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

This is the drill floor for the parent topic. The parent gave you the rules. Here we hunt down every possible shape an exam or textbook can hand you, and solve one of each — so that when you meet a new integral, you already recognise which cell of the grid it lives in.

Before anything, one honest reminder of what all our tools mean, in plain words:

Everything below is the single move: peel one factor to be $d u$ , rebuild the rest with an identity. We just do it in every disguise.

The scenario matrix

Every integral in this topic falls into exactly one row. Our job is to have solved at least one example of each.

#	Cell (which case)	Trigger you spot	$u$ = ?	Example
1	$sin^{m} cos^{n}$ , cos odd	odd power on $cos$	$u = sin x$	Ex 1
2	$sin^{m} cos^{n}$ , sin odd	odd power on $sin$	$u = cos x$	Ex 2
3	$sin^{m} cos^{n}$ , both even	no odd factor	power-reduction	Ex 3
4	$sin^{m} cos^{n}$ , both odd	both odd — a free choice!	either	Ex 4
5	$tan^{m} sec^{n}$ , sec even	even power on $sec$	$u = tan x$	Ex 5
6	$tan^{m} sec^{n}$ , tan odd	odd power on $tan$	$u = sec x$	Ex 6
7	Degenerate: pure $\int tan x d x$ (odd tan, no sec, no even fallback)	$tan$ alone	$u = cos x$	Ex 7
8	Definite integral with sign/quadrant care	limits given	as above	Ex 8
9	Word problem (RMS / average value) — real-world	"average power"	power-reduction	Ex 9
10	Exam twist: $n$ even and $m$ even in $tan / sec$ — neither shortcut applies cleanly	$tan^{2} sec^{0}$	identity only	Ex 10

Ex 1 — Cell 1: cosine power odd

Ex 2 — Cell 2: sine power odd (mind the sign!)

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

Ex 4 — Cell 4: both odd (a free choice — see it geometrically)

Here $m$ and $n$ are odd. You may peel either one. Let's see both roads give the same total, and why that must be so.

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: the degenerate corner, $\int tan x d x$

This is the case with an odd $tan$ but no $sec$ at all — Cell 6's trick has nothing to save. We fall back to the definition.

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

Now we attach numbers and evaluate, watching signs across the interval.

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

Worked example A household voltage is

$v (t) = V_{0} sin t$ . Engineers care about the average of $v^{2}$ over one full cycle (that's what heats a resistor). Compute the average value $\frac{1}{2 π} \int_{0}^{2 π} V_{0}^{2} sin^{2} t d t$ . Forecast: $sin^{2}$ is both-even → power reduction (Cell 3 machinery), then divide by the interval length.

Step 1 — power-reduce. Why? No factor peels; $sin^{2} t = \frac{1 - c o s 2 t}{2}$ turns the square into something integrable. $\int_{0}^{2 π} sin^{2} t d t = \int_{0}^{2 π} \frac{1 - c o s 2 t}{2} d t = \frac{1}{2} [t - \frac{s i n 2 t}{2}]_{0}^{2 π}$ Step 2 — evaluate the bracket. Why does the $sin 2 t$ term vanish? $sin (2 \cdot 2 π) = sin (4 π) = 0$ and $sin 0 = 0$ — a full number of cycles washes it out. $= \frac{1}{2} [(2 π - 0) - (0 - 0)] = π$ Step 3 — form the average. $avg of v^{2} = \frac{1}{2 π} \cdot V_{0}^{2} \cdot π = \frac{V _{0}^{2}}{2}$ Units / meaning: $v^{2}$ has units of volts². The RMS voltage is $avg = V_{0} / 2 \approx 0.707 V_{0}$ — exactly the famous " $230 V$ RMS from a $\approx 325 V$ peak" rule. ✓

Verify: average value $= \frac{V _{0}^{2}}{2}$ ; with $V_{0} = 1$ this is $0.5$ .

Ex 10 — Cell 10: exam twist, $\int tan^{2} x d x$ (neither shortcut fires)

Here $tan^{2} sec^{0}$ : $sec$ power $0$ is even but there's no $sec^{2}$ to save; $tan$ power $2$ is not odd. Both parity shortcuts stall. The escape is the identity alone.

Recall Self-test: name the cell before solving

For each, state the matrix row, then $u$ (or "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, with sec supplied via $tan^{2} = sec^{2} - 1$ ) → $u = sec x$ where a $sec$ appears. $\int cos^{4} x d x$ ::: 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 — the rulebook these examples drill.
u-substitution — every peel-and-rename step is this.
Pythagorean identities — the rebuild tool in Cells 1, 2, 5, 6, 10.
Power-reduction & double-angle formulas — Cells 3 and 9.
Reduction formulas — the systematic way to handle very high powers ( $sin^{10}$ etc.).
Integration by parts — needed for the odd-sec cases like $\int sec^{3} x d x$ .
Trigonometric substitution — the reverse manoeuvre for algebraic roots.