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

This is the case-atlas for the parent topic. The parent showed you the machinery. Here we drive it through every kind of pothole: each sign pattern, definite integrals whose limits force us to think about quadrants, a degenerate limit, a real-world arc, and an exam twist where the pattern is hidden.

Before symbols, one promise: every letter below is earned. $a$ is always a fixed positive number (a length — think "radius"). $x$ is our variable (a "position"). $θ$ (theta) is a Greek letter we use for an angle, measured in radians. A radian is just an angle size; $π$ radians = a straight line = $18 0^{\circ}$ .

The scenario matrix

Every trig-sub problem lands in exactly one of these cells. Our examples below cover all of them.

#	Cell (the scenario)	Which sub	Covered by
A	$a^{2} - x^{2}$ , indefinite, root in denominator	$x = a sin θ$	Ex 1
B	$a^{2} - x^{2}$ , definite integral (limits → quadrant care)	$x = a sin θ$	Ex 2
C	$a^{2} + x^{2}$ , root on top, needs $\int sec θ d θ$	$x = a tan θ$	Ex 3
D	$(a^{2} + x^{2})^{2}$ power denominator, no visible root	$x = a tan θ$	Ex 4
E	$x^{2} - a^{2}$ with $x > 0$ (branch $[0, \frac{π}{2})$ )	$x = a sec θ$	Ex 5
F	Negative $x$ region $x < - a$ (branch $[π, \frac{3 π}{2})$ ) — sign trap	$x = a sec θ$	Ex 6
G	Completing the square first (offset centre, hidden pattern)	$x - h = a sin θ$	Ex 7
H	Word problem / geometry — area of an ellipse cap (real-world)	$x = a sin θ$	Ex 8
I	Degenerate / limiting case: $a \to 0$ , does the formula survive?	—	Ex 9

Example A — root in the denominator (indefinite)

Worked example Ex 1 · Cell A —

$\int \frac{x ^{2}}{16 - x ^{2}} d x$ Forecast: the radical is "number minus $x^{2}$ ", $16 = 4^{2}$ . Guess the sub before reading on.

Step 1. Recognise $a^{2} - x^{2}$ with $a = 4$ . Set $x = 4 sin θ$ , so $d x = 4 cos θ d θ$ , and restrict $θ \in [- \frac{π}{2}, \frac{π}{2}]$ . Why this step? Only $1 - sin^{2} θ = cos^{2} θ$ (a Pythagorean identity) turns a difference of squares into a single square, killing the root.

Step 2. Simplify the root: $16 - 16 sin^{2} θ = 16 cos^{2} θ = 4 cos θ$ . Why no $\pm$ ? On $[- \frac{π}{2}, \frac{π}{2}]$ we have $cos θ \geq 0$ , so $∣4 cos θ ∣ = 4 cos θ$ .

Step 3. Assemble. Since $x^{2} = 16 sin^{2} θ$ : $\int \frac{16 s i n ^{2} θ}{4 c o s θ} \cdot 4 cos θ d θ = 16 \int sin^{2} θ d θ .$ Why this step? The $cos θ$ on top (from $d x$ ) and bottom (from the root) cancel — that cancellation is the entire payoff of the sub.

Step 4. Power-reduce: $sin^{2} θ = \frac{1 - c o s 2 θ}{2}$ (from Power-reduction & double-angle formulas): $16 \cdot \frac{1}{2} (θ - \frac{s i n 2 θ}{2}) + C = 8 θ - 4 sin 2 θ + C = 8 θ - 8 sin θ cos θ + C,$ using $sin 2 θ = 2 sin θ cos θ$ . Why this step? $sin^{2}$ has no elementary antiderivative on sight; the double-angle identity linearises it into things we can integrate term by term.

Step 5. Back-substitute via the reference triangle: $sin θ = \frac{x}{4}$ , adjacent $= 16 - x^{2}$ , so $cos θ = \frac{16 - x ^{2}}{4}$ and $θ = arcsin \frac{x}{4}$ . $8 arcsin \frac{x}{4} - 8 \cdot \frac{x}{4} \cdot \frac{16 - x ^{2}}{4} + C = 8 arcsin \frac{x}{4} - \frac{x 16 - x ^{2}}{2} + C .$

Verify: the VERIFY block differentiates $8 arcsin \frac{x}{4} - \frac{x 16 - x ^{2}}{2}$ and confirms it simplifies to exactly $\frac{x ^{2}}{16 - x ^{2}}$ . ✓

Example B — a DEFINITE integral: limits become angles

Worked example Ex 2 · Cell B —

$\int_{0}^{3} 9 - x^{2} d x$ Forecast: this is the area under a quarter of a circle of radius $3$ . Predict the number before computing.

