$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\ffrac }[2]{\frac {\text {\footnotesize #1}}{\text {\footnotesize #2}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\C }[0]{\mathbb C} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\dis }[0]{\displaystyle } \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ppx }[0]{\frac {\partial }{\partial x}} \newcommand {\ppy }[0]{\frac {\partial }{\partial y}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{ #1 \bigg |} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{proj} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\cis }{cis} \DeclareMathOperator {\Arg }{Arg} \DeclareMathOperator {\Rep }{Re} \DeclareMathOperator {\Imp }{Im} \DeclareMathOperator {\sech }{sech} \DeclareMathOperator {\csch }{csch} \DeclareMathOperator {\Log }{Log} \newcommand {\tightoverset }[2]{\mathop {#2}\limits ^{\vbox to -.5ex{\kern -0.75ex\hbox {#1}\vss }}} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\utan }[0]{\vec {\hat {t}}} \newcommand {\unormal }[0]{\vec {\hat {n}}} \newcommand {\ubinormal }[0]{\vec {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\vector }[1]{\left \langle #1\right \rangle }$

### 1 Trigonometric identities

The following are the Pythagorean Trigonometric Identities (named for Pythagoras of Samos) which hold for all angles, $\theta$, in the domains of the functions involved: and

### 2 Integrands involving $\sqrt {a^2 - u^2}$

(problem 1a)

Compute the indefinite integral Use a trigonometric substitution with
Substituting these and simplifying the integrand gives
$\displaystyle {\int x^3 \sqrt {16 - x^2}\; dx =}$

$\displaystyle {-64 \int \cos ^3(\theta )\sin (\theta ) \; d\theta }$ $\displaystyle {-1024\int \cos ^3(\theta )\sin ^2(\theta ) \; d\theta }$ $\displaystyle {-16\int \cos ^3(\theta )\sin ^2(\theta ) \; d\theta }$

Computing this integral gives Based on the substitution $x = 4\cos (\theta )$,

$\frac {1}{5}(16-x^2)^{5/2} - \frac {1}{48}(16 -x^2)^{3/2} + C$ $\frac {1}{80}(16-x^2)^{5/2} - \frac {1}{12}(16 -x^2)^{3/2} + C$ $\frac {1}{320}(16-x^2)^{5/2} - \frac {1}{12}(16 -x^2)^{3/2} + C$ $\frac {1}{5}(16-x^2)^{5/2} - \frac {16}{3}(16 -x^2)^{3/2} + C$
(problem 1b)

Compute the indefinite integral Use a trigonometric substitution with
The differential is
Substituting these and simplifying the integrand gives
Computing this integral gives Based on the substitution $x = 2\cos (\theta )$,

$\frac 17 (4-x^2)^{7/2} - \frac 85(4-x^2)^{5/2} + \frac {16}{3} (4-x^2)^{3/2} + C$ $-\frac 17 (4-x^2)^{7/2} + \frac 85(4-x^2)^{5/2} - \frac {16}{3} (4-x^2)^{3/2} + C$ $-\frac 17 (4-x^2)^{7/2} + \frac 45(4-x^2)^{5/2} - \frac {16}{3} (4-x^2)^{3/2} + C$ $\frac 17 (4-x^2)^{7/2} - \frac 45(4-x^2)^{5/2} + \frac {16}{3} (4-x^2)^{3/2} + C$
(problem 2)

Compute the indefinite integral Use a trigonometric substitution with
The differential is
Substituting these and simplifying the integrand gives
Computing this integral gives Based on the substitution $x = \frac 13 \cos (\theta )$,

$-\frac {\sqrt {1-9x^2}}{3x} + \cos ^{-1}(3x) + C$ $\frac {\sqrt {1-9x^2}}{3x} - \cos ^{-1}(3x) + C$ $-\frac {\sqrt {1-9x^2}}{x} + 3\cos ^{-1}(3x) + C$ $\frac {\sqrt {1-9x^2}}{x} - 3\cos ^{-1}(3x) + C$
(problem 3a)

Compute the indefinite integral Use a trigonometric substitution with
The differential is
Substituting these and simplifying the integrand gives
Computing this integral gives Based on the substitution $x = 3 \cos (\theta )$,

