Trigonometric substitution — x = a sin θ, a tan θ, a sec θ cases: Question bank

4.2.9 · D5 · Maths › Calculus II — Integration › Trigonometric substitution — x = a sin θ, a tan θ, a sec θ c

Definition Setup aur notation — pehle yeh padho

Is poore page mein, $a$ ek fixed positive constant ( $a > 0$ ) hai jo integrand se aata hai (jaise $9 - x^{2}$ mein $a = 3$ hai), jabki $x$ integration ka variable hai aur $θ$ woh naya angle variable hai jo hum introduce karte hain. Teeno radicals aur unke substitutions, har ek ke liye poore $θ$ -range ke saath, yeh hain:

Radical	Substitution	Root ban jaata hai	$θ$ -range (kyun)
$a^{2} - x^{2}$	$x = a sin θ$	$a cos θ$	$[- \frac{π}{2}, \frac{π}{2}]$ taaki $cos θ \geq 0$ rahe
$a^{2} + x^{2}$	$x = a tan θ$	$a sec θ$	$(- \frac{π}{2}, \frac{π}{2})$ taaki $sec θ > 0$ rahe
$x^{2} - a^{2}$	$x = a sec θ$	$a tan θ$	$[0, \frac{π}{2}) \cup [π, \frac{3 π}{2})$ taaki $tan θ \geq 0$ rahe

Har range sirf ek reason se choose ki gayi hai: yeh trig factor ko — jisme root simplify hota hai — non-negative rakhti hai, taaki $u^{2} = ∣ u ∣ = u$ ho aur koi sign trap na aaye. Figure s01 mein reference triangles dekho.

Figure — Trigonometric substitution — x = a sin θ, a tan θ, a sec θ cases

Upar ke teeno reference triangles poora "back-substitution" engine hain: tumne $x$ jo bhi side call kiya ho, woh picture tumhe $θ$ ke har trig function ko $x$ aur $a$ ke terms mein de deti hai. Cards answer karte waqt inhe saamne rakhna.

True or false — justify

a^{2} - x^{2}

ke liye

x = a sin θ

ki jagah

x = a cos θ

substitute kar sakte hain.

True —

1 - cos^{2} θ = sin^{2} θ

bhi root ko khatam kar deta hai,

a sin θ

deta hai. Sine sirf conventional choice hai kyunki yeh clean

θ

-range

[- \frac{π}{2}, \frac{π}{2}]

rakhta hai jahan

x \leftrightarrow θ

ka correspondence one-to-one hai.

a^{2} cos^{2} θ = a cos θ

hamesha valid hai.

Saadharan mein False —

u^{2} = ∣ u ∣

, toh yeh

a ∣ cos θ ∣

ke barabar hai (yaad raho

a > 0

). Yeh

a cos θ

sirf isliye banta hai kyunki chosen range

[- \frac{π}{2}, \frac{π}{2}]

cos θ \geq 0

force karti hai.

Trig substitution ke liye integrand mein radical ka hona zaroori hai.

False —

(x^{2} + 7)^{2}

jaisa denominator jisme koi visible root nahi hai woh bhi

x = 7 tan θ

maangta hai. Reason structural hai:

x^{2} + 7

ek sum

a^{2} + x^{2}

hai jisme

a = 7

hai, ek irreducible quadratic;

x = 7 tan θ

x^{2} + 7

7 sec^{2} θ

mein badal deta hai, toh poora

()^{2}

49 sec^{4} θ

ban jaata hai aur integral

cos θ

ki clean power mein collapse ho jaata hai. Surd sign kabhi bhi deciding feature nahi tha —

+

-structure deciding feature hai.

Substitute karne ke baad,

θ

ke terms mein jawab ek complete solution hai.

False — integral

x

mein tha, toh jawab bhi

x

mein hona chahiye. Reference triangle (Figure s01) ke through back-substitute karna zaroori hai;

θ

chhodna ek adhoora problem hai.

x^{2} + a^{2}

ke liye substitution

x = a sec θ

bhi kaam karta hai.

False —

a^{2} sec^{2} θ + a^{2} = a^{2} (sec^{2} θ + 1)

, jo ek perfect square nahi hai (

sec^{2} + 1

ke liye koi identity nahi hai), toh root simplify nahi hoti. Plus-structure

tan

maangti hai, kyunki

1 + tan^{2} θ = sec^{2} θ

hai.

d x

factor optional bookkeeping hai jise tum end mein restore kar sakte ho.

