Integration by parts — derivation from product rule, LIATE mnemonic: Question bank

4.2.7 · D5 · Maths › Calculus II — Integration › Integration by parts — derivation from product rule, LIATE m

True or false — justify

Integration by parts ek bilkul naya rule hai jo differentiation se related nahi hai.

False. Ye literally product rule hai jise dono sides par integrate kiya gaya hai — kuch bhi naya assume nahi hota, ye derive hota hai.

Formula

\int u d v = uv - \int v d u

hamesha ek integral ko aasaan bana deta hai.

False. Ye tabhi help karta hai jab naya integral

\int v d u

original se simpler ho;

u

aur

d v

ka galat choice ise aur mushkil bana sakta hai ya circle mein ghuma sakta hai.

Tum integration by parts ko kisi bhi do functions ke product par apply kar sakte ho.

True in principle, useful only sometimes. Rule hamesha valid hai, lekin ye tab hi karne layak hai jab ek factor differentiate karne par simplify ho aur doosra integrate karna aasaan ho.

\int ln x d x

parts se nahi ho sakta kyunki sirf ek function hai.

False. Likho

ln x = ln x \cdot 1

; hidden factor

1

ban jaata hai

d v

, toh ye ek product hi hai.

+ C

" integration by parts mein optional hai kyunki

uv

term use absorb kar leta hai.

False.

uv

ek specific function hai, arbitrary constant nahi; har indefinite integral ko phir bhi

+ C

chahiye.

Agar ek baar parts karne ke baad phir bhi ek product milta hai, toh method fail ho gaya.

False. Parts ko repeat karna expected hai — har baar ek polynomial ka degree ek kam ho jaata hai jab tak wo khatam nahi ho jaata (jaise

\int x^{2} cos x d x

ko do passes chahiye).

u

aur

d v

ki choice final answer badal sakti hai.

False. Ek valid choice kaam mushkil bana sakti hai, lekin koi bhi sahi route same antiderivative deta hai (

+ C

tak); buri choice sirf mehnat badhati hai.

Ek definite integral ke liye,

uv

term ko bhi limits par evaluate karna padta hai.

True. Definite integrals mein boundary term

[uv]_{a}^{b}

ban jaata hai; wahan limits plug in karna bhoola dena ek classic galti hai.

Substitution aur integration by parts interchangeable hain — koi bhi freely choose kar sakte ho.

False. Pehle substitution try karo (wo simpler hai); parts un products ke liye hai jo kisi bhi single substitution se nahi suljhte.

"Loop" trick (

\int e^{x} sin x d x

) isliye kaam karta hai kyunki integral literally gayab ho jaata hai.

False. Original integral wapas aata hai, aur tum use algebraically solve karte ho jaise ek unknown — wo vanish nahi hota, isolate hota hai.

Spot the error

\int x e^{x} d x

ke liye maine

u = e^{x}

set kiya kyunki

e^{x}

itni achhi tarah integrate hota hai."

Rule chahta hai ki

u

differentiate hone par simplify ho.

u = e^{x}

ke saath,

d u = e^{x} d x

kabhi simplify nahi hota aur

\int v d u

aur mushkil ho jaata hai; LIATE kehta hai A before E, toh

u = x

"Maine

\int x e^{x} d x

ko pieces

x

aur

e^{x}

mein tod diya, aur

d v = e^{x}

liya."

d v

ke saath

d x

hona chahiye:

d v = e^{x} d x

. Kyunki

v = \int d v

d x

exactly mark karta hai kya integrate ho raha hai.

\int ln x d x = x ln x - \int x \cdot x d x

d u = \frac{1}{x} d x

hai,

x d x

nahi. Integrand ban jaata hai

x \cdot \frac{1}{x} = 1

, jo deta hai

x ln x - x + C

"Ek pass ke baad mujhe

x^{2} sin x - \int 2 x sin x d x

mila, aur maine doosre pass mein minus drop kar diya."

\int v d u

ka leading

-

har pass se carry hota hai; use khona poori tail ki sign palat deta hai. Formula ko har pass mein fresh boxed likhte raho use rakhne ke liye.

\int e^{x} sin x d x

khud par loop karke wapas aa gaya, toh iska koi jawab nahi hai."

Loop hi feature hai:

I = e^{x} sin x - (e^{x} cos x + I)

deta hai

2 I = e^{x} (sin x - cos x)

, toh

I = \frac{1}{2} e^{x} (sin x - cos x) + C

"Maine

u = x

d v = cos x d x

choose kiya, lekin phir

v = - sin x

likha."

cos x

ko integrate karne par

v = + sin x

milta hai; sign

cos x

ko differentiate karne ka hai, integrate karne ka nahi. Trig ke derivative/integral signs ko mix up karna sabse common galti hai.