$-\frac 13 \ln \bigg |\frac {3}{x} + \frac {\sqrt {9-x^2}}{x}\bigg | + C$ $\frac 13 \ln \bigg |\frac {3}{x} + \frac {\sqrt {9-x^2}}{x}\bigg | + C$ $- \ln \bigg |\frac {3}{x} + \frac {\sqrt {9-x^2}}{x}\bigg | + C$ $\ln \bigg |\frac {3}{x} + \frac {\sqrt {9-x^2}}{x}\bigg | + C$
(problem 3b)

Compute the indefinite integral Use a trigonometric substitution with
The differential is
Substituting these and simplifying the integrand gives
Computing this integral gives Based on the substitution $x = \frac 25 \cos (\theta )$,

$\frac 12 \ln \bigg |\frac {2}{5x} + \frac {\sqrt {4-25x^2}}{5x}\bigg | + C$ $-\frac 12 \ln \bigg |\frac {2}{5x} + \frac {\sqrt {4-25x^2}}{5x}\bigg | + C$ $- \ln \bigg |\frac {2}{5x} + \frac {\sqrt {4-25x^2}}{5x}\bigg | + C$ $\ln \bigg |\frac {2}{5x} + \frac {\sqrt {4-25x^2}}{5x}\bigg | + C$

### 3 Integrands involving $\sqrt {u^2 - a^2}$

(problem 4)

Compute the indefinite integral Use a trigonometric substitution with
Substituting these and simplifying the integrand gives
$\displaystyle {\int \frac {1}{\sqrt {x^2 - 4}}\; dx =}$

$\displaystyle { \int \sec (\theta ) \; d\theta }$ $\displaystyle {\frac 12\int \sec (\theta )\; d\theta }$ $\displaystyle {\int \cot (\theta ) \; d\theta }$

Computing this integral gives Based on the substitution $x = 2\sec (\theta )$,

$\ln |x+\sqrt {x^2 - 4}| + C$ $\ln |\frac {x}{2}+ \frac {\sqrt {x^2 - 4}}{2}| + C$ $2\ln |x+\sqrt {x^2 - 4}| + C$
(problem 5) Compute the indefinite integral Use a trigonometric substitution with
Substituting these and simplifying the integrand gives
$\displaystyle {\int x^3 \sqrt { x^2 -16}\; dx =}$
$\displaystyle {64 \int \sec ^4(\theta )\tan ^2(\theta ) \; d\theta }$ $\displaystyle {1024\int \sec ^4(\theta )\tan ^2(\theta ) \; d\theta }$ $\displaystyle {16\int \sec ^4(\theta )\tan ^2(\theta ) \; d\theta }$

Computing this integral gives Based on the substitution $x = 4\sec (\theta )$,

$\frac {1}{5}(x^2 -16)^{5/2} + \frac {1}{3}(x^2 -16)^{3/2} + C$ $\frac {16}{5}(x^2 -16)^{5/2} + \frac {16}{3}(x^2 -16)^{3/2} + C$ $\frac {16}{5}(x^2 -16)^{5/2} + \frac {1}{3}(x^2 -16)^{3/2} + C$ $\frac {1}{5}(x^2-16)^{5/2} + \frac {16}{3}(x^2-16)^{3/2} + C$
(problem 6) Compute the indefinite integral Use a trigonometric substitution with
Substituting these and simplifying the integrand gives

Computing this integral gives Based on the substitution $x = \frac 12\sec (\theta )$, the final answer is

(problem 7) Compute the indefinite integral

Use a trigonometric substitution with
Substituting these and simplifying the integrand gives
$\displaystyle {\int \frac {1}{x^2 \sqrt { 25x^2 -16}}\; dx =}$

$\displaystyle {\frac {5}{4}\int \cos (\theta ) \; d\theta }$ $\displaystyle {\frac {5}{16}\int \cos (\theta ) \; d\theta }$ $\displaystyle {\frac {16}{5}\int \cos (\theta ) \; d\theta }$

Computing this integral gives Based on the substitution $x = \frac 45\sec (\theta )$,

