Question bank — Trigonometric integrals — sinᵐ·cosⁿ cases, tan and sec cases

Sach — ${\sin }^m{\cos }^n$ ke liye parity method ::: Sach. Odd power se ek factor alag karna hamesha ek even power chhodta hai, aur identities sirf even powers ko rewrite karti hain. Isliye ${\cos }^{99}$ bilkul ${\cos }^3$ jaisa behave karta hai: ek peel karo, baaki convert karo.

$\int {\sin }^{4}x{\cos }^{4}xdx$ ko substitution $u=\sin x$ se kiya ja sakta hai. Jhooth — both even blocks substitution ::: Jhooth. Ek $\cos$ peel karne par ${\cos }^3$ bachta hai, jo ek odd leftover hai jise ${\cos }^2=1-{\sin }^2$ poori tarah convert nahi kar sakta, aur extra $\cos dx$ ka koi matching factor nahi hai. Both-even hone par power reduction zaruri hai.

$\int {\tan }^{m}x{\sec }^{n}xdx$ mein jab $n$ even ho, to hum hamesha $u=\tan x$ lete hain. Sach — even secant → tan-u ::: Sach. $\frac{d}{dx}\tan x={\sec }^{2}x$, isliye extra ${\sec }^{2}x$ hi $du$ ban jaata hai. Kyunki $n$ even hai, ${\sec }^{n-2}x$ ek even secant power hai jise ${\sec }^2=1+{\tan }^2$ poori tarah $\tan$ mein rewrite kar deta hai.

$\int {\tan }^{2}xdx=\frac{{\tan }^{3}x}{3}+C$. Jhooth — dx, d(tan x) nahi hai ::: Jhooth. Power rule ke liye variable ka apna differential chahiye; yahan $dx\ne d(\tan x)$. Sahi tarika: ${\tan }^{2}x={\sec }^{2}x-1$, jisse $\tan x-x+C$ milta hai.

$\int \sec xdx=\ln |\sec x|+C$, $\int \tan xdx=\ln |\sec x|+C$ ki analogy se. Jhooth — sec aur tan alag tarah integrate hote hain ::: Jhooth. Yeh analogy ek trap hai. $\int \sec xdx=\ln |\sec x+\tan x|+C$ (jo $\frac{\sec x+\tan x}{\sec x+\tan x}$ se multiply karke milta hai). Sirf $\int \tan xdx$ hi $\ln |\sec x|+C$ deta hai.

$\int {\tan }^{3}x{\sec }^{4}xdx$ ke liye dono strategies (save ${\sec }^2$, ya save $\sec \tan$) kaam karti hain. Sach — dono parities favorable hain ::: Sach. $\sec$ power even aur $\tan$ power odd hai, isliye $u=\tan x$ ya $u=\sec x$ dono succeed hote hain. Jo zyada cleanly convert kare woh chuno; answers $1+{\tan }^2={\sec }^2$ use karne ke baad match karte hain.

$u$-substitution ek hi problem mein kabhi use nahi ho sakte. Jhooth — ye combine hote hain ::: Jhooth. Power reduction ke baad aksar $\int \cos 2xdx$ ya $\int {\cos }^{2}2xdx$ milta hai, aur baad wale ko phir se power reduction ya chhoti $u=2x$ mental substitution chahiye. Ye freely stack hote hain.

$\int {\sin }^{2}x{\cos }^{3}xdx$ mein $\cos$ ka odd power hai, isliye $u=\cos x$ lete hain." Galat u ka choice ::: Error: tum $u$ us function ko set kar rahe ho jo peel kiya gaya hai. Tum ek $\cos x$ bachate ho $du=\cos xdx$ ke liye, jo $u=\sin x$ force karta hai. Rule: $u$ woh function hai jiska power even rehta hai.

