Part 1: Integration, Arc Length, Average Value, Work

Set up the arc-length integral for $y=x^{3/2}$ on $0 \le x \le 4$

Formula for arc-length integral: $L = \int_a^b \sqrt{1 + (\frac {dy}{dx})²}  dx$

• $a = 0$

• $b=4$

• $\frac {dy}{dx} = y^\prime(x)$

  • $y^\prime(x) = \frac 32 x^{\frac 12}$ 

$L = \int_0^4 \sqrt {1 + (\frac 32 x^{\frac 12})²}  dx$

$L = \int_0^4 \sqrt {1 + \frac 94 x}  dx$

Set up the arc-length integral for $y=e^{2x}$ on $0\le x \le 1$

$a = 0$

$b=1$

$\frac {dy}{dx} = 2e^{2x}$

$L = \int_0^1 \sqrt {1 + (2e^{2x})²}  dx$

$L = \int_0^1 \sqrt {1 + 4e^{4x²}}  dx$

Compute the arc length of the parametric curve

$x(t) = t² - 4 \ln t$

$y(t) = 4\sqrt 2 t$

$t   \epsilon   [1,e]$

$L = \int_a^b \sqrt {(\frac {dx}{dt})² + (\frac {dy}{dt})²}  dt$  

Step 1: Find the derivatives

$x^\prime(t) = 2t - \frac 4t$

$y^\prime(t) = 4\sqrt 2$

Step 2: Square the derivatives

$(\frac {dx}{dt})² = (2t- \frac 4t)² = 4t² + \frac {16}{t²} - 16$

$(\frac {dy}{dt})² = (4\sqrt 2)²$ $= 32$

$L = \int_1^e \sqrt {(2t-\frac 4t)² + (4\sqrt 2)²}  dt$

$L = \int_1^e \sqrt {4t² - 16 + \frac {16}{t²} + 32}  dt$

$L = \int_1^e \sqrt {4t² +  \frac {16}{t²} + 16}  dt$

$L = \int_1^e \sqrt {4(t + \frac 2t)²}  dt$

$L = \int_1^e 2 \cdot (t+ \frac 2t)  dt = 2\int_1^e t + \frac 2t  dt$

$L = [2\frac {t²}2 + 4\ln|t|]_1^e =[t²+4\ln|t|]_1^e$

$L = (e² + 4) - (1)$

$L = e²+3$

Compute the arc length of

$x(t) = 3t²$

$y(t) = 2t³$

$t  \epsilon  [0,1]$

$x^\prime(t) = 6t$

$y^\prime(t) = 6t²$

$(\frac {dx}{dt})² = 36t²$

$(\frac {dy}{dt})² = 36t^4$

$L = \int_0^1  \sqrt {36t² + 36t^4}  dt$

$L = \int_0^1  6t \sqrt{1 + t²}  dt$

$L  = 6\int_0^1  t\sqrt {1 + t²}  dt$

• $u =1 + t²$

• $du = 2t  dt$

• $\frac 12 du = t  dt$

$L = 3 \int_0^1 \sqrt u  du$

$L = 3[\frac 23 \sqrt {u³}]_0^1$

$L = 3[\frac 23 \sqrt {(1+t²)³}]_0^1$

$L = 3(\frac 23 \sqrt {8} - \frac 23)$

$L = 2 \sqrt 8 - 2$

$L = 4\sqrt 2 - 2$

Compute the arc length of:

