1.4 Trigonometric Substitution

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\ffrac }[2]{\frac {\text {\footnotesize $#1$}}{\text {\footnotesize $#2$}}} \newcommand {\newrobustcmd }[0]{} \newcommand {\csshow }[1]{\begingroup \expandafter \endgroup \expandafter \show \csname #1\endcsname } \newcommand {\csmeaning }[1]{\ifcsname #1\endcsname \expandafter \meaning \csname #1\endcsname \else \detokenize {undefined}\fi } \newcommand {\ifdefmacro }[0]{} \newcommand {\ifdefparam }[0]{} \newcommand {\ifdefprotected }[0]{} \newcommand {\ifnumequal }[1]{\ifnumcomp {#1}=} \newcommand {\ifnumgreater }[1]{\ifnumcomp {#1}>} \newcommand {\ifnumless }[1]{\ifnumcomp {#1}<} \newcommand {\ifdimequal }[1]{\ifdimcomp {#1}=} \newcommand {\ifdimgreater }[1]{\ifdimcomp {#1}>} \newcommand {\ifdimless }[1]{\ifdimcomp {#1}<} \newcommand {\expandonce }[1]{\unexpanded \expandafter {#1}} \newcommand {\csexpandonce }[1]{\expandafter \expandonce \csname #1\endcsname } \newcommand {\protecting }[0]{} \newcommand {\csdef }[1]{\expandafter \def \csname #1\endcsname } \newcommand {\csedef }[1]{\expandafter \edef \csname #1\endcsname } \newcommand {\csgdef }[1]{\expandafter \gdef \csname #1\endcsname } \newcommand {\csxdef }[1]{\expandafter \xdef \csname #1\endcsname } \newcommand {\cslet }[2]{\expandafter \let \csname #1\endcsname #2} \newcommand {\letcs }[2]{\ifcsdef {#2} {\expandafter \let \expandafter #1\csname #2\endcsname } {\undef #1}} \newcommand {\csletcs }[2]{\ifcsdef {#2} {\expandafter \let \csname #1\expandafter \endcsname \csname #2\endcsname } {\csundef {#1}}} \newcommand {\csuse }[1]{\ifcsname #1\endcsname \csname #1\expandafter \endcsname \fi } \newcommand {\appto }[2]{\ifundef {#1} {\edef #1{\unexpanded {#2}}} {\edef #1{\expandonce #1\unexpanded {#2}}}} \newcommand {\eappto }[2]{\ifundef {#1} {\edef #1{#2}} {\edef #1{\expandonce #1#2}}} \newcommand {\gappto }[2]{\ifundef {#1} {\xdef #1{\unexpanded {#2}}} {\xdef #1{\expandonce #1\unexpanded {#2}}}} \newcommand {\xappto }[2]{\ifundef {#1} {\xdef #1{#2}} {\xdef #1{\expandonce #1#2}}} \newcommand {\preto }[2]{\ifundef {#1} {\edef #1{\unexpanded {#2}}} {\edef #1{\unexpanded {#2}\expandonce #1}}} \newcommand {\epreto }[2]{\ifundef {#1} {\edef #1{#2}} {\edef #1{#2\expandonce #1}}} \newcommand {\gpreto }[2]{\ifundef {#1} {\xdef #1{\unexpanded {#2}}} {\xdef #1{\unexpanded {#2}\expandonce #1}}} \newcommand {\xpreto }[2]{\ifundef {#1} {\xdef #1{#2}} {\xdef #1{#2\expandonce #1}}} \newcommand {\csappto }[1]{\expandafter \appto \csname #1\endcsname } \newcommand {\cseappto }[1]{\expandafter \eappto \csname #1\endcsname } \newcommand {\csgappto }[1]{\expandafter \gappto \csname #1\endcsname } \newcommand {\csxappto }[1]{\expandafter \xappto \csname #1\endcsname } \newcommand {\cspreto }[1]{\expandafter \preto \csname #1\endcsname } \newcommand {\csepreto }[1]{\expandafter \epreto \csname #1\endcsname } \newcommand {\csgpreto }[1]{\expandafter \gpreto \csname #1\endcsname } \newcommand {\csxpreto }[1]{\expandafter \xpreto \csname #1\endcsname } \newcommand {\csnumdef }[1]{\expandafter \numdef \csname #1\endcsname } \newcommand {\csnumgdef }[1]{\expandafter \numgdef \csname #1\endcsname } \newcommand {\csdimdef }[1]{\expandafter \dimdef \csname #1\endcsname } \newcommand {\csdimgdef }[1]{\expandafter \dimgdef \csname #1\endcsname } \newcommand {\csgluedef }[1]{\expandafter \gluedef \csname #1\endcsname } \newcommand {\csgluegdef }[1]{\expandafter \gluegdef \csname #1\endcsname } \newcommand {\mudef }[2]{\ifundef #1{\def #1{0mu}}{}\edef #1{\the \muexpr #2}} \newcommand {\csmudef }[1]{\expandafter \mudef \csname #1\endcsname } \newcommand {\csmugdef }[1]{\expandafter \mugdef \csname #1\endcsname } \newcommand {\listbreak }[0]{} \newcommand {\listadd }[2]{\ifblank {#2}{}{\appto #1{#2|}}} \newcommand {\listgadd }[2]{\ifblank {#2}{}{\gappto #1{#2|}}} \newcommand {\listcsadd }[1]{\expandafter \listadd \csname #1\endcsname } \newcommand {\listcseadd }[1]{\expandafter \listeadd \csname #1\endcsname } \newcommand {\listcsgadd }[1]{\expandafter \listgadd \csname #1\endcsname } \newcommand {\listcsxadd }[1]{\expandafter \listxadd \csname #1\endcsname } \newcommand {\dolistloop }[0]{\forlistloop \do } \newcommand {\dolistcsloop }[0]{\forlistcsloop \do } \newcommand {\AfterPreamble }[0]{\AtBeginDocument } \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 } \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

