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

WHAT hum dekhte hain: radical ke andar ka sign structure — ye akela choice dictate karta hai. (a) $16-x^2$ ye $a^2-x^2$ form hai jisme $a=4$ (kyunki $4^2=16$). → $x=4\sin \theta$, $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$, $dx=4\cos \theta d\theta$, root $\to 4\cos \theta$ (valid hai kyunki is interval par $\cos \theta \ge 0$). WHY sine: $1-{\sin }^2\theta ={\cos }^2\theta$ ek $a^2$ se shuru hone wale difference ko khatam kar deta hai. (b) $x^2+5$ ye $a^2+x^2$ form hai jisme $a=\sqrt{5}$. → $x=\sqrt{5}\tan \theta$, $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$, $dx=\sqrt{5}{\sec }^2\theta d\theta$, root $\to \sqrt{5}\sec \theta$ (valid hai kyunki is interval par $\sec \theta >0$). WHY tangent: $1+{\tan }^2\theta ={\sec }^2\theta$ ek sum ko khatam kar deta hai. (c) $x^2-9$ ye $x^2-a^2$ form hai jisme $a=3$. → $x=3\sec \theta$, $\theta \in [0,\frac{\pi }{2})$ ($x\ge 3$ lete hue), $dx=3\sec \theta \tan \theta d\theta$, root $\to 3\tan \theta$ (valid hai kyunki is interval par $\tan \theta \ge 0$). WHY secant: ${\sec }^2\theta -1={\tan }^2\theta$ ek $x^2$ se shuru hone wale difference ko khatam kar deta hai.

Step 1 — substitute. $x=4\sin \theta$, $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$, $dx=4\cos \theta d\theta$. Figure s02 mein sine triangle dekho: opp $=x$, hyp $=4$. Step 2 — root simplify karo. $\sqrt{16-x^2}=\sqrt{16-16{\sin }^2\theta }=\sqrt{16{\cos }^2\theta }=4\cos \theta$. WHY $+4\cos \theta$, koi $|\cdot |$ nahi bachta: generally $\sqrt{u^2}=|u|$ hota hai, to strictly $\sqrt{16{\cos }^2\theta }=4|\cos \theta |$. Lekin chosen range $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$ force karti hai $\cos \theta \ge 0$, isliye $|\cos \theta |=\cos \theta$ aur bars drop ho jaate hain. Ye hi wajah hai ki range restriction exist karti hai. Step 3 — assemble. $dx$ ka $4\cos \theta$ aur root ka $4\cos \theta$ cancel ho jaate hain: $\int \frac{4\cos \theta d\theta }{4\cos \theta }=\int 1d\theta =\theta +C.$ WHY $\int 1d\theta =\theta$: wo function jiska derivative constant $1$ ho, woh $\theta$ khud hai ($\frac{d}{d\theta }\theta =1$). Derivative ko ulta padhne se integral milta hai. Step 4 — back-substitute. $\sin \theta =\frac{x}{4}$ se (Figure s02), $\theta =\arcsin \frac{x}{4}$ — "wo angle jiska sine $\frac{x}{4}$ hai." $\boxed{\int \frac{dx}{\sqrt{16-x^2}}=\arcsin \frac{x}{4}+C,x\in (-4,4)}$ Edge/domain note: integrand ko $16-x^2>0$ chahiye, yani $x\in (-4,4)$; $x=\pm 4$ par root $0$ ho jaata hai aur integrand blow up ho jaata hai, isliye answer strictly andar hi valid hai. Aur $\arcsin \frac{x}{4}$ khud sirf $\frac{x}{4}\in [-1,1]$ ke liye defined hai — same interval, jo ek acha consistency check hai.

