Trigonometric substitution — x = a sin θ, a tan θ, a sec θ cases: Exercises

Figure s01 — the three reference triangles. Read each triangle from the angle $θ$ at the origin: the side across from $θ$ is the opposite (labelled "opp"), the side along the base touching $θ$ is the adjacent (labelled "adj"), and the slanted side is the hypotenuse (labelled "hyp"). For $x = a sin θ$ : opp $= x$ , hyp $= a$ , adj $= a^{2} - x^{2}$ . For $x = a tan θ$ : opp $= x$ , adj $= a$ , hyp $= a^{2} + x^{2}$ . For $x = a sec θ$ : hyp $= x$ , adj $= a$ , opp $= x^{2} - a^{2}$ . Every back-substitution just reads these sides off.

Definition Standard

$θ$ -ranges (these fix every sign — and WHY that exact branch) To write a root as a plus value (not $\pm$ , not $∣ \cdot ∣$ ), we restrict $θ$ to an interval where the relevant trig function is $\geq 0$ :

Sine case $x = a sin θ$ : as $x$ sweeps $[- a, a]$ , $sin θ = \frac{x}{a}$ sweeps $[- 1, 1]$ , and $θ \in [- \frac{π}{2}, \frac{π}{2}]$ is the branch on which sine takes each value once. There $cos θ \geq 0 \Rightarrow a^{2} - x^{2} = a cos θ$ .
Tangent case $x = a tan θ$ : as $x$ sweeps all reals, $tan θ = \frac{x}{a}$ needs $θ \in (- \frac{π}{2}, \frac{π}{2})$ (the branch where tangent is one-to-one and covers every real). There $sec θ = \frac{1}{c o s θ} > 0 \Rightarrow a^{2} + x^{2} = a sec θ$ .
Secant case $x = a sec θ$ : here $∣ x ∣ \geq a$ . For $x \geq a > 0$ we need $sec θ = \frac{x}{a} \geq 1$ , i.e. $cos θ \in (0, 1]$ , which is exactly $θ \in [0, \frac{π}{2})$ — and there $tan θ \geq 0$ . For $x \leq - a$ we need $sec θ \leq - 1$ , i.e. $cos θ \in [- 1, 0)$ ; the branch $θ \in [π, \frac{3 π}{2})$ is chosen precisely because there both $cos θ < 0$ (so $sec θ \leq - 1$ matches $x \leq - a$ ) and $tan θ \geq 0$ (so $x^{2} - a^{2} = a tan θ$ still comes out $+$ ). That double requirement — right sign of $sec$ AND non-negative $tan$ — is the whole reason $[π, \frac{3 π}{2})$ is the partner of $[0, \frac{π}{2})$ .

Every problem below has $x \geq a > 0$ , so we always sit in $[0, \frac{π}{2})$ for the secant case.

Level 1 — Recognition

Recall Solution L1·1

WHAT we look at: the sign structure inside the radical — this alone dictates the choice. (a) $16 - x^{2}$ is $a^{2} - x^{2}$ with $a = 4$ (since $4^{2} = 16$ ). → $x = 4 sin θ$ , $θ \in [- \frac{π}{2}, \frac{π}{2}]$ , $d x = 4 cos θ d θ$ , root $\to 4 cos θ$ (valid since $cos θ \geq 0$ on this interval). WHY sine: $1 - sin^{2} θ = cos^{2} θ$ kills a difference starting with $a^{2}$ . (b) $x^{2} + 5$ is $a^{2} + x^{2}$ with $a = 5$ . → $x = 5 tan θ$ , $θ \in (- \frac{π}{2}, \frac{π}{2})$ , $d x = 5 sec^{2} θ d θ$ , root $\to 5 sec θ$ (valid since $sec θ > 0$ on this interval). WHY tangent: $1 + tan^{2} θ = sec^{2} θ$ kills a sum. (c) $x^{2} - 9$ is $x^{2} - a^{2}$ with $a = 3$ . → $x = 3 sec θ$ , $θ \in [0, \frac{π}{2})$ (taking $x \geq 3$ ), $d x = 3 sec θ tan θ d θ$ , root $\to 3 tan θ$ (valid since $tan θ \geq 0$ on this interval). WHY secant: $sec^{2} θ - 1 = tan^{2} θ$ kills a difference starting with $x^{2}$ .