$\int (1-{\cos }^{2}x)\sin xdx$ mein $u=\cos x$ se $\int (1-u^2)du$ milta hai." Missing sign ::: Error: $du=-\sin xdx$, isliye $\sin xdx=-du$. Integral hai $\int (1-u^2)(-du)$. Minus chhodne se poore answer ka sign palat jaata hai.

$\int {\tan }^{4}x{\sec }^{3}xdx$: tan power even hai, isliye ${\sec }^{2}x$ save karunga aur $u=\tan x$ lunga." Even tan trigger nahi hai ::: Error: ${\sec }^{2}x$ save karne ke liye even sec power chahiye, lekin yahan $n=3$ odd hai — koi extra ${\sec }^2$ leftover nahi. Aur $\tan$ power even hai, odd nahi, isliye $\sec x\tan x$ route bhi fail hoga. Is mixed case ko Reduction formulas ya Integration by parts chahiye.

$\int {\sin }^{3}xdx=\int {\sin }^{2}x\cdot \sin xdx=\int (1-{\cos }^{2}x)\sin xdx=(1-{\cos }^{2}x)\cdot (-\cos x)+C$." Product ko termwise galat integrate kiya ::: Error: tumne substitute karne ki jagah multiply kar diya. $u=\cos x$ ke baad: $\int (1-u^2)(-du)=-u+\frac{u^3}{3}+C=-\cos x+\frac{{\cos }^{3}x}{3}+C$.

$\int {\cos }^{2}xdx=\frac{{\cos }^{3}x}{3}+C$ power rule se." Wahi dx trap jaise tan^2 mein ::: Error: $dx\ne d(\cos x)$, aur peel karne ke liye koi odd factor nahi hai. Use karo ${\cos }^{2}x=\frac{1+\cos 2x}{2}$: answer $\frac{x}{2}+\frac{\sin 2x}{4}+C$.

$\int \tan xdx=\int \frac{\sin x}{\cos x}dx$ mein $u=\sin x$ lete hain." Tan ke liye galat u ::: Error: phir $du=\cos xdx$, lekin koi extra $\cos x$ nahi hai — $\cos x$ denominator mein hai. $u=\cos x$ lo, $du=-\sin xdx$, jisse $-\ln |\cos x|+C$ milta hai.

saved factor bilkul chosen $u$ ki derivative ho, na ki sirf "koi bhi leftover"? Engine hai u-sub ::: Kyunki $u$-substitution tab hi kaam karta hai jab integrand $f(u)\frac{du}{dx}dx$ form mein ho. Saved factor literally $du$ ban jaata hai; agar woh $u$ ki derivative nahi hai, to koi valid substitution perform nahi ho sakti.

odd exponent guarantee karta hai ki method kaam karega, lekin even nahi? Odd → even leftover ::: Odd power se ek factor peel karne par even power bachti hai, aur Pythagorean identities (${\sin }^2=1-{\cos }^2$, etc.) sirf squares ko rewrite karti hain. Even power minus one odd hota hai, jo poori tarah convert nahi ho sakta.

$1+{\tan }^{2}x={\sec }^{2}x$ ko alag se memorize karne ki bajaye derive kyon karte hain? Ek identity, divided ::: Yeh bas ${\sin }^2+{\cos }^2=1$ ko ${\cos }^{2}x$ se divide karna hai. Derive karne se pata chalta hai yeh koi naya fact nahi hai, aur yaad dilata hai $\sec x=1/\cos x$, $\tan x=\sin x/\cos x$ — jab cheezein gadbad ho jaayein to useful hai.

$\int \sec xdx$ ka trick ($\frac{\sec x+\tan x}{\sec x+\tan x}$ se multiply karna) kyon kaam karta hai? Numerator = denominator ki derivative ::: Multiply karne ke baad, numerator ${\sec }^{2}x+\sec x\tan x$ bilkul $\frac{d}{dx}(\sec x+\tan x)$ hai. Isliye yeh $\int \frac{du}{u}=\ln |u|+C$ ban jaata hai jahan $u=\sec x+\tan x$.