$x(t) = \cos(2t)$

$y(t) = \sin (2t)$

$t  \epsilon  [0,\frac \pi 4]$

$x^\prime (t) = -2\sin(2t)$

$y^\prime(t) = 2\cos(2t)$

$(\frac {dx}{dt})² = (-2\sin (2t))² \Rightarrow$ $4\sin²(2t)$

$(\frac {dy}{dt})² = (2\cos(2t))² \Rightarrow 4\cos²(2t)$

$L = \int_0^{\frac \pi 4} \sqrt {4\sin²(2t) + 4\cos²(2t)}  dt$

$L = 2\int_0^{\frac \pi 4} \sqrt {\sin²(2t) + \cos²(2t)}  dt$

$L= 2 \int_0^{\frac \pi 4} 1  dt$

$L = 2[x]_0^{\frac \pi 4}$

$L = 2(\frac \pi 4 - 0)$

$L = \frac \pi 2$

Compute the arc length of

$x(t) = 4t²$

$y(t) = 8t³$

$t  \epsilon  [0,1]$

$x^\prime(t) = 8t$

$y^\prime (t) = 24t²$

$(\frac {dx}{dt})² = 64t²$

$(\frac {dy}{dt})² = 576t^4$

$L = \int_0^1 \sqrt {64t² + 576t^4}  dt$

• $64t² + 576t^4 \Rightarrow 64t²(1 +9t²)$

$L = 8\int_0^1 t\sqrt {1 + 9t²}  dt$

• $u = 1 + 9t²$

• $du = 18t  dt$

• $\frac 1{18} du= t  dt$

$L = 8 \cdot \frac 1{18} \int_1^{10} \sqrt{u}  du$

$L = \frac 49 [\frac 23 \sqrt {u³}]_1^{10}$

$L = \frac 49 (\frac 23 \sqrt {10³} - \frac 23)$

Simplified:

$L = \frac 8{27}( 10\sqrt {10} - 1)$

Set up the integral for computing the length of the curve

$x(t) = \cos t$

$y(t) = \sin (2t)$

$t  \epsilon  [0,\pi]$

$x^\prime (t) = -\sin (t)$

$y^\prime (t) = 2\cos (t)$

$(\frac {dx}{dt})² = \sin² (t)$

$(\frac {dy}{dt})² = 4\cos² (t)$

$L = \int_0^\pi \sqrt {\sin²(t) + 4\cos²(t)}  dt$

$L = \int_0^\pi \sqrt {\sin²(t) + \cos²(t) + 3\cos²(t)}  dt$

$L = \int_0^\pi \sqrt {1 + 3\cos²(t)}  dt$

Set up the arc-length integral for:

$x(t) = t²$

$y(t) = \ln t$

$t  \epsilon  [1,2]$

$x^\prime(t) = 2t$

$y^\prime (t) = \frac 1t$

$(\frac {dx}{dt})² = 4t²$

$(\frac {dy}{dt})² = \frac 1{t²}$

$L = \int_1²\sqrt {4t² + \frac 1{t²}}  dt$

For the following problems, find the average value of the function on the given interval

