Trigonometric substitution — x = a sin θ, a tan θ, a sec θ cases: Worked examples

4.2.9 · D3 · Maths › Calculus II — Integration › Trigonometric substitution — x = a sin θ, a tan θ, a sec θ c

Yeh parent topic ka case-atlas hai. Parent ne tumhe machinery dikhaayi. Yahan hum use har tarah ke pothole se guzaarte hain: har sign pattern, definite integrals jinka limits humein quadrants ke baare mein sochne par majboor karte hain, ek degenerate limit, ek real-world arc, aur ek exam twist jahan pattern chupi hoti hai.

Symbols se pehle, ek waada: neeche har letter earn kiya gaya hai. $a$ hamesha ek fixed positive number hai (ek length — socho "radius"). $x$ hamara variable hai (ek "position"). $θ$ (theta) ek Greek letter hai jo hum angle ke liye use karte hain, radians mein measure kiya jaata hai. Ek radian bas ek angle ka size hai; $π$ radians = seedhi line = $18 0^{\circ}$ .

Scenario matrix

Har trig-sub problem in cells mein se exactly ek mein jaati hai. Neeche ke hamare examples inhe saare cover karte hain.

#	Cell (scenario)	Kaun sa sub	Kaun cover karta hai
A	$a^{2} - x^{2}$ , indefinite, root denominator mein	$x = a sin θ$	Ex 1
B	$a^{2} - x^{2}$ , definite integral (limits → quadrant care)	$x = a sin θ$	Ex 2
C	$a^{2} + x^{2}$ , root upar, $\int sec θ d θ$ chahiye	$x = a tan θ$	Ex 3
D	$(a^{2} + x^{2})^{2}$ power denominator, koi visible root nahi	$x = a tan θ$	Ex 4
E	$x^{2} - a^{2}$ jab $x > 0$ (branch $[0, \frac{π}{2})$ )	$x = a sec θ$	Ex 5
F	Negative $x$ region $x < - a$ (branch $[π, \frac{3 π}{2})$ ) — sign trap	$x = a sec θ$	Ex 6
G	Pehle completing the square (offset centre, hidden pattern)	$x - h = a sin θ$	Ex 7
H	Word problem / geometry — ellipse cap ka area (real-world)	$x = a sin θ$	Ex 8
I	Degenerate / limiting case: $a \to 0$ , kya formula survive karta hai?	—	Ex 9

Example A — denominator mein root (indefinite)

Worked example Ex 1 · Cell A —

$\int \frac{x ^{2}}{16 - x ^{2}} d x$ Forecast: radical "number minus $x^{2}$ " hai, $16 = 4^{2}$ . Padhne se pehle sub guess karo.

Step 1. $a^{2} - x^{2}$ pehchano jahan $a = 4$ . Set karo $x = 4 sin θ$ , toh $d x = 4 cos θ d θ$ , aur restrict karo $θ \in [- \frac{π}{2}, \frac{π}{2}]$ . Yeh step kyun? Sirf $1 - sin^{2} θ = cos^{2} θ$ (ek Pythagorean identity) squares ke difference ko single square mein badalta hai, root khatam karta hai.

Step 2. Root simplify karo: $16 - 16 sin^{2} θ = 16 cos^{2} θ = 4 cos θ$ . $\pm$ kyun nahi? $[- \frac{π}{2}, \frac{π}{2}]$ par $cos θ \geq 0$ hai, toh $∣4 cos θ ∣ = 4 cos θ$ .

Step 3. Assemble karo. Kyunki $x^{2} = 16 sin^{2} θ$ : $\int \frac{16 s i n ^{2} θ}{4 c o s θ} \cdot 4 cos θ d θ = 16 \int sin^{2} θ d θ .$ Yeh step kyun? Upar ka $cos θ$ ( $d x$ se) aur neeche ka $cos θ$ (root se) cancel ho jaate hain — yahi sub ka poora payoff hai.

