Exercises — Integration by parts — derivation from product rule, LIATE mnemonic

Types: $x$ Algebraic hai, $\sin x$ Trigonometric hai. LIATE kehta hai A, T se pehle aata hai, isliye $u=x,dv=\sin xdx.$ $u$ ko differentiate karo aur $dv$ ko integrate karo: $du=dx,v=\int \sin xdx=-\cos x.$ Yeh choice kyun: $x$ ko differentiate karne par constant $1$ mil jaata hai — polynomial "gayab" ho jaata hai, jisse naya integral simple ho jaata hai. Agar $u=\sin x$ choose karte, toh $du=\cos xdx$ kabhi simplify nahi hota aur $dv=xdx$ se $v=\frac{x^2}{2}$ milta, jo bacha hua integral aur mushkil bana deta.

Types: $\ln x$ Logarithmic hai, $x^2$ Algebraic hai. L, A se pehle aata hai, isliye $u=\ln x,dv=x^{2}dx.$ $du=\frac{1}{x}dx,v=\int x^{2}dx=\frac{x^3}{3}.$ Kyun: $\ln x$ ka koi seedha elementary antiderivative nahi milta, lekin iska derivative $1/x$ ekdum clean hai. Isliye hum ise differentiate karte hain, kabhi integrate nahi.

"Hidden product" $\arctan x=\arctan x\cdot 1$ likhte hain. Ab do factors hain. Types: $\arctan x$ Inverse-trig hai, $1$ Algebraic hai. I, A se pehle aata hai, isliye $u=\arctan x,dv=1dx.$ $du=\frac{1}{1+x^2}dx,v=\int 1dx=x.$ Kyun: inverse-trig functions ko integrate karna mushkil hai lekin inke derivatives rational hote hain — yeh perfect $u$ material hai.

$u=x$ (A), $dv=\cos xdx$ (T), toh $du=dx$, $v=\sin x$. $\int x\cos xdx=x\sin x-\int \sin xdx=x\sin x-(-\cos x)+C=x\sin x+\cos x+C.$ Differentiate karke verify karo: $\frac{d}{dx}(x\sin x+\cos x)=\sin x+x\cos x-\sin x=x\cos x.$ ✓

$u=x$ (A), $dv=e^{2x}dx$ (E). Toh $du=dx$ aur $v=\int e^{2x}dx=\frac{1}{2}{e}^{2x}.$ $\frac{1}{2}$ kyun: chain rule se $\frac{d}{dx}{e}^{2x}=2e^{2x}$, isliye undo karne ke liye $2$ se divide karte hain. $\int x{e}^{2x}dx=x\cdot \frac{1}{2}{e}^{2x}-\int \frac{1}{2}{e}^{2x}dx=\frac{1}{2}x{e}^{2x}-\frac{1}{4}{e}^{2x}+C=\frac{1}{4}{e}^{2x}(2x-1)+C.$ Verify: $\frac{d}{dx}[\frac{1}{4}{e}^{2x}(2x-1)]=\frac{1}{4}(2e^{2x}(2x-1)+2e^{2x})=\frac{1}{4}{e}^{2x}(4x)=x{e}^{2x}.$ ✓

Hidden product: $\ln (3x)=\ln (3x)\cdot 1$. $u=\ln (3x)$, $dv=dx$ lo. $du=\frac{1}{3x}\cdot 3dx=\frac{1}{x}dx,v=x.$ $du=\frac{1}{x}dx$ kyun: chain rule se $\frac{d}{dx}\ln (3x)=\frac{1}{3x}\cdot 3=\frac{1}{x}$. (Yeh bhi: $\ln (3x)=\ln 3+\ln x$, jiska derivative clearly $\frac{1}{x}$ hai.) $\int \ln (3x)dx=x\ln (3x)-\int x\cdot \frac{1}{x}dx=x\ln (3x)-\int 1dx=x\ln (3x)-x+C.$ Verify: $\frac{d}{dx}[x\ln (3x)-x]=\ln (3x)+x\cdot \frac{1}{x}-1=\ln (3x).$ ✓

