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

Forecast: $\cos$ has odd power ($n=3$), $\sin$ has power $0$. Odd one out ($\cos$) donates $du$, so $u=\sin x$.

Step — peel one $\cos$: we split ${\cos }^{3}x={\cos }^{2}x\cdot \cos x$. The lone $\cos xdx$ is exactly $d(\sin x)$. $\int {\cos }^{2}x(\cos xdx)$ Step — convert the even leftover using ${\cos }^{2}x=1-{\sin }^{2}x$ (now everything is in $\sin$): $\int (1-{\sin }^{2}x)(\cos xdx)$ 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: there is no spare ${\sec }^{2}x$ and no spare $\sec x\tan x$ to be a $du$. So substitution can't start directly — we rewrite first with ${\tan }^{2}x={\sec }^{2}x-1$. $\int ({\sec }^{2}x-1)dx$ Now $\int {\sec }^{2}xdx=\tan x$ (because $\frac{d}{dx}\tan x={\sec }^{2}x$) and $\int 1dx=x$: $\boxed{\tan x-x+C}$

Forecast: the $\sec$ power is even ($n=2$). Save ${\sec }^{2}x$ for $du$, let $u=\tan x$. $\int {\tan }^{3}x({\sec }^{2}xdx)$ Here nothing even needs converting — the whole ${\tan }^{3}x$ is left, and ${\sec }^{2}xdx=du$. 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 is odd ($m=3$). Peel one $\sin x$; let $u=\cos x$. $\int {\sin }^{2}x{\cos }^{2}x(\sin xdx)$ Convert the even ${\sin }^{2}x$ using ${\sin }^{2}x=1-{\cos }^{2}x$: $\int (1-{\cos }^{2}x){\cos }^{2}x(\sin xdx)$ Substitute $u=\cos x$. Here $du=-\sin xdx$, so $\sin xdx=-du$ (write the minus explicitly!): $\int (1-u^2)u^2(-du)=-\int (u^2-u^4)du=-\frac{u^3}{3}+\frac{u^5}{5}+C$ Back-substitute: $\boxed{\frac{{\cos }^{5}x}{5}-\frac{{\cos }^{3}x}{3}+C}$

Forecast: both powers even → nothing peels cleanly → Power-reduction & double-angle formulas. Use ${\sin }^{2}x=\frac{1-\cos 2x}{2}$ and ${\cos }^{2}x=\frac{1+\cos 2x}{2}$: $\int \frac{1-\cos 2x}{2}\cdot \frac{1+\cos 2x}{2}dx=\frac{1}{4}\int (1-{\cos }^{2}2x)dx$ That $(1-{\cos }^{2}2x)$ is ${\sin }^{2}2x$, but power-reduce again to integrate it: ${\cos }^{2}2x=\frac{1+\cos 4x}{2}$. $\frac{1}{4}\int \left(1-\frac{1+\cos 4x}{2}\right)dx=\frac{1}{4}\int \left(\frac{1}{2}-\frac{\cos 4x}{2}\right)dx$ $=\frac{1}{4}\left(\frac{x}{2}-\frac{\sin 4x}{8}\right)+C=\boxed{\frac{x}{8}-\frac{\sin 4x}{32}+C}$

Forecast: $\tan$ odd, but there is no $\sec x\tan x$ to be $du$. Split off ${\tan }^{2}x$ and convert: $\int {\tan }^{2}x\tan xdx=\int ({\sec }^{2}x-1)\tan xdx=\int \tan x{\sec }^{2}xdx-\int \tan xdx$ First piece: $u=\tan x,du={\sec }^{2}xdx\Rightarrow \int udu=\frac{{\tan }^{2}x}{2}$. Second piece: $\int \tan xdx=-\ln |\cos x|+C=\ln |\sec x|+C$ (from $u=\cos x$). $\boxed{\frac{{\tan }^{2}x}{2}-\ln |\sec x|+C}$

Forecast: this is ${\tan }^{0}x{\sec }^{4}x$ — $\sec$ power even → save ${\sec }^{2}x$, let $u=\tan x$. $\int {\sec }^{2}x({\sec }^{2}xdx)$ Convert the saved-for-conversion ${\sec }^{2}x=1+{\tan }^{2}x$: $\int (1+{\tan }^{2}x)({\sec }^{2}xdx)$ 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 → $u=\sin x$. Peel one $\cos x$, convert ${\cos }^{2}x=1-{\sin }^{2}x$. ${\int }_0^{\pi /2}{\sin }^{2}x(1-{\sin }^{2}x)(\cos xdx)$ Change the limits too (this is a definite integral): $u=\sin x$, so $x=0\Rightarrow u=0$ and $x=\pi /2\Rightarrow u=1$. ${\int }_0^1{u}^2(1-u^2)du={\int }_0^1(u^2-u^4)du={\left[\frac{u^3}{3}-\frac{u^5}{5}\right]}_0^1$ $=\frac{1}{3}-\frac{1}{5}=\boxed{\frac{2}{15}}$

