Trigonometric integrals — sinᵐ·cosⁿ cases, tan and sec cases: Worked examples

4.2.8 · D3 · Maths › Calculus II — Integration › Trigonometric integrals — sinᵐ·cosⁿ cases, tan and sec cases

Yeh parent topic ka drill floor hai. Parent ne tumhe rules diye the. Yahan hum har possible shape ko dhundhte hain jo ek exam ya textbook tumhare saamne rakh sakta hai, aur ek-ek ka solution karte hain — taaki jab koi naya integral mile, tum turant pehchaan sako ki woh grid ke kis cell mein aata hai.

Shuru karne se pehle, ek honest reminder ki hamare saare tools ka matlab kya hai, seedhi baat mein:

Neeche sab kuch ek hi move hai: ek factor ko $d u$ banana ke liye peel karo, baaki ko identity se rebuild karo. Bas hum yeh har disguise mein karte hain.

The scenario matrix

Is topic ka har integral exactly ek row mein aata hai. Humara kaam hai ki har ek ka kam se kam ek example solve kiya ho.

#	Cell (kaun sa case)	Trigger jo tumhe dikhta hai	$u$ = ?	Example
1	$sin^{m} cos^{n}$ , cos odd	$cos$ par odd power	$u = sin x$	Ex 1
2	$sin^{m} cos^{n}$ , sin odd	$sin$ par odd power	$u = cos x$	Ex 2
3	$sin^{m} cos^{n}$ , both even	koi odd factor nahi	power-reduction	Ex 3
4	$sin^{m} cos^{n}$ , both odd	dono odd — ek free choice!	koi bhi	Ex 4
5	$tan^{m} sec^{n}$ , sec even	$sec$ par even power	$u = tan x$	Ex 5
6	$tan^{m} sec^{n}$ , tan odd	$tan$ par odd power	$u = sec x$	Ex 6
7	Degenerate: pure $\int tan x d x$ (odd tan, koi sec nahi, koi even fallback nahi)	$tan$ akela	$u = cos x$	Ex 7
8	Definite integral sign/quadrant care ke saath	limits diye hue	jaise upar	Ex 8
9	Word problem (RMS / average value) — real-world	"average power"	power-reduction	Ex 9
10	Exam twist: $tan / sec$ mein $n$ even aur $m$ even — koi bhi shortcut cleanly kaam nahi karta	$tan^{2} sec^{0}$	sirf identity	Ex 10

Ex 1 — Cell 1: cosine power odd

Ex 2 — Cell 2: sine power odd (sign ka dhyan rakho!)

Ex 3 — Cell 3: dono even (power-reduction)

Ex 4 — Cell 4: dono odd (ek free choice — geometrically dekho)

Yahan $m$ aur $n$ dono odd hain. Tum koi bhi ek peel kar sakte ho. Aao dekhen ki dono raaste same total dete hain, aur aisa kyun zaroor hona chahiye.

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

Ex 5 — Cell 5: secant power even

Ex 6 — Cell 6: tangent power odd

Ex 7 — Cell 7: degenerate corner, $\int tan x d x$

Yeh woh case hai jahan odd $tan$ hai lekin koi $sec$ bilkul nahi — Cell 6 ki trick ke paas save karne ke liye kuch nahi. Hum definition par wapas jaate hain.

Ex 8 — Cell 8: ek definite integral (limits + sign care)

Ab hum numbers lagate hain aur evaluate karte hain, interval mein signs dekhte hue.

Ex 9 — Cell 9: word problem (AC signal ka RMS)

Worked example Ek household voltage

$v (t) = V_{0} sin t$ hai. Engineers ko ek poore cycle mein $v^{2}$ ka average jaanna hota hai (yahi ek resistor ko heat karta hai). Average value compute karo $\frac{1}{2 π} \int_{0}^{2 π} V_{0}^{2} sin^{2} t d t$ . Forecast: $sin^{2}$ both-even hai → power reduction (Cell 3 machinery), phir interval length se divide karo.

Step 1 — power-reduce karo. Kyun? Koi factor nahi peelti; $sin^{2} t = \frac{1 - c o s 2 t}{2}$ square ko integrate karne layak cheez mein badal deta hai. $\int_{0}^{2 π} sin^{2} t d t = \int_{0}^{2 π} \frac{1 - c o s 2 t}{2} d t = \frac{1}{2} [t - \frac{s i n 2 t}{2}]_{0}^{2 π}$ Step 2 — bracket evaluate karo. $sin 2 t$ term kyun vanish hota hai? $sin (2 \cdot 2 π) = sin (4 π) = 0$ aur $sin 0 = 0$ — cycles ka poora number ise wash out kar deta hai. $= \frac{1}{2} [(2 π - 0) - (0 - 0)] = π$ Step 3 — average form karo. $avg of v^{2} = \frac{1}{2 π} \cdot V_{0}^{2} \cdot π = \frac{V _{0}^{2}}{2}$ Units / meaning: $v^{2}$ ke units volts² hain. RMS voltage hai $avg = V_{0} / 2 \approx 0.707 V_{0}$ — exactly woh famous " $230 V$ RMS from a $\approx 325 V$ peak" rule. ✓

Verify karo: average value $= \frac{V _{0}^{2}}{2}$ ; $V_{0} = 1$ ke saath yeh $0.5$ hai.

Ex 10 — Cell 10: exam twist, $\int tan^{2} x d x$ (koi bhi shortcut kaam nahi karta)

Yahan $tan^{2} sec^{0}$ hai: $sec$ power $0$ even hai lekin save karne ke liye koi $sec^{2}$ nahi; $tan$ power $2$ odd nahi hai. Dono parity shortcuts ruk jaate hain. Bachaav sirf identity se hai.

Recall Self-test: solve karne se pehle cell ka naam bolo

Har ek ke liye, matrix row batao, phir $u$ (ya "power-reduction"/"identity"). $\int sin^{5} x cos^{2} x d x$ ::: Cell 2 (sin odd) → $u = cos x$ . $\int sec^{4} x d x$ ::: Cell 5 (sec even) → $u = tan x$ . $\int tan^{5} x d x$ ::: Cell 6 (tan odd, $tan^{2} = sec^{2} - 1$ ke zariye sec aata hai) → $u = sec x$ jahan sec aaye. $\int cos^{4} x d x$ ::: Cell 3 (both even) → power-reduction. $\int_{0}^{π /4} sec^{2} x d x$ ::: Cell 10-style direct: $[tan x]_{0}^{π /4} = 1$ .

Connections

Parent topic — woh rulebook jise ye examples drill karte hain.
u-substitution — har peel-and-rename step yahi hai.
Pythagorean identities — Cells 1, 2, 5, 6, 10 mein rebuild tool.
Power-reduction & double-angle formulas — Cells 3 aur 9.
Reduction formulas — bahut high powers ( $sin^{10}$ etc.) handle karne ka systematic tarika.
Integration by parts — odd-sec cases jaise $\int sec^{3} x d x$ ke liye zaroori.
Trigonometric substitution — algebraic roots ke liye reverse manoeuvre.