Step 4. Power-reduce karo: $sin^{2} θ = \frac{1 - c o s 2 θ}{2}$ (Power-reduction & double-angle formulas se): $16 \cdot \frac{1}{2} (θ - \frac{s i n 2 θ}{2}) + C = 8 θ - 4 sin 2 θ + C = 8 θ - 8 sin θ cos θ + C,$ $sin 2 θ = 2 sin θ cos θ$ use kiya. Yeh step kyun? $sin^{2}$ ka seedha koi elementary antiderivative nahi hai; double-angle identity ise term by term integrate karne layak cheezein bana deti hai.

Step 5. Reference triangle se back-substitute karo: $sin θ = \frac{x}{4}$ , adjacent $= 16 - x^{2}$ , toh $cos θ = \frac{16 - x ^{2}}{4}$ aur $θ = arcsin \frac{x}{4}$ . $8 arcsin \frac{x}{4} - 8 \cdot \frac{x}{4} \cdot \frac{16 - x ^{2}}{4} + C = 8 arcsin \frac{x}{4} - \frac{x 16 - x ^{2}}{2} + C .$

Verify: VERIFY block $8 arcsin \frac{x}{4} - \frac{x 16 - x ^{2}}{2}$ ko differentiate karta hai aur confirm karta hai ki yeh exactly $\frac{x ^{2}}{16 - x ^{2}}$ mein simplify hota hai. ✓

Example B — ek DEFINITE integral: limits angles bante hain

Worked example Ex 2 · Cell B —

$\int_{0}^{3} 9 - x^{2} d x$ Forecast: yeh radius $3$ ke circle ke ek quarter ke neeche ka area hai. Compute karne se pehle number predict karo.

Step 1. $a = 3$ , $x = 3 sin θ$ , $d x = 3 cos θ d θ$ , root $\to 3 cos θ$ . Yeh step kyun? Wahi $a^{2} - x^{2}$ pattern.

Step 2 — LIMITS convert karo, back-substitute mat karo. Definite integral ke liye bounds ko $θ$ mein push karna zyaada clean hai:

$x = 0 \Rightarrow 3 sin θ = 0 \Rightarrow θ = 0$ .
$x = 3 \Rightarrow 3 sin θ = 3 \Rightarrow sin θ = 1 \Rightarrow θ = \frac{π}{2}$ .

Yeh step kyun? Agar hum bounds abhi change kar lein, toh substitution undo karna nahi padega — answer seedha number ke roop mein aa jaata hai.

Step 3. Integrand $root 9 - x^{2} se 3 cos θ \cdot d x se 3 cos θ d θ$ ban jaata hai — do cosines multiply hote hain, ek root replace karne se aur ek $d x$ se: $\int_{0}^{π /2} 3 cos θ \cdot 3 cos θ d θ = 9 \int_{0}^{π /2} cos^{2} θ d θ .$ Do cosines kyun aate hain? Substitution integral ko do jagah ek saath touch karti hai: root $9 - x^{2}$ ban jaata hai $3 cos θ$ , aur differential $d x$ ban jaata hai $3 cos θ d θ$ . Inका product $9 cos^{2} θ$ hai — isliye ek bare $cos$ nahi balki $cos^{2}$ show up hota hai.

Step 4. $cos^{2} θ = \frac{1 + c o s 2 θ}{2}$ : $9 [\frac{θ}{2} + \frac{s i n 2 θ}{4}]_{0}^{π /2} = 9 (\frac{π}{4} + 0 - 0) = \frac{9 π}{4} .$ $sin 2 θ$ term kyun vanish hota hai: $sin (π) = 0$ aur $sin 0 = 0$ — dono endpoints ise khatam kar dete hain.

Verify: VERIFY block $\int_{0}^{3} 9 - x^{2} d x$ ko symbolically re-integrate karta hai aur $\frac{9 π}{4}$ paata hai; yeh geometry se bhi match karta hai, radius $3$ ka quarter-disc: $\frac{1}{4} π (3)^{2} = \frac{9 π}{4}$ . ✓

Neeche ki figure precisely woh number dikhati hai jo humne compute kiya — shaded quarter-disc.