Recall Solution L1·2

Step 1 — substitute. $x = 4 sin θ$ , $θ \in [- \frac{π}{2}, \frac{π}{2}]$ , $d x = 4 cos θ d θ$ . See the sine triangle in Figure s02: opp $= x$ , hyp $= 4$ . Step 2 — simplify root. $16 - x^{2} = 16 - 16 sin^{2} θ = 16 cos^{2} θ = 4 cos θ$ . WHY $+ 4 cos θ$ , no $∣ \cdot ∣$ survives: in general $u^{2} = ∣ u ∣$ , so strictly $16 cos^{2} θ = 4∣ cos θ ∣$ . But the chosen range $θ \in [- \frac{π}{2}, \frac{π}{2}]$ forces $cos θ \geq 0$ , so $∣ cos θ ∣ = cos θ$ and the bars drop. This is why the range restriction exists at all. Step 3 — assemble. The $4 cos θ$ from $d x$ cancels the $4 cos θ$ from the root: $\int \frac{4 c o s θ d θ}{4 c o s θ} = \int 1 d θ = θ + C .$ WHY $\int 1 d θ = θ$ : the function whose derivative is the constant $1$ is $θ$ itself ( $\frac{d}{d θ} θ = 1$ ). Reading the derivative backwards gives the integral. Step 4 — back-substitute. From $sin θ = \frac{x}{4}$ (Figure s02), $θ = arcsin \frac{x}{4}$ — "the angle whose sine is $\frac{x}{4}$ ." $\int \frac{d x}{16 - x ^{2}} = arcsin \frac{x}{4} + C, x \in (- 4, 4)$ Edge/domain note: the integrand needs $16 - x^{2} > 0$ , i.e. $x \in (- 4, 4)$ ; at $x = \pm 4$ the root is $0$ and the integrand blows up, so the answer is only valid strictly inside. And $arcsin \frac{x}{4}$ is itself only defined for $\frac{x}{4} \in [- 1, 1]$ — the same interval, which is a good consistency check.

Level 2 — Application

Recall Solution L2·1

Step 1. $a = 3$ , $θ \in [- \frac{π}{2}, \frac{π}{2}]$ : $x = 3 sin θ$ , $d x = 3 cos θ d θ$ , $9 - x^{2} = 3 cos θ$ (bars drop because $cos θ \geq 0$ on this range), $x^{2} = 9 sin^{2} θ$ . Same sine triangle, Figure s02 with $a = 3$ . Step 2 — substitute all pieces. Recall $csc θ = \frac{1}{s i n θ}$ (defined above), so $\frac{1}{s i n ^{2} θ} = csc^{2} θ$ : $\int \frac{3 c o s θ d θ}{9 s i n ^{2} θ \cdot 3 c o s θ} = \frac{1}{9} \int \frac{d θ}{s i n ^{2} θ} = \frac{1}{9} \int csc^{2} θ d θ .$ WHY it collapsed: the $cos θ$ from $d x$ cancels the $cos θ$ from the root — leaving a standard integral in $sin$ only. Step 3. $\int csc^{2} θ d θ = - cot θ$ , so $= - \frac{1}{9} cot θ + C$ . WHY $\int csc^{2} θ d θ = - cot θ$ : differentiate $- cot θ$ . Since $cot θ = \frac{c o s θ}{s i n θ}$ , the quotient rule gives $\frac{d}{d θ} cot θ = - csc^{2} θ$ , so $\frac{d}{d θ} (- cot θ) = csc^{2} θ$ . Reading backwards is the integral. Step 4 — triangle (Figure s02). $sin θ = \frac{x}{3}$ : opposite $= x$ , hyp $= 3$ , adjacent $= 9 - x^{2}$ , so $cot θ = \frac{adj}{opp} = \frac{9 - x ^{2}}{x}$ . $= - \frac{9 - x ^{2}}{9 x} + C, 0 < ∣ x ∣ < 3$

