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

4.2.9 · D2 · Maths › Calculus II — Integration › Trigonometric substitution — x = a sin θ, a tan θ, a sec θ c

Pehli line se pehle, teen simple baatein:

• $\int \dots dx$ ka matlab hai "jaise $x$ slide karta hai, thin strips ko add karo" — yeh area under a curve hai.
• ​ hai "right triangle ki woh side jo abhi pata nahi," Pythagoras se.
• $a$ bas ek fixed positive number hai (ek size), aur $x$ hai moving input.

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Step 1 — Square root ko ek circle ki tarah dekho

KYA. $y=\sqrt{a^2-x^2}$ dekho. Dono sides ko square karo: $y^2=a^2-x^2$, yaani $x^2+y^2=a^2$. Yeh origin par centred radius $a$ ke circle ki equation hai (Pythagoras: do legs $x,y$ aur hypotenuse $a$).

KYUN. Humara integral $\int \sqrt{a^2-x^2}dx$ isliye do $x$-values ke beech us circle ki top half ke neeche ka area hai. Shape jaanna hamein batata hai ki final answer zaroor ek angle se describe ho sakta hai — kyunki circles angles se bane hote hain.

PICTURE. Red curve upper semicircle hai. Shaded blue strip ek term $\sqrt{a^2-x^2}dx$ hai — height $\sqrt{a^2-x^2}$ aur width $dx$ wala ek tall thin rectangle.

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Step 2 — Moving point ko uske ANGLE se naam do

KYA. Circle par point $(x,\sqrt{a^2-x^2})$ lo. Uski taraf radius $a$ kheencho. Joh angle woh radius vertical ($y$-axis) se banata hai use $\theta$ kaho. Tab horizontal leg hai $x=a\sin \theta$ aur vertical leg hai $\sqrt{a^2-x^2}=a\cos \theta$.

YEH substitution kyun, koi aur kyun nahi. Circle par har point ek angle se set hota hai. Length $x$ ko angle $\theta$ se replace karna ek awkward coordinate ko circle ke natural parameter se trade karta hai. Yeh exactly change of variable idea hai: woh variable chuno jo shape prefer karta hai.

PICTURE. Pink angle $\theta$ upar se khulta hai; horizontal chalk-blue leg $x=a\sin \theta$ hai; vertical pale-yellow leg $a\cos \theta$ hai.

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Step 3 — Pythagorean identity se root ko khatam karo

KYA. $x=a\sin \theta$ ko root mein substitute karo: $\sqrt{a^2-x^2}=\sqrt{a^2-a^2{\sin }^2\theta }=\sqrt{a^2\underset{={\cos }^2\theta }{\underbrace{(1-{\sin }^2\theta )}}}=\sqrt{a^2{\cos }^2\theta }=a\cos \theta .$

KYUN. Identity $1-{\sin }^2\theta ={\cos }^2\theta$ ek difference of two squares ko ek perfect square mein badal deti hai, toh square root saaf nikal jaata hai. Yahi collapse hai jis wajah se humne sine chuna.

$+a\cos \theta$ kyun, $\pm$ nahi. Generally $\sqrt{u^2}=|u|$. Yahan $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$, aur us arc par $\cos \theta \ge 0$ (vertical leg kabhi neeche nahi jaata), toh $|a\cos \theta |=a\cos \theta$.

PICTURE. Green wedge $1$ (poora radius${}^2$ scaled) ko ${\sin }^2\theta$ aur ${\cos }^2\theta$ mein split karta dikhata hai; vertical leg surviving $a\cos \theta$ ke roop mein highlight hai.

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Step 4 — Width $dx$ ko bhi convert karo

KYA. Agar $x=a\sin \theta$ hai, tab $\theta$ mein ek tiny push $x$ ko move karta hai $dx=a\cos \theta d\theta .$

KYUN. Strip ki width $x$ mein measure hoti hai, lekin ab hum $\theta$ mein count kar rahe hain. Humein ruler translate karna hoga. $x=a\sin \theta$ differentiate karna ($\sin$ ka rate of change $\cos$ hai) conversion factor $a\cos \theta$ deta hai. Yeh bhoolna number-one trig-sub error hai — $dx$ decoration nahi hai, yeh hai strip ki width.

PICTURE. Rim par ek small angle change $d\theta$ ek horizontal shift $dx=a\cos \theta d\theta$ sweep karta hai; jitna point top ke paas hoga ($\cos \theta$ bada), utna wider shift.