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

Example C — root upar, $sec$ integral aata hai

Worked example Ex 3 · Cell C —

$\int \frac{x ^{2} + 25}{x} d x$ (maano $x > 0$ ) Forecast: "number plus $x^{2}$ ", $25 = 5^{2}$ . Sub guess karo.

Step 1. $a^{2} + x^{2}$ jahan $a = 5$ : set karo $x = 5 tan θ$ , $d x = 5 sec^{2} θ d θ$ , $θ \in (- \frac{π}{2}, \frac{π}{2})$ . Yeh step kyun? $1 + tan^{2} θ = sec^{2} θ$ squares ke sum ko ek square mein badal deta hai.

Step 2. Root: $25 + 25 tan^{2} θ = 5 sec θ$ (us range par $sec θ > 0$ ke saath). Substitute karo: $\int \frac{5 s e c θ}{5 t a n θ} \cdot 5 sec^{2} θ d θ = 5 \int \frac{s e c ^{3} θ}{t a n θ} d θ .$

Step 3. Structure expose karne ke liye sin/cos mein convert karo: $\frac{sec ^{3} θ}{tan θ} = \frac{1/ cos ^{3} θ}{sin θ / cos θ} = \frac{1}{cos ^{2} θ sin θ}$ . Yeh step kyun? Sab kuch sin/cos mein hone par hum ek clean substitution ya ek known integral dhundh sakte hain.

Step 4. Split karo $\frac{1}{cos ^{2} θ sin θ} = \frac{sin ^{2} θ + cos ^{2} θ}{cos ^{2} θ sin θ} = \frac{sin θ}{cos ^{2} θ} + \frac{1}{sin θ}$ ( $sin^{2} + cos^{2} = 1$ use kiya). Yeh step kyun? Pehla piece ek perfect $u = cos θ$ sub hai; doosra standard $\int csc θ d θ$ hai. $5 \int \frac{s i n θ}{c o s ^{2} θ} d θ + 5 \int csc θ d θ = 5 \cdot \frac{1}{c o s θ} + 5 ln ∣ csc θ - cot θ ∣ + C = 5 sec θ + 5 ln ∣ csc θ - cot θ ∣ + C .$

Step 5. Reference triangle: $tan θ = \frac{x}{5}$ → opposite $x$ , adjacent $5$ , hyp $x^{2} + 25$ . Toh $sec θ = \frac{x ^{2} + 25}{5}$ , $csc θ = \frac{x ^{2} + 25}{x}$ , $cot θ = \frac{5}{x}$ . $= x^{2} + 25 + 5 ln \frac{x ^{2} + 25 - 5}{x} + C .$

Step 6 — $x < 0$ branch (isliye humne "maano $x > 0$ " kaha). $x < 0$ ke liye substitution $x = 5 tan θ$ ko ek $θ$ chahiye jahan $tan θ < 0$ ho, yaani $θ \in (- \frac{π}{2}, 0)$ , jahan $sec θ$ abhi bhi $> 0$ hai — toh root abhi bhi $5 sec θ$ hai (roots kabhi negative nahi hote), lekin ab $x = 5 tan θ < 0$ jabki $x^{2} + 25 = 5 sec θ > 0$ . Triangle ko dobara padhne par, $csc θ = \frac{x ^{2} + 25}{x}$ negative ho jaata hai, aur $5 ln ∣ csc θ - cot θ ∣$ mein absolute-value bars exactly us sign flip ko absorb kar lete hain. Yeh kyun matter karta hai: antiderivative $x^{2} + 25 + 5 ln \frac{x ^{2} + 25 - 5}{x} + C$ mein $∣ \cdot ∣$ precisely isliye likha gaya hai taaki yeh $x$ ke dono signs ke liye valid rahe — $∣ \cdot ∣$ decoration nahi hai, yahi $x > 0$ aur $x < 0$ branches ko ek formula mein stitch karta hai.