Forecast: rewrite as ${\sin }^{3}x{\cos }^{-4}x$. The $\sin$ power is odd — peel one $\sin x$, let $u=\cos x$. (Negative powers of $\cos$ are fine; parity of $\sin$ is what matters.) $\int {\sin }^{2}x{\cos }^{-4}x(\sin xdx)=\int (1-{\cos }^{2}x){\cos }^{-4}x(\sin xdx)$ 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 is odd and $\tan$ power is even ($m=0$) — neither parity rule from Case B applies. This is the classic case for Integration by parts. Write ${\sec }^{3}x=\sec x\cdot {\sec }^{2}x$ and set $u=\sec x,dv={\sec }^{2}xdx\Rightarrow du=\sec x\tan xdx,v=\tan x.$ 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 ${\tan }^{2}x={\sec }^{2}x-1$: $=\sec x\tan x-\int \sec x({\sec }^{2}x-1)dx=\sec x\tan x-\int {\sec }^{3}xdx+\int \sec xdx$ The unknown integral reappears! Call it $I=\int {\sec }^{3}xdx$. Then $I=\sec x\tan x-I+\ln |\sec x+\tan x|\Rightarrow 2I=\sec x\tan x+\ln |\sec x+\tan x|$ $\boxed{\int {\sec }^{3}xdx=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln |\sec x+\tan x|+C}$

Forecast: $\tan$ even, $\sec$ odd — again no parity rule fits directly. Convert all $\tan$ to $\sec$ and reduce to secant powers. Use ${\tan }^{2}x={\sec }^{2}x-1$: $\int ({\sec }^{2}x-1){\sec }^{3}xdx=\int {\sec }^{5}xdx-\int {\sec }^{3}xdx$ We already know $\int {\sec }^{3}xdx$ (Problem 4.1). For $\int {\sec }^{5}xdx$ use the reduction formula $\int {\sec }^{n}xdx=\frac{{\sec }^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}\int {\sec }^{n-2}xdx.$ With $n=5$: $\int {\sec }^{5}xdx=\frac{{\sec }^{3}x\tan x}{4}+\frac{3}{4}\int {\sec }^{3}xdx.$ Subtract $\int {\sec }^{3}xdx$: $\int {\tan }^{2}x{\sec }^{3}xdx=\frac{{\sec }^{3}x\tan x}{4}+\frac{3}{4}I-I=\frac{{\sec }^{3}x\tan x}{4}-\frac{1}{4}I$ where $I=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln |\sec x+\tan x|$. So $\boxed{\frac{{\sec }^{3}x\tan x}{4}-\frac{\sec x\tan x}{8}-\frac{1}{8}\ln |\sec x+\tan x|+C}$

Forecast: two valid routes! $\tan$ odd (save $\sec x\tan x$, $u=\sec x$) or $\sec$ even (save ${\sec }^{2}x$, $u=\tan x$). The $\sec$-even route is cleaner here because we get a plain polynomial in $\tan$. Save ${\sec }^{2}x$, let $u=\tan x$: $\int {\tan }^{5}x{\sec }^{2}x({\sec }^{2}xdx)=\int {\tan }^{5}x(1+{\tan }^{2}x)({\sec }^{2}xdx)$ Change limits: $u=\tan x$, so $x=0\Rightarrow u=0$, $x=\pi /4\Rightarrow u=\tan \frac{\pi }{4}=1$. ${\int }_0^1{u}^5(1+u^2)du={\int }_0^1(u^5+u^7)du={\left[\frac{u^6}{6}+\frac{u^8}{8}\right]}_0^1=\frac{1}{6}+\frac{1}{8}$ $=\frac{4}{24}+\frac{3}{24}=\boxed{\frac{7}{24}}$

Forecast: $\sin$ power odd ($m=5$) — peel one $\sin x$, let $u=\cos x$. Leftover ${\sin }^{4}x=({\sin }^{2}x{)}^2=(1-{\cos }^{2}x{)}^2$. $\int ({\sin }^{2}x{)}^2{\cos }^{4}x(\sin xdx)=\int (1-{\cos }^{2}x{)}^2{\cos }^{4}x(\sin xdx)$ Substitute $u=\cos x,\sin xdx=-du$: $-\int (1-u^2{)}^2{u}^{4}du$ Expand $(1-u^2{)}^2=1-2u^2+u^4$, so $(1-u^2{)}^2{u}^4=u^4-2u^6+u^8$: $-\int (u^4-2u^6+u^8)du=-\left(\frac{u^5}{5}-\frac{2u^7}{7}+\frac{u^9}{9}\right)+C$ $\boxed{-\frac{{\cos }^{5}x}{5}+\frac{2{\cos }^{7}x}{7}-\frac{{\cos }^{9}x}{9}+C}$

First integral. $\frac{1}{{\cos }^{2}x}={\sec }^{2}x$, and ${\sec }^{2}x$ is itself the derivative of $\tan x$. No peeling, no identity — just recognition: $\int {\sec }^{2}xdx=\boxed{\tan x+C}$ Second integral. $\frac{{\sin }^{2}x}{{\cos }^{2}x}={\tan }^{2}x$. This is Problem 1.2's engine: convert ${\tan }^{2}x={\sec }^{2}x-1$: $\int ({\sec }^{2}x-1)dx=\boxed{\tan x-x+C}$ The relationship (the edge insight): the two integrands differ by exactly $1$, since ${\sec }^{2}x-{\tan }^{2}x=1$. So their integrals differ by $\int 1dx=x$ — visible as the extra $-x$ in the second answer. Recognising ${\sec }^{2}x-{\tan }^{2}x=1$ before integrating saves all the work.