Average value equation: $\frac 1{b-a} \int_a^b f(x)  dx$

a) $f(x) = 3x² - x + 2$ on $[0,2]$

• $\frac 1{2-0} \int_0² (3x²-x+2)  dx$ 

• $\frac 12 [x³-\frac {x²}2+2x]_0²$

• $\frac 12 (8 - 2 + 4)$

• $5$

b) $f(x) = 4\cos x$ on $[0,\pi]$

• $\frac 1{\pi - 0} \int_0^\pi 4\cos x  dx$

• $\frac 4\pi [\sin x]_0^\pi$

• $\frac 4\pi (\sin(\pi) - \sin(0))$

• $\frac 4\pi (0 - 0) = 0$

c) $f(x) = 5x + 1$ on $[-1,3]$

• $\frac {1}{3 - (-1)} \int_{-1}³$$(5x+1)  dx$

• $\frac 14 [\frac 52 x² + x]_{-1}³$

• $\frac 14 ((\frac 52 (9) + 3)-(\frac 52 - 1))$

• $\frac 14 (24)$

• $6$

d) $f(x) = e^x$ on $[0,\ln 4]$

• $\frac 1{\ln 4-0}\int_0^{\ln 4} e^x  dx$

• $\frac 1{\ln 4} [e^x]_0^{\ln 4}$

• $\frac 1{\ln 4} (4 - 1)$

• $\frac {1}{\ln 4} 3$

• $\frac 3{\ln 4}$

e) $f(x) = x² + 4$ on $[-2,2]$

• $\frac 1{2 - (-2)} \int_{-2}² (x² + 4)  dx$ 

• $\frac 14 [\frac 13 x³ + 4x]_{-2}²$

• $\frac 14 ((\frac 83 + 8)-(\frac {-8}3 -8))$

• $\frac 14 (\frac {32}3 + \frac {32}{3})$

• $\frac 14 \frac {64}3$

• $\frac {64}{12} = \frac {16}3$

For $f(x) = 6 - x²$ on $[0,3]$

a) Find the average value

• $\frac {1}{3}\int_0³ (6-x²)  dx$

• $\frac 13 [6x - \frac 13 x ³]_0³$

• $\frac 13 (18 - 9)$

• $\frac 13 \cdot 9$

• $3$

b) Find c such that $f(x)$ equals the average value

• $f(x) = 6 - c² = 3$

• $6-c² = 3$

• $-c² = 3 - 6$

• $c² = 3$

• $c = \sqrt 3$

Find the average value of the function on the given interval

a) $f(x) = \ln x$ on $[1,e]$

• $\frac 1{e-1} \int_1^e \ln x  dx$

• $\frac 1{e-1} [\ln x]_1^e$

• $\frac 1{e-1} (1 - 0)$

• $\frac 1{e-1}1$

• $\frac 1{e-1}$

b) $f(x) = \frac 1{x+2}$ on $[0,2]$

• $\frac 1{2} \int_0² (\frac 1{x+2})  dx$

• $\frac 12 [\ln|x+2|]_0²$

• $\frac 12 (\ln |4| - \ln |0|)$

• $\frac 12 ()</p></li></ul><p></p><p>c)$f(x) = \sin²x$on$[0,\pi]$</p><p></p><ul><li><p>$\frac 1{\pi} \int_0^\pi (\sin² x)  dx$</p></li><li><p>$\frac 1\pi [\frac {2x-\sin(2x)}4]_0^\pi$</p></li><li><p>$\frac 1\pi (\frac {2\pi - \sin (2\pi)}{4}) = \frac 12$</p></li></ul><div data-type="horizontalRule"><hr></div><p>Tank temperature$T(t) = 20 + 3t$,$0\le t\le 4$. Find average temperature.</p><p></p><ul><li><p>$\frac 1{4} \int_0^4 (20 + 3t)  dt$</p></li><li><p>$\frac 14 [20t + \frac 32 t²]_0^4$</p></li><li><p>$\frac 14 (80 + 24)$</p></li><li><p>$26$</p></li></ul><div data-type="horizontalRule"><hr></div><p>A vertical steel cable (40 m long, 2 N/m) is lifted to the top. Compute the work required.