Recall Solution L2·2

WHY tangent: $(x^{2} + 4)^{3/2} = (x^{2} + 4)^{3}$ — a sum $a^{2} + x^{2}$ with $a = 2$ . Step 1. $x = 2 tan θ$ , $θ \in (- \frac{π}{2}, \frac{π}{2})$ , $d x = 2 sec^{2} θ d θ$ . On this interval $sec θ > 0$ , so $x^{2} + 4 = 2 sec θ \Rightarrow (x^{2} + 4)^{3/2} = (2 sec θ)^{3} = 8 sec^{3} θ$ . See the tangent triangle, Figure s03. Step 2. $\int \frac{2 s e c ^{2} θ d θ}{8 s e c ^{3} θ} = \frac{1}{4} \int \frac{d θ}{s e c θ} = \frac{1}{4} \int cos θ d θ = \frac{1}{4} sin θ + C .$ WHY $\frac{1}{s e c θ} = cos θ$ : because $sec θ$ is defined as $\frac{1}{c o s θ}$ , its reciprocal is $cos θ$ . WHY $\int cos θ d θ = sin θ$ : $\frac{d}{d θ} sin θ = cos θ$ , so integrating (the reverse of differentiating) returns $sin θ$ . Step 3 — triangle (Figure s03). $tan θ = \frac{x}{2}$ : opposite $= x$ , adjacent $= 2$ , hyp $= x^{2} + 4$ , so $sin θ = \frac{opp}{hyp} = \frac{x}{x ^{2} + 4}$ . $= \frac{x}{4 x ^{2} + 4} + C (all real x)$

Level 3 — Analysis

Recall Solution L3·1

Step 1. $a = 3$ , $θ \in [0, \frac{π}{2})$ (taking $x \geq 3$ ): $x = 3 sec θ$ , $d x = 3 sec θ tan θ d θ$ . On this interval $tan θ \geq 0$ , so $x^{2} - 9 = 9 tan^{2} θ = 3 tan θ$ (the $∣ \cdot ∣$ drops because $tan θ \geq 0$ here — see the range table above). Secant triangle, Figure s04. Step 2 — assemble. $\int 3 tan θ \cdot 3 sec θ tan θ d θ = 9 \int sec θ tan^{2} θ d θ .$ Step 3 — turn $tan^{2}$ into $sec^{2} - 1$ (via Pythagorean identities, so the pieces become integrals we can build): $9 \int sec θ (sec^{2} θ - 1) d θ = 9 \int sec^{3} θ d θ - 9 \int sec θ d θ .$ Step 4 — WHERE the two sub-integrals come from (not "just known"). For $\int sec θ d θ$ : multiply top and bottom by $(sec θ + tan θ)$ : $\int sec θ \cdot \frac{s e c θ + t a n θ}{s e c θ + t a n θ} d θ = \int \frac{s e c ^{2} θ + s e c θ t a n θ}{s e c θ + t a n θ} d θ$ . The numerator is exactly the derivative of the denominator ( $\frac{d}{d θ} (sec θ + tan θ) = sec θ tan θ + sec^{2} θ$ ), so this is $\int \frac{w ^{'}}{w} d θ = ln ∣ w ∣$ , giving $\int sec θ d θ = ln ∣ sec θ + tan θ ∣$ . For $\int sec^{3} θ d θ$ : integration by parts with $u = sec θ, d v = sec^{2} θ d θ$ gives $\int sec^{3} = sec θ tan θ - \int sec θ tan^{2} θ = sec θ tan θ - \int sec^{3} θ + \int sec θ$ . Solving this loop for $\int sec^{3} θ$ : $\int sec^{3} θ d θ = \frac{1}{2} sec θ tan θ + \frac{1}{2} ln ∣ sec θ + tan θ ∣$ . Step 5 — combine. $9 (\frac{1}{2} sec θ tan θ + \frac{1}{2} ln ∣ sec θ + tan θ ∣) - 9 ln ∣ sec θ + tan θ ∣ = \frac{9}{2} sec θ tan θ - \frac{9}{2} ln ∣ sec θ + tan θ ∣ + C .$ Step 6 — triangle (Figure s04). $sec θ = \frac{x}{3}$ , $tan θ = \frac{x ^{2} - 9}{3}$ . $= \frac{x x ^{2} - 9}{2} - \frac{9}{2} ln \frac{x + x ^{2} - 9}{3} + C$ (The $- \frac{9}{2} ln 3$ folds into $C$ ; you may also write $- \frac{9}{2} ln ∣ x + x^{2} - 9 ∣ + C$ .)

