Exercises — Trigonometric integrals — sinᵐ·cosⁿ cases, tan and sec cases

Forecast: $\cos$ ki odd power hai ($n=3$), $\sin$ ki power $0$ hai. Odd wala ($\cos$) $du$ deta hai, toh $u=\sin x$.

Step — ek $\cos$ peel karo: hum ${\cos }^{3}x={\cos }^{2}x\cdot \cos x$ mein split karte hain. Akela $\cos xdx$ exactly $d(\sin x)$ hai. $\int {\cos }^{2}x(\cos xdx)$ Step — bacha hua even part convert karo ${\cos }^{2}x=1-{\sin }^{2}x$ use karke (ab sab kuch $\sin$ mein hai): $\int (1-{\sin }^{2}x)(\cos xdx)$ Substitute $u=\sin x,du=\cos xdx$: $\int (1-u^2)du=u-\frac{u^3}{3}+C$ Back-substitute: $\boxed{\sin x-\frac{{\sin }^{3}x}{3}+C}$

Forecast: koi spare ${\sec }^{2}x$ nahi hai aur koi spare $\sec x\tan x$ nahi hai jo $du$ ban sake. Toh substitution seedha shuru nahi ho sakti — pehle ${\tan }^{2}x={\sec }^{2}x-1$ se rewrite karo. $\int ({\sec }^{2}x-1)dx$ Ab $\int {\sec }^{2}xdx=\tan x$ (kyunki $\frac{d}{dx}\tan x={\sec }^{2}x$) aur $\int 1dx=x$: $\boxed{\tan x-x+C}$

Forecast: $\sec$ power even hai ($n=2$). ${\sec }^{2}x$ ko $du$ ke liye save karo, $u=\tan x$ lo. $\int {\tan }^{3}x({\sec }^{2}xdx)$ Yahan kuch bhi convert karne ki zaroorat nahi — poora ${\tan }^{3}x$ bacha rehta hai, aur ${\sec }^{2}xdx=du$. Substitute $u=\tan x,du={\sec }^{2}xdx$: $\int u^{3}du=\frac{u^4}{4}+C=\boxed{\frac{{\tan }^{4}x}{4}+C}$

Forecast: $\sin$ power odd hai ($m=3$). Ek $\sin x$ peel karo; $u=\cos x$ lo. $\int {\sin }^{2}x{\cos }^{2}x(\sin xdx)$ Convert even ${\sin }^{2}x$ ko ${\sin }^{2}x=1-{\cos }^{2}x$ se: $\int (1-{\cos }^{2}x){\cos }^{2}x(\sin xdx)$ Substitute $u=\cos x$. Yahan $du=-\sin xdx$, toh $\sin xdx=-du$ (minus explicitly likho!): $\int (1-u^2)u^2(-du)=-\int (u^2-u^4)du=-\frac{u^3}{3}+\frac{u^5}{5}+C$ Back-substitute: $\boxed{\frac{{\cos }^{5}x}{5}-\frac{{\cos }^{3}x}{3}+C}$

Forecast: dono powers even hain → kuch bhi cleanly peel nahi hoga → Power-reduction & double-angle formulas. ${\sin }^{2}x=\frac{1-\cos 2x}{2}$ aur ${\cos }^{2}x=\frac{1+\cos 2x}{2}$ use karo: $\int \frac{1-\cos 2x}{2}\cdot \frac{1+\cos 2x}{2}dx=\frac{1}{4}\int (1-{\cos }^{2}2x)dx$ Woh $(1-{\cos }^{2}2x)$ actually ${\sin }^{2}2x$ hai, lekin isko integrate karne ke liye dobara power-reduce karo: ${\cos }^{2}2x=\frac{1+\cos 4x}{2}$. $\frac{1}{4}\int \left(1-\frac{1+\cos 4x}{2}\right)dx=\frac{1}{4}\int \left(\frac{1}{2}-\frac{\cos 4x}{2}\right)dx$ $=\frac{1}{4}\left(\frac{x}{2}-\frac{\sin 4x}{8}\right)+C=\boxed{\frac{x}{8}-\frac{\sin 4x}{32}+C}$