Verify: VERIFY block is answer ko ( $x > 0$ ke liye) differentiate karta hai aur confirm karta hai ki yeh $\frac{x ^{2} + 25}{x}$ mein collapse hota hai. ✓

Example D — power denominator, koi visible root nahi

Worked example Ex 4 · Cell D —

$\int \frac{d x}{( x ^{2} + 4 ) ^{2}}$ Forecast: koi nahi hai! Lekin $(x^{2} + 4)^{2} = (x^{2} + 4)^{4}$ . Phir bhi sub guess karo.

Step 1. $x = 2 tan θ$ , $d x = 2 sec^{2} θ d θ$ . Toh $x^{2} + 4 = 4 sec^{2} θ$ , isliye $(x^{2} + 4)^{2} = 16 sec^{4} θ$ . Yeh step kyun? Sign structure $a^{2} + x^{2}$ dictate karta hai $tan$ chahe root likha ho ya nahi; $tan$ $x^{2} + 4$ ko single $sec^{2}$ mein collapse kar deta hai.

Step 2. $\int \frac{2 s e c ^{2} θ}{16 s e c ^{4} θ} d θ = \frac{1}{8} \int \frac{d θ}{s e c ^{2} θ} = \frac{1}{8} \int cos^{2} θ d θ .$ Yeh step kyun? $1/ sec^{2} θ = cos^{2} θ$ — $sec$ ke powers cancel hokar friendly $cos^{2}$ banta hai.

Step 3. Power-reduce karo: $\frac{1}{8} \cdot \frac{1}{2} (θ + \frac{s i n 2 θ}{2}) + C = \frac{θ}{16} + \frac{s i n θ c o s θ}{16} + C .$

Step 4. Triangle: $tan θ = \frac{x}{2}$ , hyp $x^{2} + 4$ , toh $sin θ = \frac{x}{x ^{2} + 4}$ , $cos θ = \frac{2}{x ^{2} + 4}$ , $θ = arctan \frac{x}{2}$ . $= \frac{1}{16} arctan \frac{x}{2} + \frac{1}{16} \cdot \frac{2 x}{x ^{2} + 4} + C = \frac{1}{16} arctan \frac{x}{2} + \frac{x}{8 ( x ^{2} + 4 )} + C .$

Verify: VERIFY block is answer ko differentiate karta hai aur confirm karta hai ki yeh $\frac{1}{( x ^{2} + 4 ) ^{2}}$ ke barabar hai. ✓

Example E — $x^{2} - a^{2}$ jab $x > 0$

Worked example Ex 5 · Cell E —

$\int \frac{d x}{x ^{2} x ^{2} - 9}$ , maano $x > 3$ Forecast: " $x^{2}$ minus number". Kaun sa sub?

Step 1. $x^{2} - a^{2}$ , $a = 3$ : $x = 3 sec θ$ , $d x = 3 sec θ tan θ d θ$ . $x > 3$ ke liye hum branch $θ \in [0, \frac{π}{2})$ par hain, jahan $tan θ \geq 0$ . Yeh step kyun? $sec^{2} θ - 1 = tan^{2} θ$ root hata deta hai; branch choice $tan θ$ (isliye root) ko non-negative rakhti hai.

Step 2. Root: $9 sec^{2} θ - 9 = 3 tan θ$ . Aur $x^{2} = 9 sec^{2} θ$ . Substitute karo: $\int \frac{3 s e c θ t a n θ}{9 s e c ^{2} θ \cdot 3 t a n θ} d θ = \frac{1}{9} \int \frac{d θ}{s e c θ} = \frac{1}{9} \int cos θ d θ .$ Yeh step kyun? Sab kuch $cos θ$ tak cancel ho jaata hai — yahi inaam hai.

Step 3. $= \frac{1}{9} sin θ + C .$

Step 4. Triangle: $sec θ = \frac{x}{3}$ → hyp $x$ , adjacent $3$ , opposite $x^{2} - 9$ , toh $sin θ = \frac{x ^{2} - 9}{x}$ . $= \frac{x ^{2} - 9}{9 x} + C .$

Verify: VERIFY block $\frac{x ^{2} - 9}{9 x}$ ko ( $x > 3$ ke liye) differentiate karta hai aur confirm karta hai ki yeh $\frac{1}{x ^{2} x ^{2} - 9}$ ke barabar hai. ✓

Step 4 mein use kiya gaya reference triangle neeche draw kiya gaya hai.