${\sin }^m{\cos }^n$ case ke liye power reduction hi kyun ek option hai? Koi derivative-shaped spare nahi ::: Substitution ke liye ek spare factor chahiye jo $du$ ke barabar ho; even/even mein aisa koi akela factor peel karne ke liye nahi hota. Power reduction substitution ko poori tarah bypass karta hai double-angle identities se exponents ghatakar.

$\sec x\tan x$ save karne ke liye $\tan$ power odd kyun honi chahiye? Even leftover convert karne ke liye ::: Ek $\tan$ hatane par even $\tan$ power bachti hai, jise ${\tan }^2={\sec }^2-1$ $\sec$ mein convert karta hai. Phir sab kuch $\sec$ mein hai aur $du=\sec x\tan xdx$. Even $\tan$ power hoti to odd, unconvertible remainder bachta.

$m=0$ ho, jaise $\int {\cos }^{3}xdx$? m=0 even hai ::: $m=0$ even count hota hai (zero $\sin$ factors). Kyunki $\cos$ power odd hai, ek $\cos$ save karo, $u=\sin x$ lo: $\int (1-{\sin }^{2}x)\cos xdx=\sin x-\frac{{\sin }^{3}x}{3}+C$.

$\int {\sin }^{0}x{\cos }^{0}xdx$. Kya method apply hota hai? Degenerate: bas dx ::: Yeh $\int 1dx=x+C$ hai. Bilkul bhi trig nahi — technically "both even" branch apply hota hai lekin trivially; koi reduction nahi chahiye.

$\int {\sec }^{2}xdx$ — kaunsa case hai, aur kya machinery fire hoti hai? Direct antiderivative ::: $\sec$ power even hai, isliye nominally $u=\tan x$, $du={\sec }^{2}xdx$ se $\int du=\tan x+C$ milta hai. Yeh kaam karta hai lekin actually sirf ${\sec }^{2}x=\frac{d}{dx}\tan x$ directly recognize karna hai.

$\int \tan xdx$: $m=1$ (odd tan), $n=0$. Kya "save $\sec x\tan x$" kaam karta hai? n=0 sec-tan save ko todta hai ::: Nahi — koi $\sec x$ nahi hai $\sec x\tan x$ banane ke liye. $n=0$ hone par memory-anchor par aao: $\frac{\sin x}{\cos x}$ ki tarah rewrite karo aur $u=\cos x$ use karo, jisse $\ln |\sec x|+C$ milta hai.

$\int {\sin }^{2}x{\cos }^{2}xdx$ (both even, chhote powers) ke liye kya double power reduction se koi shortcut hai? Double-angle shortcut ::: Haan: $\sin x\cos x=\frac{1}{2}\sin 2x$, isliye integrand $\frac{1}{4}{\sin }^{2}2x=\frac{1-\cos 4x}{8}$ hai. Integrate karne par $\frac{x}{8}-\frac{\sin 4x}{32}+C$ — do ki jagah ek power reduction.

negative ho, jaise $\int \frac{{\cos }^{3}x}{{\sin }^{2}x}dx$? Kya parity phir bhi guide karti hai? Odd cos ab bhi peel hoti hai ::: Haan. Ise ${\sin }^{-2}x{\cos }^{3}x$ maano: $\cos$ power odd hai, ek $\cos$ save karo, $u=\sin x$: $\int \frac{1-u^2}{u^2}du=-\frac{1}{u}-u+C=-\csc x-\sin x+C$. Sirf odd factor ki parity matter karti hai.

$\int {\cos }^{100}xdx$), to raw power reduction ki jagah Reduction formulas prefer kyon karte hain? Recursion, expansion se behtar hai ::: Both-even with huge powers matlab repeated power reduction se bahut saare terms aa jaayenge. Reduction formula $\int {\cos }^{n}xdx$ ko $\int {\cos }^{n-2}xdx$ ke terms mein express karta hai, jo combinatorial mess ki jagah ek clean recursive ladder deta hai.