Step 1. $a=3$, $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$: $x=3\sin \theta$, $dx=3\cos \theta d\theta$, $\sqrt{9-x^2}=3\cos \theta$ (bars drop kyunki is range par $\cos \theta \ge 0$), $x^2=9{\sin }^2\theta$. Wahi sine triangle, Figure s02 with $a=3$. Step 2 — saare pieces substitute karo. Yaad raho $\csc \theta =\frac{1}{\sin \theta }$ (upar define kiya), isliye $\frac{1}{{\sin }^2\theta }={\csc }^2\theta$: $\int \frac{3\cos \theta d\theta }{9{\sin }^2\theta \cdot 3\cos \theta }=\frac{1}{9}\int \frac{d\theta }{{\sin }^2\theta }=\frac{1}{9}\int {\csc }^2\theta d\theta .$ WHY ye collapse hua: $dx$ ka $\cos \theta$, root ke $\cos \theta$ se cancel ho gaya — sirf $\sin$ mein ek standard integral bachti hai. Step 3. $\int {\csc }^2\theta d\theta =-\cot \theta$, to $=-\frac{1}{9}\cot \theta +C$. WHY $\int {\csc }^2\theta d\theta =-\cot \theta$: $-\cot \theta$ differentiate karo. Kyunki $\cot \theta =\frac{\cos \theta }{\sin \theta }$, quotient rule deta hai $\frac{d}{d\theta }\cot \theta =-{\csc }^2\theta$, to $\frac{d}{d\theta }(-\cot \theta )={\csc }^2\theta$. Ulta padhna integral deta hai. Step 4 — triangle (Figure s02). $\sin \theta =\frac{x}{3}$: opposite $=x$, hyp $=3$, adjacent $=\sqrt{9-x^2}$, isliye $\cot \theta =\frac{\text{adj}}{\text{opp}}=\frac{\sqrt{9-x^2}}{x}$. $\boxed{=-\frac{\sqrt{9-x^2}}{9x}+C,0<|x|<3}$

WHY tangent: $(x^2+4{)}^{3/2}=(\sqrt{x^2+4}{)}^3$ — ek sum $a^2+x^2$ hai jisme $a=2$. Step 1. $x=2\tan \theta$, $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$, $dx=2{\sec }^2\theta d\theta$. Is interval par $\sec \theta >0$, isliye $\sqrt{x^2+4}=2\sec \theta \Rightarrow (x^2+4{)}^{3/2}=(2\sec \theta {)}^3=8{\sec }^3\theta$. Tangent triangle, Figure s03 dekho. Step 2. $\int \frac{2{\sec }^2\theta d\theta }{8{\sec }^3\theta }=\frac{1}{4}\int \frac{d\theta }{\sec \theta }=\frac{1}{4}\int \cos \theta d\theta =\frac{1}{4}\sin \theta +C.$ WHY $\frac{1}{\sec \theta }=\cos \theta$: kyunki $\sec \theta$ defined hai $\frac{1}{\cos \theta }$ ke roop mein, uska reciprocal $\cos \theta$ hai. WHY $\int \cos \theta d\theta =\sin \theta$: $\frac{d}{d\theta }\sin \theta =\cos \theta$, isliye integrate karna (differentiate ka reverse) $\sin \theta$ deta hai. Step 3 — triangle (Figure s03). $\tan \theta =\frac{x}{2}$: opposite $=x$, adjacent $=2$, hyp $=\sqrt{x^2+4}$, isliye $\sin \theta =\frac{\text{opp}}{\text{hyp}}=\frac{x}{\sqrt{x^2+4}}$. $\boxed{=\frac{x}{4\sqrt{x^2+4}}+C(\text{all real }x)}$