Example F — NEGATIVE branch, jahan signs bite karte hain

Worked example Ex 6 · Cell F —

$\int_{- 4}^{- \frac{4}{3}} \frac{x ^{2} - 4}{x} d x$ (saare $x < - 2$ ) Forecast: parent ke Ex 3 jaisi hi shape, lekin ab $x$ negative hai. Kya $x^{2} - 4 = 2 tan θ$ abhi bhi sach hoga? Predict karo.

Step 1. $a = 2$ , $x = 2 sec θ$ . $x < - 2$ ke liye, $sec θ < - 1$ , toh humein branch $θ \in [π, \frac{3 π}{2})$ use karni padegi. Wahan $tan θ \geq 0$ , toh $x^{2} - 4 = 4 tan^{2} θ = 2 tan θ$ (abhi bhi $+$ — branch precisely isliye choose ki gayi taaki yeh hold kare). Yeh step kyun? $u^{2} = ∣ u ∣$ . Doosra branch choose karna guarantee karta hai $tan θ \geq 0$ , toh absolute value cleanly hat jaata hai — yahi $[π, \frac{3 π}{2})$ restriction ka poora reason hai.

Step 2 — limits convert karo, branch $[π, \frac{3 π}{2})$ par rehte hue. Is branch par har angle principal $arccos$ value plus $π$ hai (taaki quadrant III mein land ho):

$x = - 4 \Rightarrow sec θ = \frac{x}{2} = - 2 \Rightarrow cos θ = - \frac{1}{2}$ . Is cosine wala quadrant-III angle $θ_{1} = π + \frac{π}{3} = \frac{4 π}{3}$ hai.
$x = - \frac{4}{3} \Rightarrow sec θ = \frac{x}{2} = - \frac{2}{3} \Rightarrow cos θ = - \frac{3}{2}$ . Is cosine wala quadrant-III angle $θ_{2} = π + \frac{π}{6} = \frac{7 π}{6}$ hai.

$π$ kyun add karein? Principal $arccos$ angle ko quadrant II mein land karta hai ( $\frac{π}{2}$ aur $π$ ke beech), jahan $tan < 0$ hai aur hamaari root identity fail ho jaati. $+ π$ se reflect karna use hamaari chosen branch par quadrant III mein le jaata hai, jahan $tan \geq 0$ hai aur $x^{2} - 4 = 2 tan θ$ hold karta hai. Negative $x$ hi hai jo in quadrant-III angles ko force karta hai.

Step 3. $d x = 2 sec θ tan θ d θ$ aur root $2 tan θ$ ke saath: $\int \frac{2 t a n θ}{2 s e c θ} \cdot 2 sec θ tan θ d θ = \int 2 tan^{2} θ d θ = 2 \int (sec^{2} θ - 1) d θ = 2 (tan θ - θ) + C .$ $sec^{2} - 1$ kyun: $tan^{2}$ directly integrable nahi hai; $tan^{2} = sec^{2} - 1$ hai.

Step 4 — branch angles ke beech evaluate karo. Endpoint $x = - 4$ lower $x$ -limit hai aur $θ_{1} = \frac{4 π}{3}$ map karta hai; endpoint $x = - \frac{4}{3}$ upper $x$ -limit hai aur $θ_{2} = \frac{7 π}{6}$ map karta hai. Toh $\int_{- 4}^{- 4/ 3} \frac{x ^{2} - 4}{x} d x = [2 (tan θ - θ)]_{θ = 4 π /3}^{θ = 7 π /6} .$ Is branch par, $tan \frac{4 π}{3} = 3$ aur $tan \frac{7 π}{6} = \frac{1}{3}$ . Isliye