We compute integrals involving square roots of sums and differences.

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: $\sin ^2(\theta ) + \cos ^2(\theta ) = 1$ $1 + \tan ^2(\theta ) = \sec ^2(\theta )$ and $1 + \cot ^2(\theta ) = \csc ^2(\theta )$

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

example 1 Compute the indefinite integral $\int x^3 \sqrt {1-x^2} \; dx.$ The key to the solution is to replace the variable, $x$ , with a trigonometric function in order to eliminate the square root. Since $1-\cos ^2(\theta )$ and the radicand, $1-x^2$ , have the same algebraic form, we will try the substitution $x = \cos (\theta )$ . To apply this to the integral, we also need to compute the differential, $dx = -\sin (\theta ) \; d\theta$ . Substituting gives, $\int x^3 \sqrt {1-x^2} \; dx = \int \cos ^3(\theta ) \sqrt {1-\cos ^2(\theta )} [-\sin (\theta )] \; d\theta .$ The point of this substitution is that the square root can be computed by using a Pythagorean trigonometric identity. From there, we can compute the integral using the skills developed in the section on trigonometric integrals. We have $\begin{align*} \int x^3 \sqrt{1-x^2} \; dx &= \int \cos^3(\theta) \sqrt{1-\cos^2(\theta)} \left[-\sin(\theta)\right] \; d\theta\\[6pt] &= -\int \cos^3(\theta) \sqrt{\sin^2(\theta)} \sin(\theta) \; d\theta\\[6pt] &= -\int \cos^3(\theta) \sin(\theta) \sin(\theta) \; d\theta\\[6pt] &= -\int \cos^3(\theta) \sin^2(\theta) \; d\theta \; \;\;\text{(odd power of cosine)}\\[6pt] &= -\int \cos^2(\theta) \sin^2(\theta) \cos(\theta) \; d\theta\\[6pt] &= -\int \left[1 - \sin^2(\theta)\right] \sin^2(\theta) \cos(\theta) \; d\theta\\[6pt] & \; \text{let $u = \sin(\theta), \; du = \cos(\theta) \, d\theta$}\\[6pt] &=-\int (1-u^2)u^2 \; du\\[6pt] &=\int (u^2 -1)u^2 \; du\\[6pt] &=\int (u^4 - u^2) \; du\\[6pt] &=\frac{u^5}{5} - \frac{u^3}{3} + C\\[6pt] &= \frac15 \sin^5(\theta) - \frac13 \sin^3(\theta) + C. \end{align*}$ At this point we have an answer in terms of the variable $\theta$ , but we require the variable $x$ . We will use the triangle below to convert back to $x$ .

From the triangle, we see that $\sin (\theta ) = \sqrt {1 - x^2}$ , and we substitute to get the final answer: $\int x^3 \sqrt {1-x^2} \; dx = \frac 15 \sin ^5(\theta ) - \frac 13 \sin ^3(\theta ) + C = \frac 15 (1-x^2)^{5/2} - \frac 13 (1-x^2)^{3/2} + C.$

Technically, $\sqrt {\sin ^2(\theta )} = |\sin (\theta )|$ . However, in the substitution $x = \cos (\theta )$ , we can choose the angle, $\theta$ , to be in the interval $[0, \pi ]$ . This way, $\sin (\theta ) \geq 0$ , and $|\sin (\theta )| = \sin (\theta )$ .