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Step 5 — All-angle integral assemble karo

KYA. Height $=a\cos \theta$ (Step 3) aur width $=a\cos \theta d\theta$ (Step 4) sum mein daalo: $\int \sqrt{a^2-x^2}dx=\int \underset{\text{height}}{\underbrace{a\cos \theta }}\cdot \underset{\text{width}}{\underbrace{a\cos \theta d\theta }}=a^2\int {\cos }^2\theta d\theta .$

KYUN. Koi bhi square roots nahi bacha — ab humare paas ek pure trig integral hai, woh type jo hum already jaante hain. Har $x$ ek $\theta$ ban gaya hai.

PICTURE. Side-by-side: left par ugly ​ world, right par clean $a^2{\cos }^2\theta$ world, beech mein "$x=a\sin \theta$" label wala arrow.

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Step 6 — Double-angle identity se ${\cos }^2\theta$ ko flatten karo

KYA. ${\cos }^2\theta$ wiggle karta hai aur directly integrable nahi hai, toh replace karo: ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}.$ Tab $a^2\int {\cos }^2\theta d\theta =\frac{a^2}{2}\int (1+\cos 2\theta )d\theta =\frac{a^2}{2}\left(\theta +\frac{\sin 2\theta }{2}\right)+C.$

KYUN. Power-reduction identity ek square (mushkil) ko constant plus single cosine (dono trivially integrable) ke roop mein rewrite karti hai. $1$ integrate karne se $\theta$ milta hai; $\cos 2\theta$ integrate karne se $\frac{1}{2}\sin 2\theta$ milta hai.

PICTURE. Bumpy curve ${\cos }^2\theta$ (average height $\frac{1}{2}$) ko flat line $\frac{1}{2}$ plus gentle wave $\frac{1}{2}\cos 2\theta$ ke roop mein redraw kiya — same area, easier pieces.

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Step 7 — Double angle undo karo

KYA. $\sin 2\theta =2\sin \theta \cos \theta$ use karo: $\frac{a^2}{2}\left(\theta +\frac{\sin 2\theta }{2}\right)+C=\frac{a^2}{2}\theta +\frac{a^2}{2}\sin \theta \cos \theta +C.$

KYUN. $\theta$ ek angle se aaya; $\sin \theta$ aur $\cos \theta$ humare triangle ke legs hain — woh quantities jo hum seedha picture se padh sakte hain. $\sin 2\theta$ ko waapas $\sin \theta \cos \theta$ mein split karna har term ko $x$ mein translation ke liye taiyaar karta hai.

PICTURE. Single arc-angle term $\frac{a^2}{2}\theta$ (ek pie-slice area) aur product term $\frac{a^2}{2}\sin \theta \cos \theta$ (ek triangle area) circle ke do alag shaded regions ke roop mein dikhaye.

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Step 8 — Reference triangle se har angle ko waapas $x$ mein translate karo

KYA. $x=a\sin \theta$ se reference right triangle banao: opposite $=x$, hypotenuse $=a$, adjacent $=\sqrt{a^2-x^2}$. Padho: $\theta =\arcsin \frac{x}{a},\sin \theta =\frac{x}{a},\cos \theta =\frac{\sqrt{a^2-x^2}}{a}.$ Substitute karo: $\frac{a^2}{2}\underset{\arcsin \frac{x}{a}}{\underbrace{\theta }}+\frac{a^2}{2}\underset{x/a}{\underbrace{\sin \theta }}\underset{\sqrt{a^2-x^2}/a}{\underbrace{\cos \theta }}=\boxed{\frac{a^2}{2}\arcsin \frac{x}{a}+\frac{x\sqrt{a^2-x^2}}{2}+C}$

KYUN. Original question $x$ mein tha; $\theta$ mein trapped answer adhoora hai. Triangle dictionary hai. Dekho $\frac{a^2}{2}\cdot \frac{x}{a}\cdot \frac{\sqrt{a^2-x^2}}{a}$ mein do $a$'s neatly cancel hokar $\frac{x\sqrt{a^2-x^2}}{2}$ dete hain.

PICTURE. Labelled reference triangle jisme $x$, $a$, $\sqrt{a^2-x^2}$, aur $\theta =\arcsin \frac{x}{a}$ marked hain.

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Step 9 — Edges check karo (koi bhi case dikhaye bina mat chodo)

KYA & KYUN. Ek formula apne extreme inputs mein survive karna chahiye.