$u=x^2$ (A), $dv=e^{x}dx$, toh $du=2xdx$, $v=e^x$: $\int x^2{e}^{x}dx=x^2{e}^x-\int 2x{e}^{x}dx.$ Bacha hua $\int 2x{e}^{x}dx$ abhi bhi ek product hai, isliye parts dobara apply karo $u=2x$, $dv=e^{x}dx$ ke saath ($du=2dx$, $v=e^x$): $\int 2x{e}^{x}dx=2x{e}^x-\int 2e^{x}dx=2x{e}^x-2e^x.$ Combine karo (leading minus dhyan rakho): $\int x^2{e}^{x}dx=x^2{e}^x-(2x{e}^x-2e^x)+C=e^x(x^2-2x+2)+C.$ Yeh terminate kyun hota hai: har pass mein polynomial ki degree ek se ghatti hai; do passes ke baad $x$ chala jaata hai. Verify: $\frac{d}{dx}[e^x(x^2-2x+2)]=e^x(x^2-2x+2)+e^x(2x-2)=e^x{x}^2.$ ✓

Maano $I=\int e^x\cos xdx$. $u=\cos x$, $dv=e^{x}dx$ lo ($du=-\sin xdx$, $v=e^x$): $I=e^x\cos x-\int e^x(-\sin x)dx=e^x\cos x+\int e^x\sin xdx.$ Naye integral par parts apply karo $u=\sin x$, $dv=e^{x}dx$ ke saath ($du=\cos xdx$, $v=e^x$): $\int e^x\sin xdx=e^x\sin x-\int e^x\cos xdx=e^x\sin x-I.$ Wapas substitute karo: $I=e^x\cos x+e^x\sin x-I\Rightarrow 2I=e^x(\cos x+\sin x).$ $\boxed{I=\frac{1}{2}{e}^x(\cos x+\sin x)+C}$ Yeh kyun kaam karta hai: original integral $I$ wapas aata hai, isliye hum ise hamesha ke liye integrate karne ki jagah algebraically solve karte hain. ($u,dv$ ka choice consistent rakho — dono baar exponential $dv$ ho — warna loop $0=0$ mein cancel ho jaata hai.) Verify: $\frac{d}{dx}[\frac{1}{2}{e}^x(\cos x+\sin x)]=\frac{1}{2}(e^x(\cos x+\sin x)+e^x(-\sin x+\cos x))=\frac{1}{2}{e}^x(2\cos x)=e^x\cos x.$ ✓

$u=(\ln x{)}^2$ (L), $dv=xdx$. Toh $v=\frac{x^2}{2}$ aur, chain rule se, $du=2\ln x\cdot \frac{1}{x}dx=\frac{2\ln x}{x}dx.$ $\int x(\ln x{)}^{2}dx=\frac{x^2}{2}(\ln x{)}^2-\int \frac{x^2}{2}\cdot \frac{2\ln x}{x}dx=\frac{x^2}{2}(\ln x{)}^2-\int x\ln xdx.$ Ab $\int x\ln xdx$ par parts karo: $u=\ln x$, $dv=xdx$, $v=\frac{x^2}{2}$, $du=\frac{1}{x}dx$: $\int x\ln xdx=\frac{x^2}{2}\ln x-\int \frac{x^2}{2}\cdot \frac{1}{x}dx=\frac{x^2}{2}\ln x-\frac{x^2}{4}.$ Combine karo: $\int x(\ln x{)}^{2}dx=\frac{x^2}{2}(\ln x{)}^2-\frac{x^2}{2}\ln x+\frac{x^2}{4}+C.$ Verify: differentiate karne par $x(\ln x{)}^2$ milta hai (VERIFY mein check kiya).

Pehle antiderivative nikalo (parent Example 1): $\int x{e}^{x}dx=e^x(x-1)$. Fundamental Theorem of Calculus se limits ke beech evaluate karo: ${\int }_0^{1}x{e}^{x}dx=[e^x(x-1){]}_0^1=e^1(1-1)-e^0(0-1)=0-(1)(-1)=1.$ Answer: $\boxed{1}$. Definite integrals ke liye tum limits ko seedha har step mein bhi carry kar sakte ho: $[uv{]}_0^1-{\int }_0^{1}vdu$.