Recall Solution L3·2

Step 1. $a = 2$ , $θ \in [- \frac{π}{2}, \frac{π}{2}]$ : $x = 2 sin θ$ , $d x = 2 cos θ d θ$ , $4 - x^{2} = 2 cos θ$ (bars drop since $cos θ \geq 0$ ). Convert limits (WHY: cleaner than back-substituting): $x = 0 \Rightarrow sin θ = 0 \Rightarrow θ = 0$ ; $x = 2 \Rightarrow sin θ = 1 \Rightarrow θ = \frac{π}{2}$ . Step 2. $\int_{0}^{π /2} \frac{4 s i n ^{2} θ}{2 c o s θ} \cdot 2 cos θ d θ = 4 \int_{0}^{π /2} sin^{2} θ d θ .$ Step 3 — power-reduction (see Power-reduction & double-angle formulas): $sin^{2} θ = \frac{1 - c o s 2 θ}{2}$ . WHY this step: $sin^{2} θ$ has no elementary antiderivative "by sight," but the identity turns it into a constant plus a plain cosine, both of which integrate in one line. $4 \cdot \frac{1}{2} \int_{0}^{π /2} (1 - cos 2 θ) d θ = 2 [θ - \frac{1}{2} sin 2 θ]_{0}^{π /2} = 2 (\frac{π}{2} - 0) = π .$ WHY $\int cos 2 θ d θ = \frac{1}{2} sin 2 θ$ : $\frac{d}{d θ} (\frac{1}{2} sin 2 θ) = \frac{1}{2} \cdot 2 cos 2 θ = cos 2 θ$ (chain rule), read backwards. $= π$ Edge note: the upper limit $x = 2 = a$ makes the integrand blow up (root $\to 0$ ), yet the area is finite — the $θ$ -integral quietly handles this improper endpoint because $θ = \frac{π}{2}$ is a perfectly ordinary limit.

Level 4 — Synthesis

Recall Solution L4·1

Step 1 — complete the square (WHY: no bare $a^{2} \pm x^{2}$ form yet). $x^{2} + 6 x + 13 = (x + 3)^{2} + 4$ . Now it is a sum with shifted variable. Step 2 — shift then substitute. Let $u = x + 3$ , $d u = d x$ . Root $= u^{2} + 4$ , $a = 2$ : $u = 2 tan θ$ , $θ \in (- \frac{π}{2}, \frac{π}{2})$ , $d u = 2 sec^{2} θ d θ$ . On this interval $sec θ > 0$ , so $u^{2} + 4 = 2 sec θ$ . (Same tangent triangle, Figure s03, with $x \to u$ .) Step 3. $\int \frac{2 s e c ^{2} θ d θ}{2 s e c θ} = \int sec θ d θ = ln ∣ sec θ + tan θ ∣ + C .$ WHY $\int sec θ d θ = ln ∣ sec θ + tan θ ∣$ : exactly the multiply-by- $(sec θ + tan θ)$ trick derived in L3·1 Step 4 — the numerator becomes the derivative of the denominator. Step 4 — triangle (Figure s03, $x \to u$ ). $tan θ = \frac{u}{2}$ , $sec θ = \frac{u ^{2} + 4}{2}$ . $ln \frac{u ^{2} + 4 + u}{2} + C = ln u^{2} + 4 + u + C (the - ln 2 folds into C) .$ Step 5 — undo $u$ . $= ln x + 3 + x^{2} + 6 x + 13 + C (all real x, since (x + 3)^{2} + 4 > 0)$

Recall Solution L4·2

Step 1. $a = 1$ , $θ \in [- \frac{π}{2}, \frac{π}{2}]$ : $x = sin θ$ , $d x = cos θ d θ$ , $1 - x^{2} = cos θ$ (bars drop since $cos θ \geq 0$ ). Sine triangle, Figure s02 with $a = 1$ . Step 2. $\int \frac{s i n ^{2} θ}{c o s θ} cos θ d θ = \int sin^{2} θ d θ .$ Step 3 — power-reduction. $sin^{2} θ = \frac{1 - c o s 2 θ}{2}$ , so $\int sin^{2} θ d θ = \frac{1}{2} (θ - \frac{1}{2} sin 2 θ) = \frac{1}{2} (θ - sin θ cos θ) + C$ . WHY the last equality: $sin 2 θ = 2 sin θ cos θ$ (double-angle, see Power-reduction & double-angle formulas), so $\frac{1}{2} sin 2 θ = sin θ cos θ$ . Step 4 — triangle / undo (Figure s02). $θ = arcsin x$ , $sin θ = x$ , $cos θ = 1 - x^{2}$ . $= \frac{1}{2} arcsin x - \frac{x 1 - x ^{2}}{2} + C, x \in (- 1, 1)$