=2\left(\tfrac{1}{\sqrt3}-\sqrt3+\tfrac{\pi}{6}\right).$$ Numerically $\tfrac{1}{\sqrt3}-\sqrt3=-\tfrac{2}{\sqrt3}\approx-1.1547$, toh answer hai $2(-1.1547+0.5236)\approx -1.262$. **Step 5 — back-substitution sanity ($x$ mein express karo).** Kyunki $\tan\theta=\tfrac{\sqrt{x^2-4}}{|x|}$ aur yahan $x<0$ toh $|x|=-x$, antiderivative $2(\tan\theta-\theta)$ $x$-language mein $-\dfrac{2\sqrt{x^2-4}}{x}-2\,\theta(x)$ padhta hai; is $x$-form ko $x=-4$ aur $x=-\tfrac4{\sqrt3}$ par evaluate karne par wahi $2\left(\tfrac1{\sqrt3}-\sqrt3+\tfrac\pi6\right)$ milta hai. Integrand $\frac{\sqrt{x^2-4}}{x}$ $x<0$ par **negative** hai (numerator $\ge0$, denominator $<0$), aur hum positive width par left-to-right integrate karte hain, toh **negative** answer bilkul expected hai. ✓ **Verify:** VERIFY block $\int_{-4}^{-4/\sqrt3}\frac{\sqrt{x^2-4}}{x}\,dx$ ko symbolically re-compute karta hai; yeh $2\left(\tfrac{1}{\sqrt3}-\sqrt3+\tfrac{\pi}{6}\right)\approx -1.262$ ke barabar hai, aur sign confirmed negative hai. ✓ > [!mistake] Branch trap > $x=-4$ ke liye $\theta=\tfrac{2\pi}{3}$ likhna tempting hai (iska bhi $\cos=-\tfrac12$). Lekin $\tfrac{2\pi}{3}$ quadrant II mein hai jahan $\tan\theta<0$, toh $\sqrt{x^2-4}=2\tan\theta$ **galat** hoga. $x<-a$ ke liye tumhe $[\pi,\tfrac{3\pi}{2})$ mein land karna zaroori hai — isliye $\tfrac{4\pi}{3}$, naa ki $\tfrac{2\pi}{3}$.

Neeche ki figure dikhati hai kyun quadrant III correct branch hai: yeh wahi jagah hai jahan substitution ka tangent non-negative rehta hai taaki root identity survive kare.

Example G — pehle square complete karo (offset centre)

Worked example Ex 7 · Cell G —

$\int \frac{d x}{5 + 4 x - x ^{2}}$ Forecast: linear term $4 x$ hai, toh koi bare $a^{2} - x^{2}$ nahi. Substitute karne se pehle hume kya karna chahiye?

Step 1. Root ke andar square complete karo: $5 + 4 x - x^{2} = - (x^{2} - 4 x) + 5 = - ((x - 2)^{2} - 4) + 5 = 9 - (x - 2)^{2} .$ Yeh step kyun? Trig sub sirf teen canonical shapes pehchaanta hai. Square complete karna shifted quadratic ko $a^{2} - u^{2}$ mein convert karta hai jahan $u = x - 2$ , $a = 3$ .

Step 2. $u = x - 2$ lo (toh $d u = d x$ ). Ab $\int \frac{d u}{9 - u ^{2}}$ , ek standard $a^{2} - u^{2}$ jahan $a = 3$ . Yeh step kyun? Yeh ab identically parent ka memorised form hai.

Step 3. $u = 3 sin θ \Rightarrow \int \frac{3 c o s θ}{3 c o s θ} d θ = θ = arcsin \frac{u}{3}$ .

Step 4. $u = x - 2$ restore karo: $\int \frac{d x}{5 + 4 x - x ^{2}} = arcsin \frac{x - 2}{3} + C .$

Verify: VERIFY block $arcsin \frac{x - 2}{3}$ ko differentiate karta hai aur confirm karta hai $\frac{d}{d x} = \frac{1}{9 - ( x - 2 ) ^{2}} = \frac{1}{5 + 4 x - x ^{2}}$ . ✓

Example H — ek real-world geometry problem

Worked example Ex 8 · Cell H — Elliptical window cap ka area

Ek stained-glass window ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ hai (units: metres). Ek horizontal glazing bar $y = 1.5$ par chhalta hai. Bar ke upar wala cap ka area dhundho (region $y \geq 1.5$ ellipse ke andar).

