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

WHY does this work at all?

The engine under everything is u-substitution. Substitution needs the integrand to look like $\int f(u)\,\frac{du}{dx}\,dx.$ So we want a chunk that is "the inside" ( $u$ ) and another chunk that is "its derivative" ( $du$ ).

For trig, the two derivative facts that drive everything:

And the two Pythagorean identities that convert the leftover:

CASE A — $\displaystyle\int \sin^m x\,\cos^n x\,dx$

WHY parity? An odd power means you can split off one factor and still leave an even power, and even powers are exactly what the identities $\sin^2/\cos^2$ can rewrite (they only know squares).

Worked example Odd cosine:

$\displaystyle\int \sin^2x\,\cos^3x\,dx$ Why this step? $\cos$ is odd ( $n=3$ ) → save one $\cos$ , set $u=\sin x$ . $\int \sin^2x\,\cos^2x\,(\cos x\,dx)$ Replace $\cos^2x=1-\sin^2x$ (the leftover, now all in $\sin$ ): $=\int \sin^2x\,(1-\sin^2x)\,(\cos x\,dx)$ Let $u=\sin x,\ du=\cos x\,dx$ : $=\int u^2(1-u^2)\,du=\int (u^2-u^4)\,du=\frac{u^3}{3}-\frac{u^5}{5}+C$ $=\boxed{\frac{\sin^3x}{3}-\frac{\sin^5x}{5}+C}$

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

CASE B — $\displaystyle\int \tan^m x\,\sec^n x\,dx$

WHY these splits? Because $du(\tan)=\sec^2x\,dx$ needs a spare $\sec^2$ , and $du(\sec)=\sec x\tan x\,dx$ needs a spare $\sec\tan$ . The identities supply only even $\tan$ or even $\sec$ replacements, so we save the factor that leaves even leftovers.

Forecast-then-Verify drill

Common mistakes (Steel-manned)

Active-recall flashcards

\int\sin^m\cos^n

, cosine power odd — which

u

u=\sin x

; save one

\cos x

for

du

, convert rest with

\cos^2=1-\sin^2

\int\sin^m\cos^n

, both powers even — method?

Power reduction:

\sin^2=\frac{1-\cos2x}{2},\ \cos^2=\frac{1+\cos2x}{2}

\int\tan^m\sec^n

, secant power even — which

u

u=\tan x

; save

\sec^2x

, convert via

\sec^2=1+\tan^2

\int\tan^m\sec^n

, tangent power odd — which

u

u=\sec x

; save

\sec x\tan x

, convert via

\tan^2=\sec^2-1

\int\tan x\,dx=?

\ln|\sec x|+C

(i.e.

-\ln|\cos x|+C

\int\sec x\,dx=?

\ln|\sec x+\tan x|+C

(multiply by

\frac{\sec x+\tan x}{\sec x+\tan x}

\int\sin^2x\,dx=?

\frac{x}{2}-\frac{\sin2x}{4}+C

\int\tan^2x\,dx=?

\tan x-x+C

using

\tan^2=\sec^2-1

Which identity rebuilds the leftover in

\tan

\sec

when saving

\sec x\tan x

\tan^2x=\sec^2x-1

Recall Feynman: explain to a 12-year-old

Imagine the integral is a Lego tower of $\sin$ and $\cos$ bricks. To take it apart neatly, you pull off one special brick that matches the "machine" ( $du$ ). If you have an odd number of cosine bricks, pull one cosine off — now an even pile is left, and we have a magic rule ( $\cos^2=1-\sin^2$ ) that turns even cosine piles into sine piles. Then we just rename $\sin x$ as $u$ and it becomes baby algebra ( $u^2-u^4$ etc.). Same trick for $\tan$ / $\sec$ : pull off the brick that equals the derivative, convert the rest. If everything is even and nothing peels off cleanly, we use the "cut the power in half" rule (power reduction) instead.

Connections

u-substitution — the core mechanism every case reduces to.
Pythagorean identities — $\sin^2+\cos^2=1$ and $1+\tan^2=\sec^2$ .
Power-reduction & double-angle formulas — handles the both-even case.
Trigonometric substitution — the reverse idea, used for $\sqrt{a^2-x^2}$ type integrals.
Reduction formulas — systematic recursion for high $\sin^n/\cos^n$ powers.
Integration by parts — alternative route for $\int\sec^3x\,dx$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, trig integrals ka pura khel ek hi cheez par tika hai: u-substitution. Substitution ke liye humein integrand mein ek "andar wala part" ( $u$ ) aur uska "derivative" ( $du$ ) chahiye. Toh hum sirf yeh decide karte hain ki kaunsa factor peel karke $du$ banega, aur baaki ko kaunsi Pythagorean identity se convert karenge.

$\sin^m\cos^n$ ke liye rule simple hai: jiski power odd hai, usme se ek factor nikaal lo ( $du$ banane ke liye), aur jo bacha woh even power identity ( $\sin^2=1-\cos^2$ ya $\cos^2=1-\sin^2$ ) se convert ho jaata hai. Agar dono powers even hain, toh koi factor saaf $du$ nahi banta — tab power reduction ( $\sin^2=\frac{1-\cos2x}{2}$ wagairah) use karo. Yaad rakhne ka mantra: "Odd one out goes du."

$\tan^m\sec^n$ mein bhi wahi logic, bas anchors badle: $\frac{d}{dx}\tan x=\sec^2x$ aur $\frac{d}{dx}\sec x=\sec x\tan x$ . Toh agar sec ki power even hai, ek $\sec^2x$ bachao aur $u=\tan x$ rakho. Agar tan ki power odd hai, ek $\sec x\tan x$ bachao aur $u=\sec x$ rakho; baaki $\tan^2=\sec^2-1$ se convert. Aur do classic results ratne ki zaroorat nahi — derive kar lo: $\int\tan x\,dx=\ln|\sec x|$ , aur $\int\sec x\,dx=\ln|\sec x+\tan x|$ (numerator-denominator trick se).

Sabse common galti: $\int\tan^2x\,dx$ ko $\frac{\tan^3x}{3}$ likh dena — galat! Power rule yahan kaam nahi karta kyunki $dx\ne d(\tan x)$ . Sahi: $\tan^2=\sec^2-1$ , toh answer $\tan x - x + C$ . Bas parity dekho, factor peel karo, identity lagao — har sawaal mechanical ho jaayega.

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

WHY does this work at all?

CASE A — $\displaystyle\int \sin^m x\,\cos^n x\,dx$

CASE B — $\displaystyle\int \tan^m x\,\sec^n x\,dx$

Forecast-then-Verify drill

Common mistakes (Steel-manned)

Active-recall flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus II — Integration

Connections