Step 1. $a = 3$ , $x = 3 sin θ$ , $d x = 3 cos θ d θ$ , root $\to 3 cos θ$ . Why this step? Same $a^{2} - x^{2}$ pattern.

Step 2 — convert the LIMITS, don't back-substitute. For a definite integral it's cleaner to push the bounds into $θ$ :

$x = 0 \Rightarrow 3 sin θ = 0 \Rightarrow θ = 0$ .
$x = 3 \Rightarrow 3 sin θ = 3 \Rightarrow sin θ = 1 \Rightarrow θ = \frac{π}{2}$ .

Why this step? If we change the bounds now, we never have to undo the substitution — the answer pops out as a number.

Step 3. The integrand becomes $from the root 9 - x^{2} 3 cos θ \cdot from d x 3 cos θ d θ$ — two cosines multiply, one from replacing the root and one from $d x$ : $\int_{0}^{π /2} 3 cos θ \cdot 3 cos θ d θ = 9 \int_{0}^{π /2} cos^{2} θ d θ .$ Why do the two cosines appear? The substitution touches the integral in two places at once: the root $9 - x^{2}$ becomes $3 cos θ$ , and the differential $d x$ becomes $3 cos θ d θ$ . Their product is $9 cos^{2} θ$ — that is why a $cos^{2}$ (not a bare $cos$ ) shows up.

Step 4. $cos^{2} θ = \frac{1 + c o s 2 θ}{2}$ : $9 [\frac{θ}{2} + \frac{s i n 2 θ}{4}]_{0}^{π /2} = 9 (\frac{π}{4} + 0 - 0) = \frac{9 π}{4} .$ Why the $sin 2 θ$ term vanishes: $sin (π) = 0$ and $sin 0 = 0$ — both endpoints kill it.

Verify: the VERIFY block re-integrates $\int_{0}^{3} 9 - x^{2} d x$ symbolically and gets $\frac{9 π}{4}$ ; this also matches the geometry, a quarter-disc of radius $3$ : $\frac{1}{4} π (3)^{2} = \frac{9 π}{4}$ . ✓

The figure below shows exactly what number we computed — the shaded quarter-disc.

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

Example C — root on top, the $sec$ integral appears

Worked example Ex 3 · Cell C —

$\int \frac{x ^{2} + 25}{x} d x$ (take $x > 0$ ) Forecast: "number plus $x^{2}$ ", $25 = 5^{2}$ . Guess the sub.

Step 1. $a^{2} + x^{2}$ with $a = 5$ : set $x = 5 tan θ$ , $d x = 5 sec^{2} θ d θ$ , $θ \in (- \frac{π}{2}, \frac{π}{2})$ . Why this step? $1 + tan^{2} θ = sec^{2} θ$ turns a sum of squares into one square.

Step 2. Root: $25 + 25 tan^{2} θ = 5 sec θ$ (with $sec θ > 0$ on that range). Substitute: $\int \frac{5 s e c θ}{5 t a n θ} \cdot 5 sec^{2} θ d θ = 5 \int \frac{s e c ^{3} θ}{t a n θ} d θ .$

Step 3. Convert to sin/cos to expose structure: $\frac{sec ^{3} θ}{tan θ} = \frac{1/ cos ^{3} θ}{sin θ / cos θ} = \frac{1}{cos ^{2} θ sin θ}$ . Why this step? Everything in sin/cos lets us hunt for a clean substitution or a known integral.

Step 4. Split $\frac{1}{cos ^{2} θ sin θ} = \frac{sin ^{2} θ + cos ^{2} θ}{cos ^{2} θ sin θ} = \frac{sin θ}{cos ^{2} θ} + \frac{1}{sin θ}$ (used $sin^{2} + cos^{2} = 1$ ). Why this step? The first piece is a perfect $u = cos θ$ sub; the second is the standard $\int csc θ d θ$ . $5 \int \frac{s i n θ}{c o s ^{2} θ} d θ + 5 \int csc θ d θ = 5 \cdot \frac{1}{c o s θ} + 5 ln ∣ csc θ - cot θ ∣ + C = 5 sec θ + 5 ln ∣ csc θ - cot θ ∣ + C .$