Level 5 — Mastery

Recall Solution L5·1

Step 1. $a = 2$ , and since $x$ runs $2 \to 2$ (all $\geq 2 = a$ ) we sit in $θ \in [0, \frac{π}{2})$ : $x = 2 sec θ$ , $d x = 2 sec θ tan θ d θ$ , $x^{2} - 2 = 2 tan θ$ . WHY $+ 2 tan θ$ : on $[0, \frac{π}{2})$ we have $tan θ \geq 0$ , so the $∣ \cdot ∣$ drops and the root stays non-negative (range table, secant case). Secant triangle, Figure s04 with $a = 2$ . Convert limits. $x = 2 \Rightarrow sec θ = 1 \Rightarrow θ = 0$ ; $x = 2 \Rightarrow sec θ = \frac{2}{2} = 2 \Rightarrow cos θ = \frac{1}{2} \Rightarrow θ = \frac{π}{4}$ . Step 2. $\int_{0}^{π /4} \frac{2 t a n θ}{2 s e c θ} \cdot 2 sec θ tan θ d θ = 2 \int_{0}^{π /4} tan^{2} θ d θ .$ Track the constant: $\frac{2 t a n θ}{2 s e c θ} \cdot 2 sec θ tan θ = 2 tan^{2} θ$ . ✓ Step 3. $tan^{2} θ = sec^{2} θ - 1 \Rightarrow \int tan^{2} θ d θ = tan θ - θ$ . WHY $\int sec^{2} θ d θ = tan θ$ : $\frac{d}{d θ} tan θ = sec^{2} θ$ , read backwards; and $\int 1 d θ = θ$ . $2 [tan θ - θ]_{0}^{π /4} = 2 (1 - \frac{π}{4}) .$ $= 2 (1 - \frac{π}{4}) \approx 0.303$

Recall Solution L5·2

Step 1 — set up. $y^{'} = x$ , so $L = \int_{0}^{1} 1 + x^{2} d x$ . A sum $\Rightarrow$ tangent, $a = 1$ . Step 2. $x = tan θ$ , $θ \in (- \frac{π}{2}, \frac{π}{2})$ , $d x = sec^{2} θ d θ$ . On this interval $sec θ > 0$ , so $1 + x^{2} = sec θ$ . Limits: $x = 0 \to θ = 0$ ; $x = 1 \to tan θ = 1 \to θ = \frac{π}{4}$ . $L = \int_{0}^{π /4} sec θ \cdot sec^{2} θ d θ = \int_{0}^{π /4} sec^{3} θ d θ .$ Step 3 — use the $\int sec^{3}$ result (derived by parts in L3·1 Step 4): $\int sec^{3} θ d θ = \frac{1}{2} sec θ tan θ + \frac{1}{2} ln ∣ sec θ + tan θ ∣$ . At $θ = \frac{π}{4}$ : $sec θ = 2$ , $tan θ = 1$ . At $θ = 0$ : both terms $0$ . $L = \frac{1}{2} (2 \cdot 1) + \frac{1}{2} ln ∣ 2 + 1∣ = \frac{2}{2} + \frac{1}{2} ln (2 + 1) .$ $L = \frac{2}{2} + \frac{1}{2} ln (1 + 2) \approx 1.148$

Recall Solution L5·3

Way A — plain $u$ -sub (faster). Let $u = x^{2} + 4$ , $d u = 2 x d x$ (see Integration by substitution (u-sub)). $\int \frac{x d x}{u} = \frac{1}{2} \int u^{- 1/2} d u = u = x^{2} + 4 + C .$ WHY $\frac{1}{2} \int u^{- 1/2} d u = u$ : the power rule $\int u^{n} d u = \frac{u ^{n + 1}}{n + 1}$ with $n = - \frac{1}{2}$ gives $\frac{u ^{1/2}}{1/2} = 2 u$ , times the $\frac{1}{2}$ out front $= u$ . Way B — trig sub (works, but heavier). $x = 2 tan θ$ , $θ \in (- \frac{π}{2}, \frac{π}{2})$ so $sec θ > 0$ : $\int \frac{2 t a n θ}{2 s e c θ} 2 sec^{2} θ d θ = 2 \int tan θ sec θ d θ = 2 sec θ = 2 \cdot \frac{x ^{2} + 4}{2} = x^{2} + 4 + C .$ WHY $\int sec θ tan θ d θ = sec θ$ : $\frac{d}{d θ} sec θ = sec θ tan θ$ , read backwards. **WHY