</p><p></p><p>$dW =$(weight density)(length element)(distance lifted)$= 2dx \cdot (40-x)$</p><p>$W = \int_0^{40} 2(40 - x)dx$</p><p>$W = \int_0^{40} (80 - 2x) dx$</p><p>$W = [80x - x²]_0^{40}$</p><p>$W = (3200 - 1600) = 1600   Joules$</p><div data-type="horizontalRule"><hr></div><p>Trig Integrals: For the following problems, evaluate the integral</p><p></p><p>a)$\int \cos³ x  dx$</p><p></p><ul><li><p>$\int \cos²(x) \cdot \cos(x)  dx$</p><ul><li><p>$\cos²x + \sin²x = 1$</p></li><li><p>$\cos²x = 1 - \sin²x$</p></li></ul></li><li><p>$\int \cos x (1-\sin²x)  dx$</p></li><li><p>$\int \cos x - \cos x \sin²x x  dx$</p></li><li><p>$\int \cos x   dx - \int \sin²x \cos x  dx$</p><ul><li><p>$u = \sin x$</p></li><li><p>$du = \cos x  dx$</p></li></ul></li><li><p>$\sin x + C - \int u² du $</p></li><li><p>$\sin x - \frac 13 u³ + C$</p></li><li><p>$\sin x - \frac 13 (\sin x)³ + C$</p></li></ul><p></p><p>b)$\int \sin²(2x)  dx$</p><p></p><ul><li><p>$\sin²(\theta) = \frac {1-\cos(2\theta)}{2}$</p></li></ul><p></p><ul><li><p>$\int \sin²(2x)  dx \Rightarrow \int \frac {1-\cos(4x)}{2}  dx$</p></li><li><p>$\int \frac 12 + \frac {-\cos(4x)}2  dx$</p></li><li><p>$\int \frac 12  dx + \int \frac {-\cos(4x)}2  dx$</p></li><li><p>$\frac 12x + \frac 12 \int -\cos(4x)  dx$</p><ul><li><p>$u = 4x$</p></li><li><p>$du = 4dx$</p></li><li><p>$\frac 14 du = dx$</p></li></ul></li><li><p>$\frac x2 + \frac 18 \int -\cos u  du$</p></li><li><p>$\frac x2 + \frac 18(-\sin(4x)) + C$</p></li><li><p>$\frac x2 + \frac {-\sin(4x)}8 + C$</p></li></ul><p></p><p>c)$\int_0^{\frac \pi 4} \sin³x  dx$</p><p></p><ul><li><p>$\int_{0}^{\frac \pi 4} \sin³ x  dx \Rightarrow \int_0^{\frac \pi 4} \sin²x \sin x  dx$</p></li><li><p>$\Rightarrow \int_0^{\frac \pi 4} \sin x(1 - \cos² x) dx$</p></li><li><p>$\Rightarrow \int_0^{\frac \pi 4} (\sin x - \sin x\cos²x)  dx$</p></li><li><p>$\Rightarrow \int_0^{\frac \pi 4} \sin x  dx + \int_0^{\frac \pi 4} -\sin x\cos²x  dx$</p></li></ul><p></p><p>LHS:</p><ul><li><p>$\int_{0}^{\frac \pi 4} \sin x  dx \Rightarrow [-\cos x]_0^{\frac \pi 4}$</p></li><li><p>$(-\cos (\frac \pi 4) + 1) = (-\frac {\sqrt 2}2 + 1)$</p></li></ul><p></p><p>RHS:</p><ul><li><p>$\int_0^{\frac \pi 4} -\sin x \cos²x  dx$</p><ul><li><p>$u = \cos x$</p></li><li><p>$du = -\sin x  dx$</p></li></ul></li><li><p>$\int_1^{\frac {\sqrt 2}2} u²  du$</p></li><li><p>$[\frac 13 u³]_1^{\frac {\sqrt 2}2}$</p></li><li><p>$(\frac 13 (\frac {\sqrt 2}2)³ - \frac 13)$</p></li><li><p>$\frac {2\sqrt 2}{24} - \frac 13$</p></li></ul><p></p><p>Total:</p><ul><li><p>$-\frac {\sqrt 2}{2} + 1 + \frac {2 \sqrt 2}{24} - \frac 13$</p></li><li><p>$-\frac {10 \sqrt 2}{24} + \frac 23$</p></li><li><p>$\frac 23 - \frac {5\sqrt 2}{12}$</p></li></ul><p></p><p>d)$\int \sin² x \cos²x  dx$</p><p></p><ul><li><p>$\int (\sin x \cos x)²  dx$</p></li></ul><p></p><ul><li><p>$\sin(2x) = 2\sin x \cos x$</p></li><li><p>$\sin x \cos x = \frac 12 \sin(2x)$</p></li></ul><p></p><ul><li><p>$\int (\frac {\sin(2x)}2)²  dx$</p></li><li><p>$\frac 14 \int \sin²(2x)  dx$</p><ul><li><p>$\sin² \theta = \frac 12 (1-\cos (2\theta))$</p></li></ul></li><li><p>$\frac 14 \int \frac 12(1 - \cos 4x)  dx$</p></li><li><p>$\frac 18 \int (1 - \cos 4x)  dx$</p></li><li><p>$\frac 18 (\int 1  dx + \int -\cos 4x  dx) $</p></li></ul><p></p><p>LHS:</p><ul><li><p>$\int 1  dx = x + C$</p></li></ul><p></p><p>RHS:</p><ul><li><p>$\int -\cos 4x  dx \Rightarrow -\frac 14 \sin (4x) + C$</p></li></ul><p></p><p>Total:</p><ul><li><p>$\frac 18(x - \frac 14 \sin (4x)  +C)$</p></li><li><p>$\frac x8 - \frac {\sin (4x)}{32} + C$</p></li></ul><p></p><p>e)$\int \tan^5x \sec³x  dx$</p><p></p><ul><li><p>$\int \tan^5 x \sec^3 x  dx \Rightarrow \int \tan^4x\sec²x \sec x \tan x  dx$</p></li><li><p>$\Rightarrow \int (\sec² x - 1)² \sec ²x \sec x \tan x  dx$</p><ul><li><p>$u = \sec x$</p></li><li><p>$du = \sec x \tan x  dx$</p></li></ul></li><li><p>$\Rightarrow \int (u² - 1)² u²  du$</p></li><li><p>$\Rightarrow \int (u^4 -2u² + 1)u²  du$</p></li><li><p>$\int (u^6 -2u^4 + u²)  du$</p></li></ul><p></p><ul><li><p>$\Rightarrow \int u^6  du + \int -2u^4  du + \int u²  du$</p></li></ul><p></p><p>1st:</p><ul><li><p>$\frac 17 u^7 + C$</p></li><li><p>$\frac {\sec^7 x}7 + C$</p></li></ul><p></p><p>2nd:</p><ul><li><p>$-\frac 25 u^5 + C$</p></li><li><p>$-\frac {2\sec^5 x}{5} + C$</p></li></ul><p></p><p>3rd:</p><ul><li><p>$\frac 13 u³ + C$</p></li><li><p>$\frac {\sec³ x}3 + C$</p></li></ul><p></p><p>Sum:</p><ul><li><p>$\frac {\sec^7 x}{7} - \frac {2\sec^5 x}{5} + \frac {\sec³ x}{3} +C$</p></li></ul><p></p><p>f)$\int \cos^7 (3x)  dx$</p><p></p><ul><li><p>$u = 3x$</p></li><li><p>$du = 3  dx$</p></li><li><p>$\frac 13 du = dx$</p></li></ul><p></p><ul><li><p>$\frac 13 \int \cos^7(u)  du$</p></li><li><p>$\frac 13 \int \cos^6 u\cos u  du\Rightarrow \frac 13 \int (\cos² u)³ \cos u  du$</p></li><li><p>$\frac 13 \int (1 - \sin² u)³ \cos u  du$</p></li></ul><p></p><ul><li><p>$v = \sin u$</p></li><li><p>$dv = \cos u  du$</p></li></ul><p></p><ul><li><p>$\frac 13 \int (1-v²)³ dv$</p></li></ul><p></p><ul><li><p>$\frac 13 \int (1+v^4-2x²)(1-v²)  dx$</p></li><li><p>$\frac 13 \int (1 + v^4 - 2v² - v² - v^6 + 2v^4)  dx$</p></li><li><p>$\frac 13 \int (1 -v^6 + 3v^4 -3v²)  dv$</p></li></ul><p></p><ul><li><p>$\frac 13 (\int 1  dv + \int -v^6  dv + \int 3v^4  dv + \int -3v²  dv)$</p></li></ul><p></p><p>1st:</p><ul><li><p>$v + C$</p></li><li><p>$\sin u + C$</p></li><li><p>$\sin (3x) + C$</p></li></ul><p></p><p>2nd: </p><ul><li><p>$-\frac {v^7}{7} + C$</p></li><li><p>$-\frac {\sin^7u}{7} + C$</p></li><li><p>$-\frac {\sin^7(3x)}{7} + C$</p></li></ul><p></p><p>3rd:</p><ul><li><p>$\frac {3v^5}{5} + C$</p></li><li><p>$\frac {3\sin(u)^5}{5} + C$</p></li><li><p>$\frac {3\sin(3x)^5}{5} + C$</p></li></ul><p></p><p>4th:</p><ul><li><p>$-v³ + C$</p></li><li><p>$-\sin³ u + C$</p></li><li><p>$-\sin³ (3x) + C$</p></li></ul><p></p><p>Sum:</p><ul><li><p>$\frac 13(\sin(3x) - \frac {\sin^7(3x)}{7} + \frac {3\sin^5(3x)}{5} - \sin³(3x) + C)$</p></li><li><p>$\frac {\sin (3x)}{3} - \frac {\sin^7(3x)}{21} + \frac {\sin^5(3x)}{5} - \frac {\sin³(3x)}{3} + C$</p></li></ul><p></p><p>g)$\int \tan^6 x  dx$</p><p></p><ul><li><p>$\tan^6x = (\tan² x)³ = (\sec²x - 1)³$</p></li><li><p>$(\sec²x - 1)³ = \sec^6 x - 3\sec^4x + 3\sec² x - 1$</p></li></ul><p></p><p>$\int \sec^6 x  dx + \int -3\sec^4x  dx + \int 3\sec²x  dx + \int -1  dx$</p><p></p><p>1st:</p><ul><li><p>$\int \sec^6x dx$</p></li><li><p>$\int (\sec^4 x)\sec²x  dx$</p></li><li><p>$ \int (\tan²x+1)² \sec²x  dx$</p><ul><li><p>$u=\tan x$</p></li><li><p>$du = \sec² x  dx$</p></li></ul></li><li><p>$\int (u² + 1)²  du$</p></li><li><p>$\int (u^4 + 2u² + 1)  dx$</p></li><li><p>$\frac {u^5}{5} + \frac {2u³}{3} +u$</p></li><li><p>$\frac {\tan^5 x}{5} + \frac {2\tan³ x}{3} + \tan x$</p></li></ul><p></p><p>h)$\tan^4 x  dx$</p><div data-type="horizontalRule"><hr></div><p>Trig Substitution:</p><p>For the following problems, evaluate the integral.</p><p></p><p>a)$\int \sqrt{9-x²}  dx$</p><p></p><ul><li><p>$x = 3\sin(\theta)$</p></li><li><p>$dx = 3\cos(\theta)  d\theta$</p></li></ul><ul><li><p>$\sqrt {9-x²} = 3\cos \theta$</p></li></ul><p></p><ul><li><p>$\int 3\cos(\theta) \cdot 3\cos(\theta)  d\theta$</p></li><li><p>$9\int \cos²(\theta)  d\theta$</p></li><li><p>$9\int \frac {1+\cos(2\theta)}{2}  d\theta$</p></li><li><p>$\frac 92\int 1 + \cos (2\theta)  d\theta$</p></li><li><p>$\frac 92 (\theta + \frac 12 \sin (2\theta)) + C$</p></li><li><p>$\frac 92 \theta + \frac 94 \sin (2\theta) + C$</p></li></ul><p></p><ul><li><p>$\theta = \arcsin \frac x3$</p></li><li><p>$\sin(2\theta) =$$\sin \theta \cos \theta + \cos \theta \sin \theta$</p><ul><li><p>$\sin(2\theta) = 2\sin\theta\cos\theta$</p></li><li><p>$2\sin\theta\cos\theta = \frac 23 \frac {\sqrt {9-x²}}{3}$</p></li></ul></li></ul><p></p><ul><li><p>$\frac 92 \arcsin \frac x3 + \frac 94 \frac {2x}3 \frac {\sqrt {9-x²}}3 + C$</p></li><li><p>$\frac 92 \arcsin \frac x3 + \frac {x\sqrt {9-x²}}2 + C$</p></li></ul><p></p><p>b)$\int \frac {x²}{\sqrt {25-x²}}  dx$</p><p></p><ul><li><p>$\sqrt {25 - x²} \Rightarrow$</p></li><li><p>$x = 5\sin(\theta)$</p></li><li><p>$dx = 5\cos(\theta)  d\theta$</p></li><li><p>$\sqrt {25 - x²} = 5\cos(\theta)$</p></li></ul><p></p><ul><li><p>$\int \frac {25\sin²(\theta)}{5\cos(\theta)} 5\cos(\theta)  d\theta$</p></li><li><p>$\int 25 \sin²(\theta)  d\theta$</p></li></ul><p></p><ul><li><p>$25\int \sin²(\theta)  d\theta$</p><ul><li><p>$\sin²(\theta) = \frac {1-\sin(2\theta)}{2}$</p></li></ul></li><li><p>$25\int \frac {1-\cos(2\theta)}{2}  d\theta$</p></li><li><p>$\frac {25}2 \int 1 - \cos(2\theta)  d\theta$</p></li><li><p>$\frac {25}2 (\theta - \frac 12\sin(2\theta)) + C$</p></li><li><p>$\frac {25}2\theta - \frac {25}{4}\sin(2\theta) + C$</p></li></ul><p></p><ul><li><p>$\theta = \arcsin (\frac x5)$</p></li><li><p>$\sin (2\theta) = 2\sin\theta\cos\theta = \frac {2x}5 \cdot\frac {\sqrt {25-x²}}{5}$</p></li></ul><p></p><ul><li><p>$\frac {25}{2} \cdot \arcsin \frac {x}{5} - \frac {25}4 \cdot \frac {2x}5 \cdot \frac {\sqrt {25-x²}}5$</p></li><li><p>$\frac {25}{2} \arcsin (\frac x5) - \frac {x}{2} \sqrt {25-x²}$</p></li></ul><p></p><p>c)$\int \sqrt {x² - 16}  dx$</p><p></p><ul><li><p>$x = 4\sec(\theta)$</p></li><li><p>$dx = 4\sec(\theta)\tan(\theta)  d\theta$</p></li><li><p>$\sqrt {x² - 16} = 4\tan(\theta)$</p></li></ul><p></p><ul><li><p>$\int 4\tan (\theta)  4\sec(\theta)\tan(\theta)  d\theta$</p></li><li><p>$16\int \tan²(\theta)\sec(\theta)  d\theta$</p></li><li><p>$16\int (\sec²(\theta) -1)\sec (\theta)  d\theta$</p></li><li><p>$16\int (\sec³(\theta) - \sec(\theta))  d\theta$</p></li><li><p>$16\int \sec³(\theta)  d\theta + \int -\sec(\theta)  d\theta$</p></li></ul><p></p><p>LHS:</p><ul><li><p>$\int \sec³(\theta)  d\theta$</p></li><li><p>$\int \sec(\theta)\cdot \sec²(\theta)  d\theta$</p></li><li><p>$\sec \theta \tan \theta - \int \sec \theta \tan² \theta  d\theta$</p></li><li><p>$= \frac 12 (\sec (\theta) \tan (\theta) + \ln |\sec (\theta) + \tan (\theta)|) + C$</p></li></ul><p></p><p>RHS:</p><ul><li><p>$\int -\sec(\theta)  d\theta$</p></li><li><p>$-\ln |\sec(\theta) + \tan (\theta) | + C$</p></li></ul><p></p><p>Combine:</p><ul><li><p>$16(\frac 12(\sec(\theta)\tan(\theta) + \ln | \sec(\theta) + \tan(\theta)|) - \ln |\sec(\theta + \tan(\theta)|

• $8\sec(\theta)\tan(\theta) + 8\ln |\sec(\theta) + \tan(\theta)| - 16\ln |\sec(\theta) + \tan(\theta)| + C$

• $8\sec(\theta)\tan(\theta) - 8\ln|\sec(\theta) + \tan(\theta)| + C$

• $8\frac x4 \frac {\sqrt {x² - 16}}{4}-8\ln |\frac x4 + \frac {\sqrt {x²-16}}{4}| +C$

• $\frac {x\sqrt{x²-16}}2 - 8\ln|\frac {x+\sqrt {x²-16}}{4}| + C$