Exercises — U-substitution — technique, change of limits for definite integrals

Choose karo $u=x^3+7$. Kyun? $(x^3+7)$ 4th power ke andar baitha hai, aur uski derivative $3x^2$ hai — jo bilkul saamne hi khadi hai. Differentiate: $du=3x^{2}dx$. Kyun? Isse poora clump $3x^{2}dx$ simply $du$ ban jaata hai. Replace: $\int u^{4}du$. Har $x$ chala gaya — including $dx$. Integrate: $\frac{u^5}{5}+C$. Back-substitute (indefinite hai, toh $x$ wapas laao): $\boxed{\frac{(x^3+7{)}^5}{5}+C}$.

Choose karo $u=5x$. Kyun? Cosine ka argument $5x$ hai; yeh jo inner linear function hai, usi ko rename karna hai. Differentiate: $du=5dx\Rightarrow dx=\frac{1}{5}du$. Kyun? Hamaare paas sirf bare $dx$ hai, toh hum use $u$ mein express karte hain. Replace: $\int \cos u\cdot \frac{1}{5}du=\frac{1}{5}\int \cos udu$. Integrate: $\frac{1}{5}\sin u+C$. Back-substitute: $\boxed{\frac{1}{5}\sin (5x)+C}$.

Choose karo $u=x^2+1$. Kyun? Uski derivative $2x$ bilkul numerator hai — classic "$\int \frac{u^{\prime }}{u}$" shape, jo ek logarithm produce karta hai. Differentiate: $du=2xdx$ — poora numerator-times-$dx$. Replace: $\int \frac{1}{u}du$. Integrate: $\ln |u|+C$. Absolute value kyun? $\int \frac{1}{u}du=\ln |u|$ negative $u$ ke liye bhi kaam karta hai; yahaan $x^2+1>0$ hamesha hai, toh $|u|=u$. Back-substitute: $\boxed{\ln (x^2+1)+C}$.

Choose karo $u=x^2-4$, toh $du=2xdx$. Kyun? Inner function $x^2-4$ hai; uski derivative $2x$ hai, aur hamaare paas $x$ hai — proportional, equal nahi. Constant fix karo: hamaare paas $xdx$ hai lekin chahiye $2xdx$, toh $xdx=\frac{1}{2}du$. Yeh legal hai kyunki $\frac{1}{2}$ ek constant hai aur $\int$ constants pe linear hai. Replace: $\int u^3\cdot \frac{1}{2}du=\frac{1}{2}\int u^{3}du$. Integrate: $\frac{1}{2}\cdot \frac{u^4}{4}=\frac{u^4}{8}$. Back-substitute: $\boxed{\frac{(x^2-4{)}^4}{8}+C}$.

Choose karo $u=x^2$, toh $du=2xdx\Rightarrow xdx=\frac{1}{2}du$. Limits change karo (Kyun? variable ab $u$ hai, toh purane numbers $0,1$ ab range describe nahi karte):

• $x=0\Rightarrow u=0^2=0$
• $x=1\Rightarrow u=1^2=1$

Neeche diya figure padho. Left panel (blue) original region hai: $y=x{e}^{x^2}$ ke neeche trapped area jab $x$ $0$ se $1$ tak jaata hai. Curve $x=0$ ke paas flat start hota hai (kyunki factor $x$ wahaan height ko khatam kar deta hai) aur $x=1$ ke paas upar shoot karta hai. Right panel (orange) substitution ke baad ka same total area hai: ab hum $y=\frac{1}{2}{e}^u$ plot karte hain jab $u$ $0$ se $1$ tak jaata hai. Substitution ne horizontal axis ko stretch aur reshape kar diya hai — jo points $x=0$ ke paas bunched the woh $u$ mein spread out ho jaate hain — phir bhi shaded area identical hai. Iska matlab yahi hai "u-sub area conserve karta hai": $\frac{1}{2}$ aur $du=2xdx$ ki bookkeeping axis ke warped hone ko exactly compensate karti hai. Dono endpoints $0$ aur $1$ yahaan sirf isliye same rehte hain kyunki $g(0)=0$ aur $g(1)=1$; region ka interior phir bhi transform hota hai.