Step 1. $a=3$, $\theta \in [0,\frac{\pi }{2})$ ($x\ge 3$ lete hue): $x=3\sec \theta$, $dx=3\sec \theta \tan \theta d\theta$. Is interval par $\tan \theta \ge 0$, isliye $\sqrt{x^2-9}=\sqrt{9{\tan }^2\theta }=3\tan \theta$ ($|\cdot |$ drop ho jaata hai kyunki yahan $\tan \theta \ge 0$ hai — upar range table dekho). Secant triangle, Figure s04. Step 2 — assemble. $\int 3\tan \theta \cdot 3\sec \theta \tan \theta d\theta =9\int \sec \theta {\tan }^2\theta d\theta .$ Step 3 — ${\tan }^2$ ko ${\sec }^2-1$ mein badlo (Pythagorean identities ke through, taaki pieces aise integrals ban jaayein jo hum build kar sakein): $9\int \sec \theta ({\sec }^2\theta -1)d\theta =9\int {\sec }^3\theta d\theta -9\int \sec \theta d\theta .$ Step 4 — donon sub-integrals KAHAN se aate hain (ye "just known" nahi hain). $\int \sec \theta d\theta$ ke liye: upar aur neeche $(\sec \theta +\tan \theta )$ se multiply karo: $\int \sec \theta \cdot \frac{\sec \theta +\tan \theta }{\sec \theta +\tan \theta }d\theta =\int \frac{{\sec }^2\theta +\sec \theta \tan \theta }{\sec \theta +\tan \theta }d\theta$. Numerator exactly denominator ka derivative hai ($\frac{d}{d\theta }(\sec \theta +\tan \theta )=\sec \theta \tan \theta +{\sec }^2\theta$), isliye ye $\int \frac{w^{\prime }}{w}d\theta =\ln |w|$ hai, jo deta hai $\int \sec \theta d\theta =\ln |\sec \theta +\tan \theta |$. $\int {\sec }^3\theta d\theta$ ke liye: integration by parts with $u=\sec \theta ,dv={\sec }^2\theta d\theta$ se milta hai $\int {\sec }^3=\sec \theta \tan \theta -\int \sec \theta {\tan }^2\theta =\sec \theta \tan \theta -\int {\sec }^3\theta +\int \sec \theta$. Is loop ko $\int {\sec }^3\theta$ ke liye solve karne par: $\int {\sec }^3\theta d\theta =\frac{1}{2}\sec \theta \tan \theta +\frac{1}{2}\ln |\sec \theta +\tan \theta |$. Step 5 — combine. $9\left(\frac{1}{2}\sec \theta \tan \theta +\frac{1}{2}\ln |\sec \theta +\tan \theta |\right)-9\ln |\sec \theta +\tan \theta |=\frac{9}{2}\sec \theta \tan \theta -\frac{9}{2}\ln |\sec \theta +\tan \theta |+C.$ Step 6 — triangle (Figure s04). $\sec \theta =\frac{x}{3}$, $\tan \theta =\frac{\sqrt{x^2-9}}{3}$. $\boxed{=\frac{x\sqrt{x^2-9}}{2}-\frac{9}{2}\ln \left|\frac{x+\sqrt{x^2-9}}{3}\right|+C}$ ($-\frac{9}{2}\ln 3$ ko $C$ mein fold kar lo; tum $-\frac{9}{2}\ln |x+\sqrt{x^2-9}|+C$ bhi likh sakte ho.)

Step 1. $a=2$, $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$: $x=2\sin \theta$, $dx=2\cos \theta d\theta$, $\sqrt{4-x^2}=2\cos \theta$ (bars drop kyunki $\cos \theta \ge 0$). Limits convert karo (WHY: back-substitute karne se zyada clean hai): $x=0\Rightarrow \sin \theta =0\Rightarrow \theta =0$; $x=2\Rightarrow \sin \theta =1\Rightarrow \theta =\frac{\pi }{2}$. Step 2. ${\int }_0^{\pi /2}\frac{4{\sin }^2\theta }{2\cos \theta }\cdot 2\cos \theta d\theta =4{\int }_0^{\pi /2}{\sin }^2\theta d\theta .$ Step 3 — power-reduction (dekho Power-reduction & double-angle formulas): ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$. WHY ye step: ${\sin }^2\theta$ ka koi elementary antiderivative "by sight" nahi hai, lekin ye identity ise ek constant plus ek plain cosine mein badal deta hai, dono ek line mein integrate ho jaate hain. $4\cdot \frac{1}{2}{\int }_0^{\pi /2}(1-\cos 2\theta )d\theta =2{\left[\theta -\frac{1}{2}\sin 2\theta \right]}_0^{\pi /2}=2\left(\frac{\pi }{2}-0\right)=\pi .$ WHY $\int \cos 2\theta d\theta =\frac{1}{2}\sin 2\theta$: $\frac{d}{d\theta }\left(\frac{1}{2}\sin 2\theta \right)=\frac{1}{2}\cdot 2\cos 2\theta =\cos 2\theta$ (chain rule), ulta padhne par. $\boxed{=\pi }$ Edge note: upper limit $x=2=a$ par integrand blow up karta hai (root $\to 0$), phir bhi area finite hai — $\theta$-integral is improper endpoint ko quietly handle kar leta hai kyunki $\theta =\frac{\pi }{2}$ ek bilkul ordinary limit hai.