False —

d x = a cos θ d θ

(etc.) integrand ka hissa hai aur aksar crucial factors cancel karta hai; ise skip karna integral ko bilkul badal deta hai.

\int \frac{d x}{a ^{2} + x ^{2}} = \frac{1}{a} arctan \frac{x}{a} + C

bina poore trig substitution ke derive nahi ho sakta.

False. Yeh seedha known derivative

\frac{d}{d x} arctan \frac{x}{a} = \frac{a}{a ^{2} + x ^{2}}

se mil sakta hai, koi substitution zaroori nahi. Trig sub (

x = a tan θ

, ise

\frac{1}{a} \int d θ

mein collapse karna) ek clean route hai, aur yeh explain karta hai

\frac{1}{a}

kahan se aata hai —

a^{2} sec^{2} θ

par

d x = a sec^{2} θ d θ

cancellation — lekin yeh required nahi hai.

Spot the error

9 - x^{2} = 3 cos θ

, aur kyunki yeh ek length hai toh yeh positive hai, isliye koi range check zaroori nahi."

Positivity automatically 'yeh ek length hai' se nahi aati — yeh range

θ \in [- \frac{π}{2}, \frac{π}{2}]

se aati hai jo

cos θ \geq 0

force karti hai. Range restriction actual justification hai, aur tumhe yeh state karna hi hoga.

x^{2} - 25

ke liye maine

x = 5 sin θ

set kiya, jisse

25 sin^{2} θ - 25 = 5 sin^{2} θ - 1

mila."

Error galat sub hai —

sin^{2} θ - 1 \leq 0

, toh imaginary root milti hai.

x^{2} - a^{2}

structure ko

x = 5 sec θ

chahiye, kyunki

sec^{2} θ - 1 = tan^{2} θ \geq 0

hai.

\int 4 - x^{2} d x = 4 \int cos^{2} θ d θ

after

x = 2 sin θ

Yahan

a = 2

hai: root

2 cos θ

deta hai aur

d x = 2 cos θ d θ

, toh product

2 cos θ \cdot 2 cos θ = 4 cos^{2} θ

hai — actually sahi hai. Trap: verify karo ki coefficient genuinely

a^{2} = 2^{2} = 4

hai na ki koi number fudge karke "sahi dikhaya" gaya hai.

x = tan θ

ke baad, mujhe

tan θ - θ + C

mila, aur kyunki

tan θ = x^{2} - 1

hai toh main done hoon."

Mixed cases —

tan θ = x

tan

sub se aaya, lekin

tan θ = x^{2} - 1

sec

sub se belong karta hai. Triangle apne khud ke substitution se banana hoga, kisi doosre ka borrow nahi karna (Figure s01 mein dono triangles compare karo).

θ = arcsin \frac{x}{a}

sabhi real

x

ke liye valid hai."

Sirf

∣ x ∣ \leq a

ke liye.

arcsin

[- 1, 1]

par defined hai, jo

a^{2} - x^{2}

ke real hone ki condition

∣ x ∣ \leq a

se exactly match karta hai — domain aur substitution agree karte hain, yahi point hai.

\int \frac{d x}{x ^{2} x ^{2} + 4}

ke liye main

x^{2}

ko constant ki tarah bahar nikal sakta hoon."

x^{2}

ek variable hai, constant nahi — substitution ke baad yeh

4 tan^{2} θ

ban jaata hai aur integral ke andar rehta hai. Sirf genuine constants (

a^{2} = 4

) bahar ja sakte hain.

Why questions

Pythagorean identities trig substitution ko kaam kyun karaate hain?

Yeh do squares ke sum ya difference ko single squared term mein convert kar dete hain, toh

single square

bina koi root baaki rakhe ek bare trig function ban jaata hai.

a^{2} + x^{2}

ke liye

tan

(na ki

sin

sec

) sahi sub kyun hai?

Sirf

1 + tan^{2} θ = sec^{2} θ

ek plus of squares ko ek square mein badal sakta hai;

sin

cos

differences handle karte hain, aur

sec

x^{2} - a^{2}

handle karta hai.

sec

case mein ek interval ki jagah split range

[0, \frac{π}{2}) \cup [π, \frac{3 π}{2})

kyun hoti hai?

x = a sec θ

ko dono

x \geq a

aur

x \leq - a

reach karne chahiye, aur humein har piece par

tan θ \geq 0

chahiye taaki root

a tan θ

positive rahe; ek continuous interval dono nahi kar sakta.

