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

True — parity method for ${\sin }^m{\cos }^n$ ::: True. Splitting off one factor from an odd power always leaves an even power, and identities only rewrite even powers. So ${\cos }^{99}$ behaves exactly like ${\cos }^3$: peel one, convert the rest.

$\int {\sin }^{4}x{\cos }^{4}xdx$ can be done by the substitution $u=\sin x$. False — both even blocks substitution ::: False. Peeling one $\cos$ leaves ${\cos }^3$, an odd leftover that ${\cos }^2=1-{\sin }^2$ cannot fully convert, and the spare $\cos dx$ has no matching factor. Both-even forces power reduction.

$\int {\tan }^{m}x{\sec }^{n}xdx$ with $n$ even, we always let $u=\tan x$. True — even secant → tan-u ::: True. $\frac{d}{dx}\tan x={\sec }^{2}x$, so a spare ${\sec }^{2}x$ becomes $du$. Since $n$ is even, ${\sec }^{n-2}x$ is an even secant power that ${\sec }^2=1+{\tan }^2$ rewrites entirely in $\tan$.

$\int {\tan }^{2}xdx=\frac{{\tan }^{3}x}{3}+C$. False — dx is not d(tan x) ::: False. The power rule needs the variable's own differential; here $dx\ne d(\tan x)$. Correct route: ${\tan }^{2}x={\sec }^{2}x-1$, giving $\tan x-x+C$.

$\int \sec xdx=\ln |\sec x|+C$, by analogy with $\int \tan xdx=\ln |\sec x|+C$. False — sec and tan integrate differently ::: False. That analogy is a trap. $\int \sec xdx=\ln |\sec x+\tan x|+C$ (from multiplying by $\frac{\sec x+\tan x}{\sec x+\tan x}$). Only $\int \tan xdx$ gives $\ln |\sec x|+C$.

$\int {\tan }^{3}x{\sec }^{4}xdx$ both strategies (save ${\sec }^2$, or save $\sec \tan$) work. True — both parities favourable ::: True. $\sec$ power is even and $\tan$ power is odd, so either $u=\tan x$ or $u=\sec x$ succeeds. Pick whichever converts more cleanly; answers match after using $1+{\tan }^2={\sec }^2$.

$u$-substitution can never be used in the same problem. False — they combine ::: False. After power reduction you often meet $\int \cos 2xdx$ or $\int {\cos }^{2}2xdx$, and the latter needs another power reduction or a small $u=2x$ mental substitution. They stack freely.

$\cos$ has the odd power in $\int {\sin }^{2}x{\cos }^{3}xdx$, so let $u=\cos x$." Wrong choice of u ::: Error: you set $u$ to the peeled function. You save one $\cos x$ for $du=\cos xdx$, which forces $u=\sin x$. Rule: $u$ is the function whose power stays even.

$\int (1-{\cos }^{2}x)\sin xdx$ with $u=\cos x$ gives $\int (1-u^2)du$." Missing sign ::: Error: $du=-\sin xdx$, so $\sin xdx=-du$. The integral is $\int (1-u^2)(-du)$. Dropping the minus flips the whole answer's sign.

$\int {\tan }^{4}x{\sec }^{3}xdx$: tan power even, so I'll save ${\sec }^{2}x$ and let $u=\tan x$." Even tan is not the trigger ::: Error: saving ${\sec }^{2}x$ requires an even sec power, but here $n=3$ is odd — no spare ${\sec }^2$ leftover. And $\tan$ power is even, not odd, so the $\sec x\tan x$ route fails too. This mixed case needs Reduction formulas or Integration by parts.

$\int {\sin }^{3}xdx=\int {\sin }^{2}x\cdot \sin xdx=\int (1-{\cos }^{2}x)\sin xdx=(1-{\cos }^{2}x)\cdot (-\cos x)+C$." Integrated a product termwise wrongly ::: Error: you multiplied instead of substituting. After $u=\cos x$: $\int (1-u^2)(-du)=-u+\frac{u^3}{3}+C=-\cos x+\frac{{\cos }^{3}x}{3}+C$.

$\int {\cos }^{2}xdx=\frac{{\cos }^{3}x}{3}+C$ by the power rule." Same dx trap as tan^2 ::: Error: $dx\ne d(\cos x)$, and there's no odd factor to peel. Use ${\cos }^{2}x=\frac{1+\cos 2x}{2}$: answer $\frac{x}{2}+\frac{\sin 2x}{4}+C$.