• $x=0$. $\arcsin 0=0$ aur triangle term $0$ hai → total area $0$. Correct: koi strip nahi, koi area nahi. ✓
• $x=a$. $\arcsin \frac{a}{a}=\arcsin 1=\frac{\pi }{2}$, aur $\sqrt{a^2-a^2}=0$, toh answer $=\frac{a^2}{2}\cdot \frac{\pi }{2}=\frac{\pi a^2}{4}$ — exactly poore circle $\pi a^2$ ka ek quarter. ✓ Poora quarter-disc.
• $x=-a$. $\arcsin (-1)=-\frac{\pi }{2}$, root $=0$, answer $=-\frac{\pi a^2}{4}$; sign sirf accumulation ki direction track karta hai. ✓
• Domain limit. $\sqrt{a^2-x^2}$ ko $|x|\le a$ chahiye — uske bahar semicircle exist hi nahi karta, $\arcsin$ ke domain $[-1,1]$ se match karta hai. Substitution ki range $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$ precisely yahi cover karti hai. ✓

PICTURE. Quarter disc shaded, $x=a$ par sector wedge aur vanishing triangle ke saath.

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Ek-picture summary

Sab kuch ek board par: semicircle, strip, angle $\theta$, reference triangle, aur final area sector + triangle mein split.

Recall Feynman retelling — ek dost ko batao

Ek square root $\sqrt{a^2-x^2}$ secretly radius $a$ ke circle ki height hai. Toh un heights ko add karna jaise $x$ slide karta hai bas ek circular arc ke neeche area measure karna hai. Circles ko side-lengths se describe kiya jaana pasand nahi; unhe angles pasand hain. Toh hum apne point ko us angle $\theta$ se rename karte hain jo uska radius banata hai — ab $x=a\sin \theta$ aur height $a\cos \theta$ hai. Pythagorean identity square root ko poori tarah nigal jaati hai. Hum strip width bhi angle steps mein re-measure karte hain, $dx=a\cos \theta d\theta$. Poora integral $a^2\int {\cos }^2\theta d\theta$ ban jaata hai, jise hum double-angle trick se constant plus cosine mein flatten karte hain — dono easy. Integrate karne se ek angle piece milta hai ($\frac{a^2}{2}\theta$) aur ek product piece ($\frac{a^2}{2}\sin \theta \cos \theta$). Finally hum $x=a\sin \theta$ se triangle draw karte hain aur angles ko waapas $x$ mein translate karte hain: angle piece pie-slice hai, product piece radius ke neeche triangle hai, aur milke woh exactly woh circular area hain jise hum measure karna shuru kiye the. Humne area ko angle mein disguise kiya, easily solve kiya, aur undisguise kar diya.

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Recall checkpoints

Recall Quick self-test

$y=\sqrt{a^2-x^2}$ kaun si shape hai? ::: radius $a$ ke circle ki top half $x=a\sin \theta$ substitute kyun karte hain? ::: circle par ek point naturally apne angle se set hota hai; sine root ko $1-{\sin }^2={\cos }^2$ se khatam karta hai $dx$ ki jagah kya aata hai? ::: $a\cos \theta d\theta$ — substitution differentiate karo $a\cos \theta$ par $\pm$ kyun nahi? ::: $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$ $\cos \theta \ge 0$ rakhta hai Double-angle identity kyun use karte hain? ::: ${\cos }^2\theta$ directly integrable nahi hai; yeh $\frac{1}{2}(1+\cos 2\theta )$ ban jaata hai Do final terms geometrically matlab kya rakhte hain? ::: sector area + triangle area = swept circular region $x=a$ check karo: ::: $\frac{\pi a^2}{4}$ milta hai, ek quarter circle ✓

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Connections

• Parent topic (Hinglish) — woh full three-case table jisme yeh page zoom karta hai.
• Pythagorean identities — Step 3 ka engine.
• Integration by substitution (u-sub) — trig sub ek clever change of variable hai.
• Power-reduction & double-angle formulas — Steps 6–7.
• Reference right triangle method — Step 8, back-translation dictionary.
• Arc length and surface area — jahan yeh circular-area integrals dobara aate hain.
• Hyperbolic substitution — $\sqrt{a^2+x^2}$ / $\sqrt{x^2-a^2}$ ka cousin $\sinh ,\cosh$ ke saath.