Replace: ${\int }_0^1{e}^u\cdot \frac{1}{2}du=\frac{1}{2}{\int }_0^1{e}^{u}du$. Integrate & evaluate: $\frac{1}{2}[e^u{]}_0^1=\frac{1}{2}(e^1-e^0)=\frac{1}{2}(e-1)$. $\boxed{\frac{e-1}{2}\approx 0.859}$.

Choose karo $u=\ln x$, toh $du=\frac{1}{x}dx$. Kyun? Factor $\frac{1}{x}dx$ exactly $du$ hai, aur $\ln x$ akela appear hota hai — perfect "$u$ aur $du$" pair. Limits change karo:

• $x=1\Rightarrow u=\ln 1=0$
• $x=2\Rightarrow u=\ln 2\approx 0.693$

Replace: ${\int }_0^{\ln 2}udu$. Integrate & evaluate: $[\frac{u^2}{2}{]}_0^{\ln 2}=\frac{(\ln 2{)}^2}{2}$. $\boxed{\frac{(\ln 2{)}^2}{2}\approx 0.240}$.

Choose karo $u=x+1$, toh $du=dx$. Kyun? Awkward part root hai; uska inside $x+1$ hai. Problem: extra factor $x$ ek variable hai, constant nahi — hum use bahar nahi nikal sakte. Isliye, kyunki $u=x+1$ ka matlab hai $x=u-1$, hum us ko bhi replace karte hain. Replace: $\int (u-1)\sqrt{u}du=\int (u^{3/2}-u^{1/2})du$. Integrate: $\frac{u^{5/2}}{5/2}-\frac{u^{3/2}}{3/2}=\frac{2}{5}{u}^{5/2}-\frac{2}{3}{u}^{3/2}$. Back-substitute: $\boxed{\frac{2}{5}(x+1{)}^{5/2}-\frac{2}{3}(x+1{)}^{3/2}+C}$.

Pehle rewrite karo: $\tan x=\frac{\sin x}{\cos x}$. Kyun? Ab denominator $\cos x$ ki derivative $-\sin x$ hai, jo (sign tak) numerator hai — "$\int \frac{u^{\prime }}{u}$" shape appear ho jaata hai. Choose karo $u=\cos x$, toh $du=-\sin xdx\Rightarrow \sin xdx=-du$. Replace: $\int \frac{1}{u}\cdot (-du)=-\int \frac{1}{u}du$. Integrate: $-\ln |u|+C$. Back-substitute: $\boxed{-\ln |\cos x|+C}$ (equivalently $\ln |\sec x|+C$). Absolute value ab kyun rehta hai? $\cos x$ kuch intervals pe negative hota hai, toh $|u|$ genuinely zaroori hai.

Choose karo $u=\sin x$, toh $du=\cos xdx$. Kyun? $\cos xdx$ exactly $du$ hai, $\sin x=u$ akela bachta hai. Limits change karo:

• $x=0\Rightarrow u=\sin 0=0$
• $x=\pi /2\Rightarrow u=\sin (\pi /2)=1$

Replace: ${\int }_0^{1}udu$. Integrate & evaluate: $[\frac{u^2}{2}{]}_0^1=\frac{1}{2}$. $\boxed{\frac{1}{2}}$. Doosre choice ko cross-check karo: $u=\cos x$ se $du=-\sin xdx$ milta hai, limits $1\to 0$, integral $-{\int }_1^{0}udu={\int }_0^{1}udu=\frac{1}{2}$. Same answer — ek achhi sanity habit.