Step 5. Reference triangle: $tan θ = \frac{x}{5}$ → opposite $x$ , adjacent $5$ , hyp $x^{2} + 25$ . So $sec θ = \frac{x ^{2} + 25}{5}$ , $csc θ = \frac{x ^{2} + 25}{x}$ , $cot θ = \frac{5}{x}$ . $= x^{2} + 25 + 5 ln \frac{x ^{2} + 25 - 5}{x} + C .$

Step 6 — the $x < 0$ branch (why we said "take $x > 0$ "). For $x < 0$ the substitution $x = 5 tan θ$ needs a $θ$ with $tan θ < 0$ , i.e. $θ \in (- \frac{π}{2}, 0)$ , where $sec θ$ is still $> 0$ — so the root is still $5 sec θ$ (roots are never negative), but now $x = 5 tan θ < 0$ while $x^{2} + 25 = 5 sec θ > 0$ . Re-reading the triangle, $csc θ = \frac{x ^{2} + 25}{x}$ becomes negative, and the absolute-value bars in $5 ln ∣ csc θ - cot θ ∣$ absorb exactly that sign flip. Why this matters: the antiderivative $x^{2} + 25 + 5 ln \frac{x ^{2} + 25 - 5}{x} + C$ is written with $∣ \cdot ∣$ precisely so it stays valid for both signs of $x$ — the $∣ \cdot ∣$ is not decoration, it is what stitches the $x > 0$ and $x < 0$ branches into one formula.

Verify: the VERIFY block differentiates this answer (for $x > 0$ ) and confirms it collapses to $\frac{x ^{2} + 25}{x}$ . ✓

Example D — a power denominator, no visible root

Worked example Ex 4 · Cell D —

$\int \frac{d x}{( x ^{2} + 4 ) ^{2}}$ Forecast: there's no ! But $(x^{2} + 4)^{2} = (x^{2} + 4)^{4}$ . Guess the sub anyway.

Step 1. $x = 2 tan θ$ , $d x = 2 sec^{2} θ d θ$ . Then $x^{2} + 4 = 4 sec^{2} θ$ , so $(x^{2} + 4)^{2} = 16 sec^{4} θ$ . Why this step? The sign structure $a^{2} + x^{2}$ dictates $tan$ whether or not a root is written; $tan$ makes $x^{2} + 4$ collapse to a single $sec^{2}$ .

Step 2. $\int \frac{2 s e c ^{2} θ}{16 s e c ^{4} θ} d θ = \frac{1}{8} \int \frac{d θ}{s e c ^{2} θ} = \frac{1}{8} \int cos^{2} θ d θ .$ Why this step? $1/ sec^{2} θ = cos^{2} θ$ — the powers of $sec$ cancel to a friendly $cos^{2}$ .

Step 3. Power-reduce: $\frac{1}{8} \cdot \frac{1}{2} (θ + \frac{s i n 2 θ}{2}) + C = \frac{θ}{16} + \frac{s i n θ c o s θ}{16} + C .$

Step 4. Triangle: $tan θ = \frac{x}{2}$ , hyp $x^{2} + 4$ , so $sin θ = \frac{x}{x ^{2} + 4}$ , $cos θ = \frac{2}{x ^{2} + 4}$ , $θ = arctan \frac{x}{2}$ . $= \frac{1}{16} arctan \frac{x}{2} + \frac{1}{16} \cdot \frac{2 x}{x ^{2} + 4} + C = \frac{1}{16} arctan \frac{x}{2} + \frac{x}{8 ( x ^{2} + 4 )} + C .$

Verify: the VERIFY block differentiates this answer and confirms it equals $\frac{1}{( x ^{2} + 4 ) ^{2}}$ . ✓

Example E — $x^{2} - a^{2}$ with $x > 0$

The reference triangle we used in Step 4 is drawn below.

Example F — the NEGATIVE branch, where signs bite

Worked example Ex 6 · Cell F —

$\int_{- 4}^{- \frac{4}{3}} \frac{x ^{2} - 4}{x} d x$ (all $x < - 2$ ) Forecast: same shape as Ex 3 in the parent, but now $x$ is negative. Will $x^{2} - 4 = 2 tan θ$ still be true? Predict.

Step 1. $a = 2$ , $x = 2 sec θ$ . For $x < - 2$ , $sec θ < - 1$ , so we must use the branch $θ \in [π, \frac{3 π}{2})$ . There $tan θ \geq 0$ , so $x^{2} - 4 = 4 tan^{2} θ = 2 tan θ$ (still $+$ — the branch was chosen precisely so this holds). Why this step? $u^{2} = ∣ u ∣$ . Choosing the second branch guarantees $tan θ \geq 0$ , so the absolute value drops cleanly — this is the whole reason for the $[π, \frac{3 π}{2})$ restriction.

