko tame karta hai.
Result. $\int_{0}^{1} ln x d x = (- 1) - 0 = - 1$

Verify karo: $ε \to 0^{+} lim (ε ln ε - ε) = 0$ , aur value hai $- 1$ . ✓

Cell H — Word problem (physics: oscillating force se kaam)

Worked example Example 8 — Spring-plus-position force se kiya gaya kaam

Ek particle $x = 0$ se $x = π$ metres tak ek line par move karta hai. Uus par force (newtons mein) hai $F (x) = x sin x$ . Kaam hai $W = \int_{0}^{π} F (x) d x$ . $W$ joules mein nikalo. Forecast: force $x$ ke saath badhta hai lekin $sin x$ se oscillate karta hai; $[0, π]$ par $sin x \geq 0$ hai, isliye expect karo $W > 0$ .

Parts set up karo. $u = x$ (A), $d v = sin x d x \Rightarrow d u = d x$ , $v = - cos x$ . Yeh step kyun? $x$ simplify hokar $1$ ho jaata hai; $sin x$ cleanly integrate ho jaata hai.
Limits ke saath apply karo. $W = [- x cos x]_{0}^{π} + \int_{0}^{π} cos x d x$ Plus kyun? Leading term hai $- \int v d u = - \int (- cos x) d x = + \int cos x d x$ .
Boundary term. $[- x cos x]_{0}^{π} = - π cos π + 0 = - π (- 1) = π$ .
Remaining integral. $\int_{0}^{π} cos x d x = [sin x]_{0}^{π} = 0 - 0 = 0$ .
Combine karo. $W = π + 0 = π \approx 3.14$ joules.

Verify karo (units + sign): force (N) × distance (m) = joules ✓; $W = π > 0$ forecast se match karta hai kyunki $sin x \geq 0$ $[0, π]$ par. Numerically $W = π$ . ✓

Cell I — Exam twist: wrong- $u$ trap

Worked example Example 9 —

$\int x sec^{2} x d x$ , aur kyun tempting choice fail ho jaati hai Forecast: $sec^{2} x$ " $tan x$ ka derivative" hai, toh ise integrate karna easy hai — kya yeh $u$ hona chahiye ya $d v$ ?

Trap ( $u = sec^{2} x$ ): phir $d v = x d x$ , $v = \frac{x ^{2}}{2}$ , $d u = 2 sec^{2} x tan x d x$ . Naya integral $\int \frac{x ^{2}}{2} \cdot 2 sec^{2} x tan x d x$ worse hai — $x$ ki higher power, nastier trig. Dead end. Yeh sahi kyun lagta hai: " $sec^{2}$ nicely integrate hota hai, toh ise $u$ banao" — lekin $u$ woh hai jo tum differentiate karte ho, aur $sec^{2} x$ differentiate karne par yeh uglier ban jaata hai.

Fix (LIATE: A before T $\Rightarrow u = x$ ):

$u = x$ , $d v = sec^{2} x d x \Rightarrow d u = d x$ , $v = tan x$ . Yeh step kyun? $x$ simplify hokar $1$ ho jaata hai; $sec^{2} x$ integrate hokar $tan x$ deta hai.
Apply karo. $\int x sec^{2} x d x = x tan x - \int tan x d x$
Finish karo. $\int tan x d x = - ln ∣ cos x ∣$ , toh $\int x sec^{2} x d x = x tan x + ln ∣ cos x ∣ + C$ $ln ∣ cos x ∣$ kyun? $tan x = \frac{sin x}{cos x}$ ; numerator $- \frac{d}{d x} cos x$ hai, jo denominator ka log deta hai.

Verify karo: $\frac{d}{d x} [x tan x + ln ∣ cos x ∣] = tan x + x sec^{2} x + \frac{- sin x}{cos x} = tan x + x sec^{2} x - tan x = x sec^{2} x$ . ✓

Matrix ka Recap

Recall Kaun si trick, kaun si signature ke liye?

Poly × exp, one pass ::: poly ko ek baar differentiate karo (Ex 1) Single log / inverse-trig function ::: $f (x) \cdot 1$ likho, taaki $d v = d x$ ho (Ex 2, 5) Degree $n$ ki poly ::: parts $n$ baar repeat karo / Tabular integration (DI method) use karo (Ex 3) Integral wapas aata hai ::: ise $I$ naam do, algebraically solve karo (Ex 4) Definite integral ::: $[uv]_{a}^{b}$ carry karo, dono ends evaluate karo (Ex 6) Singular endpoint ::: limit $ε \to 0^{+}$ use karo (Ex 7) "Easy to integrate" factor ::: woh $d v$ hai, $u$ NAHI (Ex 9)

Connections

Parent topic — rule aur uska derivation.
Integration by substitution — Ex 2 aur Ex 5 mein bache hue integral ke andar use hota hai.
Definite integrals aur Fundamental Theorem of Calculus — Ex 6–8 mein boundary term.
Tabular integration (DI method) — repeated-parts case ke liye shortcut (Ex 3).
Reduction formulae — repeated parts kya ban jaata hai jab power ek general $n$ ho.