Forecast: $x$ ke liye solve karne par $number - y^{2}$ milega. $y$ mein sub guess karo.

Step 1 — set up karo. Ellipse se, $x = \pm 4 1 - \frac{y ^{2}}{9} = \pm \frac{4}{3} 9 - y^{2}$ . Height $y$ par cap $x = - \frac{4}{3} 9 - y^{2}$ se $x = + \frac{4}{3} 9 - y^{2}$ tak phaila hai, width $2 \cdot \frac{4}{3} 9 - y^{2}$ hai. Toh $A = \int_{1.5}^{3} \frac{8}{3} 9 - y^{2} d y .$ Yeh step kyun? Area = bar $y = 1.5$ se ellipse top $y = 3$ tak horizontal strip widths ka sum.

Step 2. $9 - y^{2}$ pehchano, $a = 3$ : $y = 3 sin θ$ , $d y = 3 cos θ d θ$ , root $\to 3 cos θ$ . Limits: $y = 1.5 \Rightarrow sin θ = \frac{1}{2} \Rightarrow θ = \frac{π}{6}$ ; $y = 3 \Rightarrow sin θ = 1 \Rightarrow θ = \frac{π}{2}$ . Yeh step kyun? $9 - y^{2}$ $a^{2} - y^{2}$ pattern hai; limits convert karna back-substitution se bachata hai.

Step 3. Root $3 cos θ$ times $d y = 3 cos θ d θ$ do cosines deta hai → ek $cos^{2}$ : $A = \frac{8}{3} \int_{π /6}^{π /2} 3 cos θ \cdot 3 cos θ d θ = 24 \int_{π /6}^{π /2} cos^{2} θ d θ .$

Step 4. $cos^{2} θ = \frac{1 + c o s 2 θ}{2}$ , aur $\int cos^{2} θ d θ = \frac{θ}{2} + \frac{s i n 2 θ}{4}$ :

=24\left[\left(\frac\pi4+\frac{\sin\pi}{4}\right)-\left(\frac{\pi}{12}+\frac{\sin(\pi/3)}{4}\right)\right].$$ $\sin\pi=0$ aur $\sin\tfrac\pi3=\tfrac{\sqrt3}{2}$ ke saath: $$A=24\left(\frac\pi4-\frac\pi{12}-\frac{\sqrt3}{8}\right)=24\cdot\frac\pi6-24\cdot\frac{\sqrt3}{8}=4\pi-3\sqrt3\approx 7.371\ \text{m}^2.$$ *Yeh step kyun?* Power-reduction hi $\cos^2$ integrate karne ka ek maatra tarika hai; endpoints phir seedhe plug in hote hain. **Verify (units + magnitude):** poori ellipse ka area $\pi(4)(3)=12\pi\approx 37.7\text{ m}^2$ hai; hamaara cap ek top slice hai, toh kuch m² sensible lagta hai. Units: $\text{m}\times\text{m}=\text{m}^2$. VERIFY block re-integrate karta hai aur exact value $4\pi-3\sqrt3$ aur numeric $7.371$ confirm karta hai. ✓

Example I — degenerate limit $a \to 0$

Worked example Ex 9 · Cell I — kya

$\int_{0}^{a} a^{2} - x^{2} d x$ $a \to 0$ par sensibly behave karta hai? Forecast: geometrically yeh integral radius $a$ ka quarter-disc hai. Jab $a$ $0$ tak shrink hota hai, disc ek point tak shrink hoti hai — area $0$ jaana chahiye. Predict karo kitni tez se.