Root hatane ke liye substitute karo: $u=\sqrt{x}$ lo, toh $x=u^2$ aur $dx=2udu$. Kyun? $e^{\sqrt{x}}$ ke andar root koi bhi simple antiderivative block karta hai; use rename karne se exponent clear ho jaata hai. Replace: $\int e^u\cdot 2udu=2\int u{e}^{u}du$. Ab Integration by Parts. Yeh rule kya hai? Yeh product rule ko reverse karta hai: $\frac{d}{dx}(pq)=p{q}^{\prime }+q{p}^{\prime }$ se, dono sides integrate karne par milta hai $\int pdq=pq-\int qdp.$ Yahaan parts kyun aur zyaada u-sub kyun nahi? Integrand $u{e}^u$ ek do unrelated pieces ka product hai (polynomial $u$ times exponential $e^u$); na toh ek doosre ki derivative hai, toh koi bhi u-sub ise collapse nahi kar sakta. Parts products ke liye tool hai. $p=u$ aur $dq=e^{u}du$ kyun choose karein (aur ulta kyun nahi)? Hum chahte hain ki surviving integral $\int qdp$ simpler ho. $p=u$ lene se $dp=du$ milta hai — $u$ differentiate karne se woh constant ban jaata hai, polynomial khatam ho jaata hai. $dq=e^{u}du$ lena safe hai kyunki $e^u$ wापस $e^u$ mein integrate hota hai (woh kabhi worse nahi hota). Toh $p=u,dq=e^{u}du\Rightarrow dp=du,q=e^u$, milta hai $\int u{e}^{u}du=u{e}^u-\int e^{u}du=u{e}^u-e^u.$ Assemble karo: $2(u{e}^u-e^u)=2e^u(u-1)$. Back-substitute $u=\sqrt{x}$: $\boxed{2e^{\sqrt{x}}(\sqrt{x}-1)+C}$.

Notice karo: $1-x^2$ ki derivative $-2x$ hai, aur hamaare paas upar $x$ hai — plain u-sub bina kisi bhi trig ke ise handle kar leta hai. Choose karo $u=1-x^2$, toh $du=-2xdx\Rightarrow xdx=-\frac{1}{2}du$. Replace: $\int \frac{1}{\sqrt{u}}\cdot (-\frac{1}{2})du=-\frac{1}{2}\int u^{-1/2}du$. Integrate: $-\frac{1}{2}\cdot \frac{u^{1/2}}{1/2}=-u^{1/2}$. Back-substitute: $\boxed{-\sqrt{1-x^2}+C}$. Trig substitution kyun nahi? Trig sub $\int \frac{1}{\sqrt{1-x^2}}$ ke liye hai (upar koi $x$ nahi). Yahaan stray $x$ ise pure u-sub banata hai — hamesha pehle u-sub try karo.

Choose karo $u=1+\tan x$, toh $du={\sec }^{2}xdx$. Kyun? $\tan x$ ki derivative ${\sec }^{2}x$ hai, jo numerator mein bilkul baitha hai — "$\int \frac{u^{\prime }}{u}$" logarithm pattern. Limits change karo:

• $x=0\Rightarrow u=1+\tan 0=1$
• $x=\pi /4\Rightarrow u=1+\tan (\pi /4)=1+1=2$

Replace: ${\int }_1^2\frac{1}{u}du$. Integrate & evaluate: $[\ln |u|{]}_1^2=\ln 2-\ln 1=\ln 2$. $\boxed{\ln 2\approx 0.693}$.

Ek layer peel karo: $u=\ln (\ln x)$ lo. Outermost kyun? Uski derivative, Chain Rule se, $\frac{1}{\ln x}\cdot \frac{1}{x}=\frac{1}{x\ln x}$ hai — exactly integrand ka baaki hissa. Differentiate: $du=\frac{1}{x\ln x}dx$. Replace: $\int \frac{1}{u}du$. Integrate: $\ln |u|+C$. Back-substitute: $\boxed{\ln |\ln (\ln x)|+C}$.