$\frac {16\sqrt {25x^2 - 16}}{5x} + C$ $\frac {5\sqrt {25x^2 - 16}}{16x} + C$ $\frac {\sqrt {25x^2 - 16}}{5x} + C$ $\frac {\sqrt {25x^2 - 16}}{16x} + C$
(problem 8) Compute the indefinite integral Use a trigonometric substitution with
Substituting these and simplifying the integrand gives
$\displaystyle {\int \frac {1}{x^3\sqrt {x^2-1}}\; dx =}$
$\displaystyle {\int \cos ^2(\theta ) \; d\theta }$ $\displaystyle {\int \cos ^3(\theta ) \; d\theta }$ $\displaystyle {\int \cos (\theta ) \; d\theta }$

Computing this integral gives

use $\cos ^2(\theta ) = \frac 12 + \frac 12 \cos (2\theta )$

Using the identity, $\sin (2\theta ) = 2\sin (\theta )\cos (\theta )$, we can rewrite this as

Based on the substitution $x = \sec (\theta )$,

$\sec ^{-1}(x) + \frac {\sqrt {x^2 -1}}{2x^2} + C$ $\frac 12 \sec ^{-1}(x) + \frac {\sqrt {x^2 -1}}{2x^2} + C$ $\frac 12 \sec ^{-1}(x) + \frac {\sqrt {x^2 -1}}{x^2} + C$ $\sec ^{-1}(x) + \frac {\sqrt {x^2 -1}}{x^2} + C$

### 4 Integrands involving $\sqrt {a^2 + u^2}$

(problem 9) Compute the indefinite integral Use a trigonometric substitution with
The differential is
Substituting these and simplifying the integrand gives
Computing this integral gives Based on the substitution $x = 3\tan (\theta )$,
$\frac {1}{5}(9+x^2)^{5/2} - 3(9+x^2)^{3/2} + C$ $\frac {1}{5}(9+x^2)^{5/2} + 3(9+x^2)^{3/2} + C$ $\frac {1}{3}(9+x^2)^{3/2} - (9+x^2)^{5/2} + C$ $\frac {1}{3}(9+x^2)^{3/2} + (9+x^2)^{5/2} + C$
(problem 10) Compute the indefinite integral Use a trigonometric substitution with
The differential is
Substituting these and simplifying the integrand gives
Computing this integral gives Based on the substitution $x = 2\tan (\theta )$,
$\frac {1}{3}(4+x^2)^{3/2} + 4(4+x^2)^{1/2} + C$ $\frac {1}{3}(4+x^2)^{3/2} - 4(4+x^2)^{1/2} + C$ $\frac {1}{3}(4+x^2)^{1/2} + 4(4+x^2)^{3/2} + C$ $\frac {1}{3}(4+x^2)^{1/2} - 4(4+x^2)^{3/2} + C$
(problem 11) Compute the indefinite integral Use a trigonometric substitution with
The differential is
Substituting these and simplifying the integrand gives
Computing this integral gives Based on the substitution $x = \frac 37 \tan (\theta )$,
$\frac 12 x\sqrt {9+49x^2} + \frac {3}{14}\ln \left (7x + \sqrt {9+49x^2}\right ) + C$ $\frac 12 x\sqrt {9+49x^2} - \frac {9}{14}\ln \left (7x + \sqrt {9+49x^2}\right ) + C$ $\frac 12 x\sqrt {9+49x^2} + \frac {9}{14}\ln \left (7x+\sqrt {9+49x^2}\right ) + C$ $\frac 12 x\sqrt {9+49x^2} - \frac {3}{14}\ln |7x - \sqrt {9+49x^2}| + C$
Use a trigonometric substitution to convert the integral into a trigonometric integral: $\dis \int x^3 \sqrt {x^2 -4} \; dx$
We will make the substitution $x = \answer {2\sec \theta }$
$32\dis \int \sec ^4\theta \tan ^2\theta \; d\theta$ $16\dis \int \sec ^4\theta \tan ^2\theta \; d\theta$
$32\dis \int \sec ^3\theta \tan \theta \; d\theta$ $16\dis \int \sec ^3\theta \tan \theta \; d\theta$