Step 1 — integral generally karo. $x = a sin θ$ use karke exactly Ex 2 ki tarah (limits $0 \to \frac{π}{2}$ , root $\to a cos θ$ , $d x \to a cos θ d θ$ ): $\int_{0}^{a} a^{2} - x^{2} d x = a^{2} \int_{0}^{π /2} cos^{2} θ d θ = a^{2} \cdot \frac{π}{4} = \frac{π a ^{2}}{4} .$ Yeh step kyun? Wahi $a^{2} - x^{2}$ machinery; $a^{2}$ front mein factor out ho jaata hai, $a$ -dependence isolate ho jaati hai.

Step 2 — degenerate check $a \to 0^{+}$ . $\frac{π a ^{2}}{4} \to 0$ , aur yeh quadratically karta hai ( $a^{2}$ ki tarah) — $a$ halve karna area quarter kar deta hai, exactly jaisa shrinking disc ko karna chahiye. Koi blow-up nahi, koi undefined $arcsin$ nahi: iska argument $x / a$ poore interval mein $[- 1, 1]$ mein rehta hai. Yeh step kyun? Degenerate inputs wahan hain jahan formulas secretly zero se divide karte hain. Yahan answer $a$ mein ek clean polynomial hai, toh yeh limit survive karta hai.

Step 3 — doosri degeneracy: $x \to a$ neeche se. Indefinite formula $\frac{a ^{2}}{2} arcsin \frac{x}{a} + \frac{x a ^{2} - x ^{2}}{2}$ mein, jab $x \to a$ term $\frac{x a ^{2} - x ^{2}}{2} \to 0$ aur $arcsin \frac{x}{a} \to arcsin 1 = \frac{π}{2}$ — dono finite. Toh even though integrand ka slope $\frac{d}{d x} a^{2} - x^{2}$ $x = a$ par blow up karta hai (ek vertical tangent), area finite hai. Yeh kyun matter karta hai: integrand ka vertical tangent area ko infinite nahi banata — sirf vertical asymptote (integrand $\to \infty$ ) hi aisa karta. Yahan integrand khud bounded rehta hai ( $0 \leq a^{2} - x^{2} \leq a$ ), toh hum safe hain.

Verify: VERIFY block symbolically confirm karta hai $\int_{0}^{a} a^{2} - x^{2} d x = \frac{π a ^{2}}{4}$ , ki yeh $a \to 0$ par $\to 0$ hota hai, aur $a = 2$ par yeh $π$ ke barabar hai (quarter of $π \cdot 2^{2}$ ). ✓

Recap — kaun sa cell, kaun sa move

Recall Har scenario ke liye one-line reflex

Difference $a^{2} - x^{2}$ ::: $x = a sin θ$ ; limits → $arcsin$ , quarter-disc geometry. Sum $a^{2} + x^{2}$ (root upar) ::: $x = a tan θ$ ; $sec^{3} / csc$ integral ke liye dhyan rakho. Power $(a^{2} + x^{2})^{n}$ koi root nahi ::: phir bhi $x = a tan θ$ ; $sec$ ke powers cancel hokar $cos$ powers bante hain. Difference $x^{2} - a^{2}$ , $x > a$ ::: $x = a sec θ$ , branch $[0, \frac{π}{2})$ . Difference $x^{2} - a^{2}$ , $x < - a$ ::: $x = a sec θ$ , branch $[π, \frac{3 π}{2})$ — sign trap! Linear term present hai ::: pehle square complete karo, phir $u = x - h$ sub karo. Definite integral ::: LIMITS ko $θ$ mein convert karo; back-substitution skip karo. Degenerate $a \to 0$ ya endpoint ::: check karo formula finite rehta hai; vertical tangent ≠ infinite area.

Connections

Parent topic (Hinglish) — core machinery.
Pythagorean identities — har root khatam karta hai.
Integration by substitution (u-sub) — andar ke $u$ -subs (Ex 3, Ex 7).
Power-reduction & double-angle formulas — har $cos^{2} / sin^{2}$ step.
Reference right triangle method — saari back-substitutions.
Partial fractions — alternative jab koi root nahi hota.
Arc length and surface area — jahan yeh $a^{2} \pm x^{2}$ integrals naturally aate hain.
Hyperbolic substitution — $x^{2} \pm a^{2}$ ke liye ek parallel tool.