Step 2 — convert the limits, staying on the branch $[π, \frac{3 π}{2})$ . On this branch each angle is the principal $arccos$ value plus $π$ (so it lands in quadrant III):

$x = - 4 \Rightarrow sec θ = \frac{x}{2} = - 2 \Rightarrow cos θ = - \frac{1}{2}$ . The quadrant-III angle with this cosine is $θ_{1} = π + \frac{π}{3} = \frac{4 π}{3}$ .
$x = - \frac{4}{3} \Rightarrow sec θ = \frac{x}{2} = - \frac{2}{3} \Rightarrow cos θ = - \frac{3}{2}$ . The quadrant-III angle with this cosine is $θ_{2} = π + \frac{π}{6} = \frac{7 π}{6}$ .

Why add $π$ ? The principal $arccos$ lands an angle in quadrant II (between $\frac{π}{2}$ and $π$ ), where $tan < 0$ and our root identity would fail. Reflecting through by $+ π$ moves it to quadrant III on our chosen branch, where $tan \geq 0$ and $x^{2} - 4 = 2 tan θ$ holds. The negative $x$ is what forces these quadrant-III angles.

Step 3. With $d x = 2 sec θ tan θ d θ$ and root $2 tan θ$ : $\int \frac{2 t a n θ}{2 s e c θ} \cdot 2 sec θ tan θ d θ = \int 2 tan^{2} θ d θ = 2 \int (sec^{2} θ - 1) d θ = 2 (tan θ - θ) + C .$ Why $sec^{2} - 1$ : $tan^{2}$ isn't directly integrable; $tan^{2} = sec^{2} - 1$ is.

Step 4 — evaluate between the branch angles. The endpoint $x = - 4$ is the lower $x$ -limit and maps to $θ_{1} = \frac{4 π}{3}$ ; the endpoint $x = - \frac{4}{3}$ is the upper $x$ -limit and maps to $θ_{2} = \frac{7 π}{6}$ . So $\int_{- 4}^{- 4/ 3} \frac{x ^{2} - 4}{x} d x = [2 (tan θ - θ)]_{θ = 4 π /3}^{θ = 7 π /6} .$ On this branch, $tan \frac{4 π}{3} = 3$ and $tan \frac{7 π}{6} = \frac{1}{3}$ . Therefore

=2\left(\tfrac{1}{\sqrt3}-\sqrt3+\tfrac{\pi}{6}\right).$$ Numerically $\tfrac{1}{\sqrt3}-\sqrt3=-\tfrac{2}{\sqrt3}\approx-1.1547$, so the answer is $2(-1.1547+0.5236)\approx -1.262$. **Step 5 — back-substitution sanity (express in $x$).** Since $\tan\theta=\tfrac{\sqrt{x^2-4}}{|x|}$ and here $x<0$ so $|x|=-x$, the antiderivative $2(\tan\theta-\theta)$ reads $-\dfrac{2\sqrt{x^2-4}}{x}-2\,\theta(x)$ in $x$-language; evaluating this $x$-form at $x=-4$ and $x=-\tfrac4{\sqrt3}$ reproduces the same $2\left(\tfrac1{\sqrt3}-\sqrt3+\tfrac\pi6\right)$. The integrand $\frac{\sqrt{x^2-4}}{x}$ is **negative** on $x<0$ (numerator $\ge0$, denominator $<0$), and we integrate left-to-right over positive width, so a **negative** answer is exactly expected. ✓ **Verify:** the VERIFY block re-computes $\int_{-4}^{-4/\sqrt3}\frac{\sqrt{x^2-4}}{x}\,dx$ symbolically; it equals $2\left(\tfrac{1}{\sqrt3}-\sqrt3+\tfrac{\pi}{6}\right)\approx -1.262$, and the sign is confirmed negative. ✓ > [!mistake] The branch trap > For $x=-4$ it is tempting to write $\theta=\tfrac{2\pi}{3}$ (also has $\cos=-\tfrac12$). But $\tfrac{2\pi}{3}$ lies in quadrant II where $\tan\theta<0$, so $\sqrt{x^2-4}=2\tan\theta$ would be **false**. For $x<-a$ you must land in $[\pi,\tfrac{3\pi}{2})$ — hence $\tfrac{4\pi}{3}$, not $\tfrac{2\pi}{3}$.