Pehle substitute karo taaki awkward $\sqrt{x}$ hat jaaye. $t=\sqrt{x}$ lo, toh $x=t^2$ aur $dx=2tdt$. $\int \cos (\sqrt{x})dx=\int \cos t\cdot 2tdt=2\int t\cos tdt.$ Ab yeh Exercise 2.1 wali shape hai. $u=t$, $dv=\cos tdt$ ke saath ($du=dt$, $v=\sin t$): $2\int t\cos tdt=2(t\sin t-\int \sin tdt)=2(t\sin t+\cos t).$ $t=\sqrt{x}$ back-substitute karo: $\int \cos (\sqrt{x})dx=2\sqrt{x}\sin (\sqrt{x})+2\cos (\sqrt{x})+C.$ Substitution pehle kyun: parts akele $\cos (\sqrt{x})$ handle nahi kar sakta — koi clean $u/dv$ split nahi hai. Substitution ise ek plain polynomial-times-trig product mein badal deta hai, jise parts aasani se solve kar leta hai. Yeh "Integration by substitution se parts set up karo" wala pattern hai. Verify: differentiate karne par $\cos (\sqrt{x})$ milta hai (VERIFY mein check kiya).

$u=\arctan x$ (I), $dv=xdx$. Toh $v=\frac{x^2}{2}$, $du=\frac{1}{1+x^2}dx$: $\int x\arctan xdx=\frac{x^2}{2}\arctan x-\int \frac{x^2}{2}\cdot \frac{1}{1+x^2}dx=\frac{x^2}{2}\arctan x-\frac{1}{2}\int \frac{x^2}{1+x^2}dx.$ Bacha hua integral rational hai. $\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}$ likh kar split karo: $\frac{1}{2}\int \left(1-\frac{1}{1+x^2}\right)dx=\frac{1}{2}(x-\arctan x).$ Isliye $\int x\arctan xdx=\frac{x^2}{2}\arctan x-\frac{x}{2}+\frac{1}{2}\arctan x+C.$ Verify: derivative $x\arctan x$ ke barabar hai (VERIFY mein check kiya).

$u=x^n$ (A), $dv=e^{x}dx$ lo, toh $du=n{x}^{n-1}dx$, $v=e^x$: $I_n=\int x^n{e}^{x}dx=x^n{e}^x-\int e^x\cdot n{x}^{n-1}dx=x^n{e}^x-n\int x^{n-1}{e}^{x}dx=x^n{e}^x-n{I}_{n-1}.$ Yeh reduction formula hai: har application power ek se ghata deta hai. Ab $n=3$ se unroll karo, $I_0=\int e^{x}dx=e^x$ use karke: $I_1=x{e}^x-I_0=x{e}^x-e^x=e^x(x-1),$ $I_2=x^2{e}^x-2I_1=x^2{e}^x-2e^x(x-1)=e^x(x^2-2x+2),$ $I_3=x^3{e}^x-3I_2=x^3{e}^x-3e^x(x^2-2x+2)=e^x(x^3-3x^2+6x-6).$ $\boxed{\int x^3{e}^{x}dx=e^x(x^3-3x^2+6x-6)+C.}$ Verify: differentiate karne par $x^3{e}^x$ milta hai (VERIFY mein check kiya).

Exercise 2.1 se antiderivative: $\int x\sin xdx=-x\cos x+\sin x$. (Woh $x\sin x+\cos x$ wala tha? Nahi — woh $x\cos x$ tha. $x\sin x$ ke liye: $u=x$, $dv=\sin xdx$, $v=-\cos x$, milta hai $-x\cos x+\int \cos xdx=-x\cos x+\sin x$.) $[0,\pi ]$ par evaluate karo: ${\int }_0^{\pi }x\sin xdx=[-x\cos x+\sin x{]}_0^{\pi }.$ $x=\pi$ par: $-\pi \cos \pi +\sin \pi =-\pi (-1)+0=\pi$. $x=0$ par: $-0\cdot \cos 0+\sin 0=0$. ${\int }_0^{\pi }x\sin xdx=\pi -0=\boxed{\pi }.✓$

Tabular integration repeated parts ka ek table mein organised form hai: $x^3$ ko Differentiate karo $0$ tak, $e^x$ ko Integrate karo (jo $e^x$ hi rehta hai), aur alternating signs $+,-,+,-$ ke saath diagonals ke along products padho:

Har D-row ko ek step neeche wali I-row se multiply karo, sign apply karo: $+x^3{e}^x-3x^2{e}^x+6x{e}^x-6e^x+C=e^x(x^3-3x^2+6x-6)+C.$ 5.1 se match kyun karta hai: tabular integration wahi hai reduction formula baar baar apply karna — alternating signs exactly wahi $-n{I}_{n-1}$ minus signs hain jo stack hote rehte hain. Yeh tab kaam karta hai jab ek factor finitely many steps mein differentiate hokar $0$ ho jaata hai (polynomials) aur doosra har baar cleanly integrate hota hai.