Forecast: $\tan$ odd hai, lekin $du$ banne ke liye koi $\sec x\tan x$ nahi hai. ${\tan }^{2}x$ alag karo aur convert karo: $\int {\tan }^{2}x\tan xdx=\int ({\sec }^{2}x-1)\tan xdx=\int \tan x{\sec }^{2}xdx-\int \tan xdx$ Pehla piece: $u=\tan x,du={\sec }^{2}xdx\Rightarrow \int udu=\frac{{\tan }^{2}x}{2}$. Doosra piece: $\int \tan xdx=-\ln |\cos x|+C=\ln |\sec x|+C$ ($u=\cos x$ se). $\boxed{\frac{{\tan }^{2}x}{2}-\ln |\sec x|+C}$

Forecast: yeh ${\tan }^{0}x{\sec }^{4}x$ hai — $\sec$ power even hai → ${\sec }^{2}x$ save karo, $u=\tan x$ lo. $\int {\sec }^{2}x({\sec }^{2}xdx)$ Convert bacha hua ${\sec }^{2}x=1+{\tan }^{2}x$: $\int (1+{\tan }^{2}x)({\sec }^{2}xdx)$ Substitute $u=\tan x,du={\sec }^{2}xdx$: $\int (1+u^2)du=u+\frac{u^3}{3}+C=\boxed{\tan x+\frac{{\tan }^{3}x}{3}+C}$

Forecast: $\cos$ odd hai → $u=\sin x$. Ek $\cos x$ peel karo, ${\cos }^{2}x=1-{\sin }^{2}x$ convert karo. ${\int }_0^{\pi /2}{\sin }^{2}x(1-{\sin }^{2}x)(\cos xdx)$ Limits bhi change karo (yeh definite integral hai): $u=\sin x$, toh $x=0\Rightarrow u=0$ aur $x=\pi /2\Rightarrow u=1$. ${\int }_0^1{u}^2(1-u^2)du={\int }_0^1(u^2-u^4)du={\left[\frac{u^3}{3}-\frac{u^5}{5}\right]}_0^1$ $=\frac{1}{3}-\frac{1}{5}=\boxed{\frac{2}{15}}$

Forecast: ${\sin }^{3}x{\cos }^{-4}x$ ki tarah rewrite karo. $\sin$ power odd hai — ek $\sin x$ peel karo, $u=\cos x$ lo. ($\cos$ ki negative powers theek hain; $\sin$ ki parity hi matter karti hai.) $\int {\sin }^{2}x{\cos }^{-4}x(\sin xdx)=\int (1-{\cos }^{2}x){\cos }^{-4}x(\sin xdx)$ Substitute $u=\cos x,\sin xdx=-du$: $\int (1-u^2)u^{-4}(-du)=-\int (u^{-4}-u^{-2})du$ $=-\left(\frac{u^{-3}}{-3}-\frac{u^{-1}}{-1}\right)+C=\frac{1}{3u^3}-\frac{1}{u}+C$ $=\boxed{\frac{1}{3{\cos }^{3}x}-\frac{1}{\cos x}+C}=\frac{{\sec }^{3}x}{3}-\sec x+C$

Forecast: $\sec$ power odd hai aur $\tan$ power even hai ($m=0$) — Case B ka koi bhi parity rule apply nahi hota. Yeh classic case hai Integration by parts ka. ${\sec }^{3}x=\sec x\cdot {\sec }^{2}x$ likho aur set karo $u=\sec x,dv={\sec }^{2}xdx\Rightarrow du=\sec x\tan xdx,v=\tan x.$ Parts, $\int udv=uv-\int vdu$: $\int {\sec }^{3}xdx=\sec x\tan x-\int \tan x\cdot \sec x\tan xdx=\sec x\tan x-\int \sec x{\tan }^{2}xdx$ Convert ${\tan }^{2}x={\sec }^{2}x-1$: $=\sec x\tan x-\int \sec x({\sec }^{2}x-1)dx=\sec x\tan x-\int {\sec }^{3}xdx+\int \sec xdx$ Unknown integral dobara aa gaya! Use $I=\int {\sec }^{3}xdx$ kaho. Toh $I=\sec x\tan x-I+\ln |\sec x+\tan x|\Rightarrow 2I=\sec x\tan x+\ln |\sec x+\tan x|$ $\boxed{\int {\sec }^{3}xdx=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln |\sec x+\tan x|+C}$

