Trigonometric Substitution

$4 - 9x^2 = 2^2 - (3x)^2$: $u = 3x$, $a = 2$.

$a^2 - u^2$ means sine: $3x = 2\sin\theta$.

$x = \tfrac23\sin\theta$, and $4 - 9x^2 = 4 - 4\sin^2\theta = 4\cos^2\theta$.

$x^2 + 7 = x^2 + (\sqrt7)^2$: $u = x$, $a = \sqrt7$.

$a^2 + u^2$ means tangent: $x = \sqrt7\tan\theta$.

$x^2 + 7 = 7\tan^2\theta + 7 = 7\sec^2\theta$, so $(x^2 + 7)^{3/2} = 7\sqrt7\sec^3\theta$.

$4x^2 - 49 = (2x)^2 - 7^2$: $u = 2x$, $a = 7$.

$u^2 - a^2$ means secant: $2x = 7\sec\theta$.

$x = \tfrac72\sec\theta$, and $4x^2 - 49 = 49\sec^2\theta - 49 = 49\tan^2\theta$.

$x = 3\sin\theta$, $dx = 3\cos\theta\,d\theta$, $\sqrt{9 - x^2} = 3\cos\theta$: $\int\dfrac{9\sin^2\theta}{3\cos\theta}\cdot 3\cos\theta\,d\theta = \int 9\sin^2\theta\,d\theta$.

$9\int\tfrac12(1 - \cos 2\theta)\,d\theta = \tfrac92\theta - \tfrac94\sin 2\theta = \tfrac92\theta - \tfrac92\sin\theta\cos\theta$.

$\sin\theta = \tfrac{x}{3}$: opposite $x$, hypotenuse $3$, adjacent $\sqrt{9 - x^2}$, so $\cos\theta = \tfrac{\sqrt{9 - x^2}}{3}$.

$\tfrac92\arcsin\tfrac{x}{3} - \tfrac92\cdot\tfrac{x}{3}\cdot\tfrac{\sqrt{9 - x^2}}{3}$.

$x = \tan\theta$, $dx = \sec^2\theta\,d\theta$, $(1 + x^2)^{5/2} = \sec^5\theta$: $\int\dfrac{\sec^2\theta}{\sec^5\theta}\,d\theta = \int\cos^3\theta\,d\theta$.

Odd power: $\int(1 - \sin^2\theta)\cos\theta\,d\theta = \sin\theta - \tfrac13\sin^3\theta$.

Limits instead of a triangle: $x = 0$ gives $\theta = 0$; $x = 1$ gives $\theta = \tfrac{\pi}{4}$.

$\dfrac{\sqrt2}{2} - \dfrac13\cdot\dfrac{\sqrt2}{4} = \dfrac{\sqrt2}{2} - \dfrac{\sqrt2}{12}$.

$x = \sec\theta$, $dx = \sec\theta\tan\theta\,d\theta$, $(x^2 - 1)^{3/2} = \tan^3\theta$: $\int\dfrac{\sec\theta\tan\theta}{\tan^3\theta}\,d\theta = \int\dfrac{\sec\theta}{\tan^2\theta}\,d\theta$.

$\dfrac{\sec\theta}{\tan^2\theta} = \dfrac{\cos\theta}{\sin^2\theta}$; with $w = \sin\theta$, $\int w^{-2}\,dw = -\dfrac{1}{\sin\theta}$.

$\sec\theta = x$: hypotenuse $x$, adjacent $1$, opposite $\sqrt{x^2 - 1}$, so $\sin\theta = \dfrac{\sqrt{x^2 - 1}}{x}$.

$-\dfrac{1}{\sin\theta} = -\dfrac{x}{\sqrt{x^2 - 1}}$.

$8 + 2x - x^2 = 9 - (x^2 - 2x + 1) = 9 - (x - 1)^2$.

$u = x - 1$, $a = 3$, form $a^2 - u^2$: $x - 1 = 3\sin\theta$.

$dx = 3\cos\theta\,d\theta$ and the root is $3\cos\theta$: $\int d\theta = \theta$.

$\theta = \arcsin\dfrac{x - 1}{3}$.

$x^2 + 4x + 13 = (x + 2)^2 + 9$.

$u = x + 2$, $a = 3$, form $a^2 + u^2$: $x + 2 = 3\tan\theta$.

$dx = 3\sec^2\theta\,d\theta$ and $(9\sec^2\theta)^{3/2} = 27\sec^3\theta$: $\int\dfrac{3\sec^2\theta}{27\sec^3\theta}\,d\theta = \tfrac19\int\cos\theta\,d\theta = \tfrac19\sin\theta$.

$\tan\theta = \tfrac{x + 2}{3}$, hypotenuse $\sqrt{x^2 + 4x + 13}$, so $\sin\theta = \dfrac{x + 2}{\sqrt{x^2 + 4x + 13}}$.

$x^2 - 6x + 5 = (x - 3)^2 - 4$.

$u = x - 3$, $a = 2$, form $u^2 - a^2$: $x - 3 = 2\sec\theta$.

$dx = 2\sec\theta\tan\theta\,d\theta$ and the root is $2\tan\theta$: $\int\sec\theta\,d\theta = \log|\sec\theta + \tan\theta|$.

$\sec\theta = \tfrac{x - 3}{2}$, $\tan\theta = \tfrac{\sqrt{x^2 - 6x + 5}}{2}$: $\log\left|\dfrac{x - 3 + \sqrt{x^2 - 6x + 5}}{2}\right|$. The $-\log 2$ joins $C$.

$y = \pm\tfrac34\sqrt{16 - x^2}$; by symmetry $A = 4\int_0^4\tfrac34\sqrt{16 - x^2}\,dx = 3\int_0^4\sqrt{16 - x^2}\,dx$.

$x = 4\sin\theta$, $dx = 4\cos\theta\,d\theta$, root $= 4\cos\theta$; $x: 0 \to 4$ is $\theta: 0 \to \tfrac{\pi}{2}$: $A = 3\int_0^{\pi/2}16\cos^2\theta\,d\theta$.

$48\int_0^{\pi/2}\tfrac12(1 + \cos 2\theta)\,d\theta = 24\Big[\theta + \tfrac12\sin 2\theta\Big]_0^{\pi/2}$.

$24\cdot\tfrac{\pi}{2} = 12\pi$.

The height is $2\sqrt{4 - x^2}$: $A = \int_1^2 2\sqrt{4 - x^2}\,dx$.

$x = 2\sin\theta$; $x = 1$ gives $\theta = \tfrac{\pi}{6}$, $x = 2$ gives $\theta = \tfrac{\pi}{2}$: $A = \int_{\pi/6}^{\pi/2}8\cos^2\theta\,d\theta$.

$4\int_{\pi/6}^{\pi/2}(1 + \cos 2\theta)\,d\theta = 4\Big[\theta + \tfrac12\sin 2\theta\Big]_{\pi/6}^{\pi/2}$.

$4\left(\tfrac{\pi}{2} - \tfrac{\pi}{6} - \tfrac{\sqrt3}{4}\right) = \tfrac{4\pi}{3} - \sqrt3$.