"Ek definite integral ke liye maine formula apply kiya lekin limits sirf

\int v d u

term par rakhi."

Dono boundary term aur remaining integral limits par evaluate hote hain:

\int_{a}^{b} u d v = [uv]_{a}^{b} - \int_{a}^{b} v d u

Why questions

LIATE Logarithmic aur Inverse-trig ko

u

ke roop mein pehle kyun rakhta hai?

Inhe integrate karna bahut mushkil hai lekin inke derivatives clean aur simplifying hote hain, isliye ye ideal factors hain jise tum differentiate karte ho.

LIATE Exponential ko last kyun rakhta hai?

e^{x}

effortlessly khud mein integrate ho jaata hai, isliye ye ideal

d v

hai — wo factor jo aasaan integrate hota hai.

Derivation product rule se kyun shuru hoti hai, kuch naya invent karne ki bajay?

Integration differentiation ko undo karta hai, toh koi bhi differentiation identity, jab integrate ki jaaye, ek integration rule deti hai — ek trusted identity reuse karna guessing se behtar hai.

\int \frac{d}{d x} (uv) d x = uv

kyun hai (koi integral sign nahi bacha)?

Fundamental Theorem of Calculus kehta hai ki ek derivative ko integrate karne par original function wapas milta hai.

Polynomials ke liye hum chahte hain ki

u

eventually vanish kare, kyun?

Har pass polynomial ko differentiate karta hai, uska degree kam karta hai; ek baar constant phir

0

ho jaane par remaining integral trivial ho jaata hai aur recursion ruk jaata hai.

d v

kuch aisa kyun hona chahiye jo tum actually integrate kar sako?

Kyunki formula likhne ke liye hi tumhe

v = \int d v

chahiye; agar tum

v

nahi nikaal sakte, toh us choice ke saath parts apply nahi ho sakta.

"Hidden product"

ln x = ln x \cdot 1

ek legitimate move kyun hai?

1

se multiply karne par function mein kuch nahi badalta lekin ek integrable factor (

d v = 1 d x

) expose ho jaata hai taaki parts applicable ho sake.

Repeated parts ko kabhi kabhi tabular integration se kyun organise kiya ja sakta hai?

Jab ek factor finite steps mein differentiate hokar zero ho jaata hai, alternating

uv

terms ek fixed pattern follow karte hain jo ek table automatically bookkeep karta hai.

Edge cases

Agar tum ek polynomial factor ko differentiate karke zero kar do — kya process khatam ho jaata hai?

Haan; ek baar

d u = 0

ho jaane par trailing

\int v d u = 0

hai, toh koi integral nahi bachta aur tum seedha jawab padh lete ho.

Kya hota hai agar tum baar baar parts lagate raho aur kabhi simplify nahi hota (jaise

\int e^{x} sin x

par galat choice)?

Tum forever loop karte raho jab tak tum returning integral ko recognize karke algebraically solve nahi karte, ya apna

u

d v

choice swap nahi karte.

Kya hoga agar dono factors differentiate hone par mushkil ho jaayein (jaise do trig functions)?

LIATE phir bhi ek rule deta hai (T before... jo bhi ho), lekin aksar ek trig identity ya substitution parts se smarter hota hai — parts obligatory nahi hai.

Kya integration by parts kaam karta hai jab

u

v

har jagah differentiable nahi ho?

Derivation assume karti hai ki dono differentiable hain; jahan koi factor smooth nahi hai, formula ka justification toot jaata hai aur tumhe un points ko alag se handle karna padta hai.

Agar tumhara choose kiya hua "

v

" ka apna

+ C

ho — kya farak padta hai?

Koi bhi antiderivative

v

kaam karta hai; arbitrary constant final result mein cancel ho jaata hai, isliye convention hai ki sabse simple

v

lo (constant

= 0

Kya

\int u d v = uv - \int v d u

kabhi antiderivative guess karne se bura ho sakta hai?

Haan — simple integrals ya jinhe ek single substitution solve kar de, unke liye parts pointless kaam badhata hai; ye stubborn products ke liye tool hai, pehla resort nahi.

Connections

Product rule (differentiation) — wo identity jis par har trap trace hoti hai.
Fundamental Theorem of Calculus — kyun derivative cleanly integrate ho jaata hai.
Integration by substitution — wo technique jo parts se pehle try karni chahiye.
Reduction formulae — repeated parts, recursion ke roop mein package ki gayi.
Tabular integration (DI method) — many-pass parts ke liye bookkeeping.
Definite integrals — jahan boundary term $[uv]_{a}^{b}$ daant kaatta hai.