The figure below shows why quadrant III is the correct branch: it is the only place where the substitution's tangent stays non-negative so the root identity survives.

Example G — complete the square first (offset centre)

Worked example Ex 7 · Cell G —

$\int \frac{d x}{5 + 4 x - x ^{2}}$ Forecast: there's a linear term $4 x$ , so no bare $a^{2} - x^{2}$ . What must we do before substituting?

Step 1. Complete the square inside the root: $5 + 4 x - x^{2} = - (x^{2} - 4 x) + 5 = - ((x - 2)^{2} - 4) + 5 = 9 - (x - 2)^{2} .$ Why this step? Trig sub only recognises the three canonical shapes. Completing the square converts the shifted quadratic into $a^{2} - u^{2}$ with $u = x - 2$ , $a = 3$ .

Step 2. Let $u = x - 2$ (so $d u = d x$ ). Now $\int \frac{d u}{9 - u ^{2}}$ , a standard $a^{2} - u^{2}$ with $a = 3$ . Why this step? This is now identically the parent's memorised form.

Step 3. $u = 3 sin θ \Rightarrow \int \frac{3 c o s θ}{3 c o s θ} d θ = θ = arcsin \frac{u}{3}$ .

Step 4. Restore $u = x - 2$ : $\int \frac{d x}{5 + 4 x - x ^{2}} = arcsin \frac{x - 2}{3} + C .$

Verify: the VERIFY block differentiates $arcsin \frac{x - 2}{3}$ and confirms $\frac{d}{d x} = \frac{1}{9 - ( x - 2 ) ^{2}} = \frac{1}{5 + 4 x - x ^{2}}$ . ✓

Example H — a real-world geometry problem

Worked example Ex 8 · Cell H — Area of an elliptical window cap

A stained-glass window is the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ (units: metres). A horizontal glazing bar runs at $y = 1.5$ . Find the area of the cap above the bar (the region $y \geq 1.5$ inside the ellipse).

Forecast: solving for $x$ will produce $number - y^{2}$ . Guess the sub in $y$ .

Step 1 — set up. From the ellipse, $x = \pm 4 1 - \frac{y ^{2}}{9} = \pm \frac{4}{3} 9 - y^{2}$ . At height $y$ the cap stretches from $x = - \frac{4}{3} 9 - y^{2}$ to $x = + \frac{4}{3} 9 - y^{2}$ , a width of $2 \cdot \frac{4}{3} 9 - y^{2}$ . So $A = \int_{1.5}^{3} \frac{8}{3} 9 - y^{2} d y .$ Why this step? Area = sum of horizontal strip widths from the bar $y = 1.5$ up to the ellipse top $y = 3$ .

Step 2. Recognise $9 - y^{2}$ , $a = 3$ : $y = 3 sin θ$ , $d y = 3 cos θ d θ$ , root $\to 3 cos θ$ . Limits: $y = 1.5 \Rightarrow sin θ = \frac{1}{2} \Rightarrow θ = \frac{π}{6}$ ; $y = 3 \Rightarrow sin θ = 1 \Rightarrow θ = \frac{π}{2}$ . Why this step? $9 - y^{2}$ is the $a^{2} - y^{2}$ pattern; converting the limits avoids back-substitution.

Step 3. Root $3 cos θ$ times $d y = 3 cos θ d θ$ gives two cosines → a $cos^{2}$ : $A = \frac{8}{3} \int_{π /6}^{π /2} 3 cos θ \cdot 3 cos θ d θ = 24 \int_{π /6}^{π /2} cos^{2} θ d θ .$

Step 4. $cos^{2} θ = \frac{1 + c o s 2 θ}{2}$ , and $\int cos^{2} θ d θ = \frac{θ}{2} + \frac{s i n 2 θ}{4}$ :

=24\left[\left(\frac\pi4+\frac{\sin\pi}{4}\right)-\left(\frac{\pi}{12}+\frac{\sin(\pi/3)}{4}\right)\right].$$ With $\sin\pi=0$ and $\sin\tfrac\pi3=\tfrac{\sqrt3}{2}$: $$A=24\left(\frac\pi4-\frac\pi{12}-\frac{\sqrt3}{8}\right)=24\cdot\frac\pi6-24\cdot\frac{\sqrt3}{8}=4\pi-3\sqrt3\approx 7.371\ \text{m}^2.$$ *Why this step?* Power-reduction is the only way to integrate $\cos^2$; the endpoints then plug straight in. **Verify (units + magnitude):** the whole ellipse area is $\pi(4)(3)=12\pi\approx 37.7\text{ m}^2$; our cap is a top slice, so a few m² is sensible. Units: $\text{m}\times\text{m}=\text{m}^2$. The VERIFY block re-integrates and confirms the exact value $4\pi-3\sqrt3$ and the numeric $7.371$. ✓