Pehle domain check karo. Root $\sqrt{x^3+1}$ real tab hoga jab $x^3+1\ge 0$, yaani $x\ge -1$. Hamaara interval $[0,3]$ aaram se $x\ge -1$ ke andar baitha hai, toh integrand poore raaste real aur continuous hai — radical ke neeche koi gap nahi. Choose karo $u=x^3+1$, toh $du=3x^{2}dx\Rightarrow x^{2}dx=\frac{1}{3}du$. Limits change karo:

• $x=0\Rightarrow u=0^3+1=1$
• $x=3\Rightarrow u=3^3+1=28$

(Kyunki hamaare interval pe $x\ge 0$ hai, $u=x^3+1\ge 1>0$ hai, toh $\sqrt{u}$ naye variable mein bhi hamesha defined hai.) Replace: ${\int }_1^{28}\sqrt{u}\cdot \frac{1}{3}du=\frac{1}{3}{\int }_1^{28}{u}^{1/2}du$. Integrate: $\frac{1}{3}\cdot \frac{u^{3/2}}{3/2}=\frac{2}{9}{u}^{3/2}$. Evaluate: $\frac{2}{9}(28^{3/2}-1^{3/2})=\frac{2}{9}(28\sqrt{28}-1)$. Kyunki $28\sqrt{28}=28\cdot 2\sqrt{7}=56\sqrt{7}$, yeh hai $\frac{2}{9}(56\sqrt{7}-1)$. $\boxed{\frac{2(56\sqrt{7}-1)}{9}\approx 32.68}$.

Choose karo $u=x^4$, toh $du=4x^{3}dx\Rightarrow x^{3}dx=\frac{1}{4}du$. Limits change karo — squaring dekho:

• $x=-1\Rightarrow u=(-1{)}^4=1$
• $x=1\Rightarrow u=1^4=1$

Dono endpoints $u=1$ pe map ho gaye! Replace: $\frac{1}{4}{\int }_1^1{e}^{u}du=0$. $\boxed{0}$. Yeh sense kyun karta hai? $x^3{e}^{x^4}$ ek odd function hai ($x\to -x$ replace karo: $(-x{)}^3{e}^{(-x{)}^4}=-x^3{e}^{x^4}$), aur symmetric interval $[-1,1]$ pe ek odd function ka integral zero hota hai. Substitution ka dono limits ek hi value pe collapse karna woh algebra hai jo yeh symmetry dikha rahi hai, na koi mistake.

Top rewrite karo: $e^{2x}=e^x\cdot e^x=(e^x{)}^2$. Yeh rewrite kyun? Hum $u=e^x+1$ set karne ki plan kar rahe hain, jiska differential $du=e^{x}dx$ hai. $e^{2x}$ ko $e^x$ ke do factors mein split karne se ek $e^x$ $dx$ ke saath pair up ho kar $du$ ban jaata hai, jabki doosra $e^x$ $u-1$ ban jaata hai — saari cheez naye variable mein cleanly convert ho jaati hai. Choose karo $u=e^x+1$, toh $e^x=u-1$ aur $du=e^{x}dx$. Replace: integrand $\frac{e^x\cdot e^x}{e^x+1}dx=\frac{(u-1)}{u}du$ (ek $e^x$ $du$ ban gaya, doosra $u-1$ ban gaya). $\int \frac{u-1}{u}du=\int (1-\frac{1}{u})du=u-\ln |u|.$ Back-substitute $u=e^x+1$: $\boxed{(e^x+1)-\ln (e^x+1)+C}$. Kyunki constant $1$ $C$ mein merge ho jaata hai, yeh $e^x-\ln (e^x+1)+C$ ke barabar hai.

• Kya mera $u$ inner function hai jiska derivative present hai? ✔
• Kya maine har $x$-piece convert kiya, including $dx\to du$? ✔
• Constant mismatch → ek number se fix karo; variable mismatch → $x=g^{-1}(u)$ solve karo. ✔
• Definite hai? Limits change karo ya back-substitute karo — kabhi dono ek saath nahi. ✔
• Kisi bhi radical/log ke neeche domain check kiya, aur naye limits plug karke compute kiye? ✔
• Kya maine signs, logs mein absolute values, aur symmetry sanity-check kiya? ✔