$\int \tan xdx=\int \frac{\sin x}{\cos x}dx$, let $u=\sin x$." Wrong u for tan ::: Error: then $du=\cos xdx$, but there's no spare $\cos x$ — the $\cos x$ is in the denominator. Let $u=\cos x$, $du=-\sin xdx$, giving $-\ln |\cos x|+C$.

saved factor be exactly the derivative of the chosen $u$, not just "any leftover"? The engine is u-sub ::: Because $u$-substitution only works when the integrand is $f(u)\frac{du}{dx}dx$. The saved factor literally becomes $du$; if it isn't $u$'s derivative, there's no valid substitution to perform.

odd exponent guarantee the method works, but an even one doesn't? Odd → even leftover ::: Peeling one factor from an odd power leaves an even power, and the Pythagorean identities (${\sin }^2=1-{\cos }^2$, etc.) only rewrite squares. An even power minus one is odd, which can't be fully converted.

$1+{\tan }^{2}x={\sec }^{2}x$ instead of memorizing it separately? One identity, divided ::: It's just ${\sin }^2+{\cos }^2=1$ divided through by ${\cos }^{2}x$. Deriving it shows it's not a new fact, and reminds you $\sec x=1/\cos x$, $\tan x=\sin x/\cos x$ — useful when things go wrong.

$\int \sec xdx$ trick (multiply by $\frac{\sec x+\tan x}{\sec x+\tan x}$) work? Numerator = derivative of denominator ::: After multiplying, the numerator ${\sec }^{2}x+\sec x\tan x$ is exactly $\frac{d}{dx}(\sec x+\tan x)$. So it becomes $\int \frac{du}{u}=\ln |u|+C$ with $u=\sec x+\tan x$.

only option for the both-even ${\sin }^m{\cos }^n$ case? No derivative-shaped spare ::: Substitution needs a spare factor equal to $du$; even/even has no such lone factor to peel. Power reduction sidesteps substitution entirely by lowering exponents via double-angle identities.

$\sec x\tan x$ require the $\tan$ power to be odd? Even leftover to convert ::: Removing one $\tan$ leaves an even $\tan$ power, which ${\tan }^2={\sec }^2-1$ converts to $\sec$. Then everything is in $\sec$ with $du=\sec x\tan xdx$. An even $\tan$ power would leave an odd, unconvertible remainder.

$m=0$, e.g. $\int {\cos }^{3}xdx$? m=0 is even ::: $m=0$ counts as even (zero $\sin$ factors). Since $\cos$ power is odd, save one $\cos$, let $u=\sin x$: $\int (1-{\sin }^{2}x)\cos xdx=\sin x-\frac{{\sin }^{3}x}{3}+C$.

$\int {\sin }^{0}x{\cos }^{0}xdx$. Does the method apply? Degenerate: just dx ::: That's $\int 1dx=x+C$. No trig at all — the "both even" branch technically applies but trivially; no reduction needed.

$\int {\sec }^{2}xdx$ — which case, and does the machinery even fire? Direct antiderivative ::: $\sec$ power is even, so nominally $u=\tan x$, $du={\sec }^{2}xdx$ gives $\int du=\tan x+C$. It works but is really just recognizing ${\sec }^{2}x=\frac{d}{dx}\tan x$ directly.

$\int \tan xdx$ under Case B parity: $m=1$ (odd tan), $n=0$. Does "save $\sec x\tan x$" work? n=0 breaks the sec-tan save ::: No — there's no $\sec x$ to form $\sec x\tan x$. With $n=0$ you drop to the memory-anchor: rewrite as $\frac{\sin x}{\cos x}$ and use $u=\cos x$, giving $\ln |\sec x|+C$.

$\int {\sin }^{2}x{\cos }^{2}xdx$ (both even, small powers), is there a shortcut over double power reduction? Double-angle shortcut ::: Yes: $\sin x\cos x=\frac{1}{2}\sin 2x$, so the integrand is $\frac{1}{4}{\sin }^{2}2x=\frac{1-\cos 4x}{8}$. Integrate to $\frac{x}{8}-\frac{\sin 4x}{32}+C$ — one power reduction instead of two.

negative, e.g. $\int \frac{{\cos }^{3}x}{{\sin }^{2}x}dx$? Does parity still guide you? Odd cos still peels ::: Yes. Treat it as ${\sin }^{-2}x{\cos }^{3}x$: $\cos$ power is odd, save one $\cos$, $u=\sin x$: $\int \frac{1-u^2}{u^2}du=-\frac{1}{u}-u+C=-\csc x-\sin x+C$. Parity of the odd factor is all that matters.

$\int {\cos }^{100}xdx$), why do we prefer Reduction formulas over raw power reduction? Recursion beats expansion ::: Both-even with huge powers means repeated power reduction explodes into many terms. A reduction formula expresses $\int {\cos }^{n}xdx$ in terms of $\int {\cos }^{n-2}xdx$, giving a clean recursive ladder instead of a combinatorial mess.