Example I — the degenerate limit $a \to 0$

Worked example Ex 9 · Cell I — does

$\int_{0}^{a} a^{2} - x^{2} d x$ behave sensibly as $a \to 0$ ? Forecast: geometrically this integral is a quarter-disc of radius $a$ . As $a$ shrinks to $0$ , the disc shrinks to a point — the area should go to $0$ . Predict how fast.

Step 1 — do the integral generally. Using $x = a sin θ$ exactly as in Ex 2 (limits $0 \to \frac{π}{2}$ , root $\to a cos θ$ , $d x \to a cos θ d θ$ ): $\int_{0}^{a} a^{2} - x^{2} d x = a^{2} \int_{0}^{π /2} cos^{2} θ d θ = a^{2} \cdot \frac{π}{4} = \frac{π a ^{2}}{4} .$ Why this step? Same $a^{2} - x^{2}$ machinery; the $a^{2}$ factors out front, isolating the $a$ -dependence.

Step 2 — the degenerate check $a \to 0^{+}$ . $\frac{π a ^{2}}{4} \to 0$ , and it does so quadratically (like $a^{2}$ ) — halving $a$ quarters the area, exactly as a shrinking disc should. No blow-up, no undefined $arcsin$ : its argument $x / a$ stays in $[- 1, 1]$ throughout the interval. Why this step? Degenerate inputs are where formulas secretly divide by zero. Here the answer is a clean polynomial in $a$ , so it survives the limit.

Step 3 — the other degeneracy: $x \to a$ from below. In the indefinite formula $\frac{a ^{2}}{2} arcsin \frac{x}{a} + \frac{x a ^{2} - x ^{2}}{2}$ , as $x \to a$ the term $\frac{x a ^{2} - x ^{2}}{2} \to 0$ and $arcsin \frac{x}{a} \to arcsin 1 = \frac{π}{2}$ — both finite. So even though the integrand's slope $\frac{d}{d x} a^{2} - x^{2}$ blows up (a vertical tangent) at $x = a$ , the area is finite. Why this matters: a vertical tangent of the integrand does not make the area infinite — only a vertical asymptote (integrand $\to \infty$ ) would. Here the integrand itself stays bounded ( $0 \leq a^{2} - x^{2} \leq a$ ), so we're safe.

Verify: the VERIFY block confirms $\int_{0}^{a} a^{2} - x^{2} d x = \frac{π a ^{2}}{4}$ symbolically, that it $\to 0$ as $a \to 0$ , and that at $a = 2$ it equals $π$ (a quarter of $π \cdot 2^{2}$ ). ✓

Recap — which cell, which move

Recall One-line reflex per scenario

Difference $a^{2} - x^{2}$ ::: $x = a sin θ$ ; limits → $arcsin$ , quarter-disc geometry. Sum $a^{2} + x^{2}$ (root on top) ::: $x = a tan θ$ ; watch for a $sec^{3} / csc$ integral. Power $(a^{2} + x^{2})^{n}$ no root ::: still $x = a tan θ$ ; powers of $sec$ cancel to $cos$ powers. Difference $x^{2} - a^{2}$ , $x > a$ ::: $x = a sec θ$ , branch $[0, \frac{π}{2})$ . Difference $x^{2} - a^{2}$ , $x < - a$ ::: $x = a sec θ$ , branch $[π, \frac{3 π}{2})$ — sign trap! Linear term present ::: complete the square first, then sub $u = x - h$ . Definite integral ::: convert the LIMITS to $θ$ ; skip back-substitution. Degenerate $a \to 0$ or endpoint ::: check the formula stays finite; vertical tangent ≠ infinite area.

Connections

Parent topic (Hinglish) — the core machinery.
Pythagorean identities — kills each root.
Integration by substitution (u-sub) — the inner $u$ -subs (Ex 3, Ex 7).
Power-reduction & double-angle formulas — every $cos^{2} / sin^{2}$ step.
Reference right triangle method — all back-substitutions.
Partial fractions — the alternative when there is no root.
Arc length and surface area — where these $a^{2} \pm x^{2}$ integrals arise naturally.
Hyperbolic substitution — a parallel tool for $x^{2} \pm a^{2}$ .