Step 1 — complete the square (WHY: abhi tak koi bare $a^2\pm x^2$ form nahi hai). $x^2+6x+13=(x+3{)}^2+4$. Ab ye shifted variable ke saath ek sum hai. Step 2 — shift phir substitute. $u=x+3$, $du=dx$ lo. Root $=\sqrt{u^2+4}$, $a=2$: $u=2\tan \theta$, $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$, $du=2{\sec }^2\theta d\theta$. Is interval par $\sec \theta >0$, isliye $\sqrt{u^2+4}=2\sec \theta$. (Wahi tangent triangle, Figure s03, $x\to u$ ke saath.) Step 3. $\int \frac{2{\sec }^2\theta d\theta }{2\sec \theta }=\int \sec \theta d\theta =\ln |\sec \theta +\tan \theta |+C.$ WHY $\int \sec \theta d\theta =\ln |\sec \theta +\tan \theta |$: exactly wo $(\sec \theta +\tan \theta )$ se multiply karne wala trick jo L3·1 Step 4 mein derive kiya — numerator denominator ka derivative ban jaata hai. Step 4 — triangle (Figure s03, $x\to u$). $\tan \theta =\frac{u}{2}$, $\sec \theta =\frac{\sqrt{u^2+4}}{2}$. $\ln \left|\frac{\sqrt{u^2+4}+u}{2}\right|+C=\ln \left|\sqrt{u^2+4}+u\right|+C(\text{the }-\ln 2\text{ folds into }C).$ Step 5 — $u$ undo karo. $\boxed{=\ln \left|x+3+\sqrt{x^2+6x+13}\right|+C(\text{all real }x,\text{since }(x+3{)}^2+4>0)}$

Step 1. $a=1$, $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$: $x=\sin \theta$, $dx=\cos \theta d\theta$, $\sqrt{1-x^2}=\cos \theta$ (bars drop kyunki $\cos \theta \ge 0$). Sine triangle, Figure s02 with $a=1$. Step 2. $\int \frac{{\sin }^2\theta }{\cos \theta }\cos \theta d\theta =\int {\sin }^2\theta d\theta .$ Step 3 — power-reduction. ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$, isliye $\int {\sin }^2\theta d\theta =\frac{1}{2}(\theta -\frac{1}{2}\sin 2\theta )=\frac{1}{2}(\theta -\sin \theta \cos \theta )+C$. WHY last equality: $\sin 2\theta =2\sin \theta \cos \theta$ (double-angle, dekho Power-reduction & double-angle formulas), isliye $\frac{1}{2}\sin 2\theta =\sin \theta \cos \theta$. Step 4 — triangle / undo (Figure s02). $\theta =\arcsin x$, $\sin \theta =x$, $\cos \theta =\sqrt{1-x^2}$. $\boxed{=\frac{1}{2}\arcsin x-\frac{x\sqrt{1-x^2}}{2}+C,x\in (-1,1)}$