End mein reference right triangle kyun banate hain?

Substitution ne

θ

ko sides ke ratio ke through define kiya (jaise

sin θ = x / a

), toh triangle (Figure s01) hum logon ko substitution undo karne ke liye

x

ke terms mein

θ

ke har trig function read karne deta hai. Dekho Reference right triangle method.

Inteegrals mein power-reduction itni baar kyun aati hai?

Root khatam karne ke baad aksar

cos^{2} θ

sin^{2} θ

bachta hai, jo directly integrate nahi hota; double-angle identity inhe

\frac{1 \pm c o s 2 θ}{2}

mein linearise kar deti hai. Dekho Power-reduction & double-angle formulas.

\frac{1}{( x ^{2} + a ^{2} ) ^{2}}

ke liye partial fractions ke over trig sub prefer kyun karein?

Partial fractions reals par irreducible quadratic power ko split nahi kar sakta;

x = a tan θ

ise ek clean

cos^{2}

-type integral mein badal deta hai. Compare karo Partial fractions.

Wahi triangle idea arc-length problems mein kyun dobara aata hai?

Arc length integrals

1 + (f^{'})^{2}

produce karte hain, exactly

a^{2} + x^{2}

wali shape, toh trig (ya Hyperbolic substitution) sub natural tool hai. Dekho Arc length and surface area.

Edge cases

x = a

(endpoint) par

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

ka kya hota hai?

Root

0

hai aur

arcsin \frac{a}{a} = \frac{π}{2}

hai, toh formula

\frac{a ^{2}}{2} \cdot \frac{π}{2}

deta hai — ek finite value; integrand bina kisi blow-up ke smoothly zero ko touch karta hai.

Agar

a^{2} - x^{2}

mein

a = 0

ho toh kya?

Yeh

- x^{2}

mein degenerate ho jaata hai, jo sirf

x = 0

par real hai; substitution collapse ho jaata hai kyunki koi triangle banana possible nahi. Trig sub assume karta hai

a > 0

, isliye humne upar

a > 0

fix kiya tha.

x^{2} - a^{2}

ke liye, exactly

x = a

par kya galat hota hai?

Root

0

hai aur

θ = 0

hai (kyunki

sec 0 = 1

), allowed range ki boundary; integrand

\frac{x ^{2} - a ^{2}}{x}

theek hai, lekin

\frac{1}{x ^{2} - a ^{2}}

wahan singular hoga.

Kya trig sub

x^{2} + a^{2}

x < 0

hone par handle kar sakta hai?

Haan —

x = a tan θ

jisme

θ \in (- \frac{π}{2}, 0)

negative

x

cover karta hai, aur

sec θ > 0

poore range mein rehta hai, toh root bina kisi sign issue ke positive rehta hai.

Kya trig sub

x^{2} + a^{2}

karne ka sirf ek hi tarika hai?

Nahi — hyperbolic substitution

x = a sinh t

1 + sinh^{2} t = cosh^{2} t

use karta hai aur aksar messy back-substitution se bachata hai; yeh ek genuine alternative hai. Dekho Hyperbolic substitution.

Agar quadratic already

a^{2} \pm x^{2}

form mein nahi hai, jaise

x^{2} + 4 x + 13

, toh kya karein?

Pehle complete the square karo:

x^{2} + 4 x + 13 = (x + 2)^{2} + 9

, phir

x + 2 = 3 tan θ

substitute karo (effectively

a = 3

). Trig sub ko standard form chahiye; algebra use se set hoti hai.

\int \frac{d x}{a ^{2} - x ^{2}}

ka jawab

arcsin

kyun hai lekin

\int \frac{d x}{x ^{2} - a ^{2}}

log kyun deta hai?

sin

-case

\int d θ = θ = arcsin

mein collapse ho jaata hai, lekin

sec

-case

\int sec θ d θ

bachata hai, jiska antiderivative ek logarithm hai — alag residual integrals, alag function families.

Connections

Pythagorean identities — kyun har case simplify hota hai.
Integration by substitution (u-sub) — change-of-variable ka backbone.
Reference right triangle method — back-substitution tool (Figure s01 mein drawn hai).
Power-reduction & double-angle formulas — baaki bache $sin^{2}, cos^{2}$ ke liye.
Partial fractions — alternative jab quadratic reducible ho.
Arc length and surface area — jahan yeh roots naturally aate hain.
Hyperbolic substitution — parallel toolkit.