Forecast: $\tan$ even hai, $\sec$ odd hai — phir se koi parity rule directly fit nahi hota. Saare $\tan$ ko $\sec$ mein convert karo aur secant powers tak reduce karo. ${\tan }^{2}x={\sec }^{2}x-1$ use karo: $\int ({\sec }^{2}x-1){\sec }^{3}xdx=\int {\sec }^{5}xdx-\int {\sec }^{3}xdx$ $\int {\sec }^{3}xdx$ hum pehle se jaante hain (Problem 4.1). $\int {\sec }^{5}xdx$ ke liye reduction formula use karo: $\int {\sec }^{n}xdx=\frac{{\sec }^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}\int {\sec }^{n-2}xdx.$ $n=5$ ke saath: $\int {\sec }^{5}xdx=\frac{{\sec }^{3}x\tan x}{4}+\frac{3}{4}\int {\sec }^{3}xdx.$ $\int {\sec }^{3}xdx$ subtract karo: $\int {\tan }^{2}x{\sec }^{3}xdx=\frac{{\sec }^{3}x\tan x}{4}+\frac{3}{4}I-I=\frac{{\sec }^{3}x\tan x}{4}-\frac{1}{4}I$ jahan $I=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln |\sec x+\tan x|$. Toh $\boxed{\frac{{\sec }^{3}x\tan x}{4}-\frac{\sec x\tan x}{8}-\frac{1}{8}\ln |\sec x+\tan x|+C}$

Forecast: do valid routes hain! $\tan$ odd (save $\sec x\tan x$, $u=\sec x$) ya $\sec$ even (save ${\sec }^{2}x$, $u=\tan x$). $\sec$-even route yahan zyada clean hai kyunki $\tan$ mein plain polynomial milta hai. ${\sec }^{2}x$ save karo, $u=\tan x$ lo: $\int {\tan }^{5}x{\sec }^{2}x({\sec }^{2}xdx)=\int {\tan }^{5}x(1+{\tan }^{2}x)({\sec }^{2}xdx)$ Limits change karo: $u=\tan x$, toh $x=0\Rightarrow u=0$, $x=\pi /4\Rightarrow u=\tan \frac{\pi }{4}=1$. ${\int }_0^1{u}^5(1+u^2)du={\int }_0^1(u^5+u^7)du={\left[\frac{u^6}{6}+\frac{u^8}{8}\right]}_0^1=\frac{1}{6}+\frac{1}{8}$ $=\frac{4}{24}+\frac{3}{24}=\boxed{\frac{7}{24}}$

Forecast: $\sin$ power odd hai ($m=5$) — ek $\sin x$ peel karo, $u=\cos x$ lo. Bacha hua ${\sin }^{4}x=({\sin }^{2}x{)}^2=(1-{\cos }^{2}x{)}^2$. $\int ({\sin }^{2}x{)}^2{\cos }^{4}x(\sin xdx)=\int (1-{\cos }^{2}x{)}^2{\cos }^{4}x(\sin xdx)$ Substitute $u=\cos x,\sin xdx=-du$: $-\int (1-u^2{)}^2{u}^{4}du$ $(1-u^2{)}^2=1-2u^2+u^4$ expand karo, toh $(1-u^2{)}^2{u}^4=u^4-2u^6+u^8$: $-\int (u^4-2u^6+u^8)du=-\left(\frac{u^5}{5}-\frac{2u^7}{7}+\frac{u^9}{9}\right)+C$ $\boxed{-\frac{{\cos }^{5}x}{5}+\frac{2{\cos }^{7}x}{7}-\frac{{\cos }^{9}x}{9}+C}$

Pehla integral. $\frac{1}{{\cos }^{2}x}={\sec }^{2}x$, aur ${\sec }^{2}x$ khud $\tan x$ ka derivative hai. Koi peeling nahi, koi identity nahi — bas recognition: $\int {\sec }^{2}xdx=\boxed{\tan x+C}$ Doosra integral. $\frac{{\sin }^{2}x}{{\cos }^{2}x}={\tan }^{2}x$. Yeh Problem 1.2 ka engine hai: ${\tan }^{2}x={\sec }^{2}x-1$ convert karo: $\int ({\sec }^{2}x-1)dx=\boxed{\tan x-x+C}$ Relationship (edge insight): dono integrands exactly $1$ se differ karte hain, kyunki ${\sec }^{2}x-{\tan }^{2}x=1$. Toh unke integrals $\int 1dx=x$ se differ karte hain — doosre answer mein extra $-x$ ke roop mein visible hai. ${\sec }^{2}x-{\tan }^{2}x=1$ ko integrate karne se pehle pehchanna saara kaam bacha leta hai.