Step 1. $a=\sqrt{2}$, aur kyunki $x$, $\sqrt{2}\to 2$ run karta hai (sab $\ge \sqrt{2}=a$) hum $\theta \in [0,\frac{\pi }{2})$ mein hain: $x=\sqrt{2}\sec \theta$, $dx=\sqrt{2}\sec \theta \tan \theta d\theta$, $\sqrt{x^2-2}=\sqrt{2}\tan \theta$. WHY $+\sqrt{2}\tan \theta$: $[0,\frac{\pi }{2})$ par $\tan \theta \ge 0$ hai, isliye $|\cdot |$ drop ho jaata hai aur root non-negative rehta hai (range table, secant case). Secant triangle, Figure s04 with $a=\sqrt{2}$. Limits convert karo. $x=\sqrt{2}\Rightarrow \sec \theta =1\Rightarrow \theta =0$; $x=2\Rightarrow \sec \theta =\frac{2}{\sqrt{2}}=\sqrt{2}\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}\Rightarrow \theta =\frac{\pi }{4}$. Step 2. ${\int }_0^{\pi /4}\frac{\sqrt{2}\tan \theta }{\sqrt{2}\sec \theta }\cdot \sqrt{2}\sec \theta \tan \theta d\theta =\sqrt{2}{\int }_0^{\pi /4}{\tan }^2\theta d\theta .$ Constant track karo: $\frac{\sqrt{2}\tan \theta }{\sqrt{2}\sec \theta }\cdot \sqrt{2}\sec \theta \tan \theta =\sqrt{2}{\tan }^2\theta$. ✓ Step 3. ${\tan }^2\theta ={\sec }^2\theta -1\Rightarrow \int {\tan }^2\theta d\theta =\tan \theta -\theta$. WHY $\int {\sec }^2\theta d\theta =\tan \theta$: $\frac{d}{d\theta }\tan \theta ={\sec }^2\theta$, ulta padhne par; aur $\int 1d\theta =\theta$. $\sqrt{2}[\tan \theta -\theta {]}_0^{\pi /4}=\sqrt{2}\left(1-\frac{\pi }{4}\right).$ $\boxed{=\sqrt{2}\left(1-\frac{\pi }{4}\right)\approx 0.303}$

Step 1 — set up. $y^{\prime }=x$, isliye $L={\int }_0^1\sqrt{1+x^2}dx$. Ek sum $\Rightarrow$ tangent, $a=1$. Step 2. $x=\tan \theta$, $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$, $dx={\sec }^2\theta d\theta$. Is interval par $\sec \theta >0$, isliye $\sqrt{1+x^2}=\sec \theta$. Limits: $x=0\to \theta =0$; $x=1\to \tan \theta =1\to \theta =\frac{\pi }{4}$. $L={\int }_0^{\pi /4}\sec \theta \cdot {\sec }^2\theta d\theta ={\int }_0^{\pi /4}{\sec }^3\theta d\theta .$ Step 3 — $\int {\sec }^3$ result use karo (L3·1 Step 4 mein parts se derive kiya): $\int {\sec }^3\theta d\theta =\frac{1}{2}\sec \theta \tan \theta +\frac{1}{2}\ln |\sec \theta +\tan \theta |$. $\theta =\frac{\pi }{4}$ par: $\sec \theta =\sqrt{2}$, $\tan \theta =1$. $\theta =0$ par: donon terms $0$. $L=\frac{1}{2}(\sqrt{2}\cdot 1)+\frac{1}{2}\ln |\sqrt{2}+1|=\frac{\sqrt{2}}{2}+\frac{1}{2}\ln (\sqrt{2}+1).$ $\boxed{L=\frac{\sqrt{2}}{2}+\frac{1}{2}\ln (1+\sqrt{2})\approx 1.148}$

Way A — plain $u$-sub (faster). $u=x^2+4$, $du=2xdx$ lo (dekho Integration by substitution (u-sub)). $\int \frac{xdx}{\sqrt{u}}=\frac{1}{2}\int u^{-1/2}du=\sqrt{u}=\sqrt{x^2+4}+C.$ WHY $\frac{1}{2}\int u^{-1/2}du=\sqrt{u}$: power rule $\int u^{n}du=\frac{u^{n+1}}{n+1}$ ke saath $n=-\frac{1}{2}$ lagate hain to milta hai $\frac{u^{1/2}}{1/2}=2\sqrt{u}$, aage wale $\frac{1}{2}$ se multiply karo $=\sqrt{u}$. Way B — trig sub (kaam karta hai, lekin heavier). $x=2\tan \theta$, $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$ isliye $\sec \theta >0$: $\int \frac{2\tan \theta }{2\sec \theta }2{\sec }^2\theta d\theta =2\int \tan \theta \sec \theta d\theta =2\sec \theta =2\cdot \frac{\sqrt{x^2+4}}{2}=\sqrt{x^2+4}+C.$ WHY $\int \sec \theta \tan \theta d\theta =\sec \theta$: $\frac{d}{d\theta }\sec \theta =\sec \theta \tan \theta$, ulta padhne par. **WHY