(problem 1a)

Compute the indefinite integral $\int x^3 \sqrt {16-x^2} \; dx.$ Use a trigonometric substitution with
$x = \answer {4\cos (\theta )} \;\; dx = \answer {-4\sin (\theta )} \; d\theta$ 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 $\int \cos ^3(\theta )\sin ^2(\theta ) \; d\theta = \answer {-(1/5)\sin ^5(\theta ) + (1/3)\sin ^3(\theta )} + C.$ Based on the substitution $x = 4\cos (\theta )$ ,
$\sin (\theta ) = \answer {\sqrt {16 - x^2}/4}$ The final answer is $\int x^3\sqrt {16 - x^2}\; dx =$

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

(problem 1b)

Compute the indefinite integral $\int x^5 \sqrt {4-x^2} \; dx.$ Use a trigonometric substitution with
$x = \answer {2\cos (\theta )}$ The differential is
$dx = \answer {-2\sin (\theta )} \; d\theta$ Substituting these and simplifying the integrand gives
$\int x^5\sqrt {4 - x^2}\; dx = -128\int \answer {\cos ^5(\theta )\sin ^2(\theta )} \; d\theta$ Computing this integral gives $\int \cos ^5(\theta )\sin ^2(\theta ) \; d\theta = \answer {-(2/5)\sin ^5(\theta ) + (1/3)\sin ^3(\theta ) +\frac 17 \sin ^7(\theta )} + C.$ Based on the substitution $x = 2\cos (\theta )$ ,
$\sin (\theta ) = \answer {\sqrt {4 - x^2}/2}$ The final answer is $\int x^5\sqrt {4 - x^2}\; dx =$

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

example 2 Compute the indefinite integral $\int \frac {\sqrt {4-x^2}}{x^2} \; dx$ We choose the substitution $x = 2\cos (\theta )$ so that $dx = -2\sin (\theta ) \, d\theta$ . The integral becomes $\begin{align*} \int \frac{\sqrt{4-x^2}}{x^2} \; dx &= \int \frac{\sqrt{4-4\cos^2(\theta)}}{4\cos^2(\theta)} \left[-2\sin(\theta)\right] \; d\theta\\[6pt] &= -\frac12\int \frac{\sqrt{4\sin^2(\theta)}}{\cos^2(\theta)} \sin(\theta) \; d\theta\\[6pt] &= -\frac12\int \frac{2\sin(\theta)}{\cos^2(\theta)} \sin(\theta) \; d\theta\\[6pt] &= -\int \frac{\sin^2(\theta)}{\cos^2(\theta)} \; d\theta\\[6pt] &= -\int \tan^2(\theta) \; d\theta\\[6pt] &= -\int \left[\sec^2(\theta) - 1\right] \; d\theta\\[6pt] &= \int \left[1 - \sec^2(\theta)\right] \; d\theta\\[6pt] &= \theta-\tan(\theta) + C. \end{align*}$

From the triangle above, we can see that $\tan (\theta ) = \frac {\sqrt {4-x^2}}{x}$ so that: $\int \frac {\sqrt {4-x^2}}{x^2} \; dx = \theta -\tan (\theta ) + C = \cos ^{-1}\left (\frac {x}{2}\right ) -\frac {\sqrt {4-x^2}}{x} + + C.$

(problem 2)

Compute the indefinite integral $\int \frac {\sqrt {1-9x^2}}{x^2} \; dx.$ Use a trigonometric substitution with
$x = \answer {\frac 13\cos (\theta )}$ The differential is
$dx = \answer {-\frac 13\sin (\theta )} \; d\theta$ Substituting these and simplifying the integrand gives
$\int \frac {\sqrt {1-9x^2}}{x^2} \; dx = -3\int \answer {\tan ^2(\theta )} d\theta$ Computing this integral gives $\int \tan ^2(\theta ) \; d\theta = \answer {\tan (\theta ) - \theta } + C.$ Based on the substitution $x = \frac 13 \cos (\theta )$ ,
$\tan (\theta ) = \answer {\sqrt {1-9x^2}/(3x)}$ The final answer is $\int \frac {\sqrt {1-9x^2}}{x^2} \; dx =$

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

example 3 Compute the indefinite integral $\int \frac {1}{x\sqrt {9-4x^2}} \; dx$ We choose the substitution $x = \frac 32\cos (\theta )$ so that $dx = -\frac 32\sin (\theta ) \, d\theta$ . The integral becomes $\begin{align*} \int \frac{1}{x\sqrt{9-4x^2}}\; dx &= \int \frac{1}{\frac32\cos(\theta)\sqrt{9-9\cos^2(\theta)}} \left[-\frac32\sin(\theta)\right]\; d\theta\\[6pt] &= -\int \frac{\sin(\theta)}{\cos(\theta)\sqrt{9\sin^2(\theta)}} \; d\theta\\[6pt] &= -\int \frac{\sin(\theta)}{\cos(\theta)\cdot 3\sin(\theta)} \; d\theta\\[6pt] &= -\frac13\int \frac{1}{\cos(\theta)} \; d\theta\\[6pt] &= -\frac13\int \sec(\theta) \; d\theta\\[6pt] &= -\frac13 \ln|\sec(\theta) + \tan(\theta)| + C. \end{align*}$

From the triangle above, we can see that $\sec (\theta ) = \frac {3}{2x}$ and $\tan (\theta ) = \frac {\sqrt {9-4x^2}}{2x}$ so that: $\int \frac {1}{x\sqrt {9-4x^2}} \; dx = -\frac 13 \ln |\sec (\theta ) + \tan (\theta )| + C = - \frac 13\ln \bigg |\frac {3 + \sqrt {9-4x^2}}{2x}\bigg | + C.$

(problem 3a)

Compute the indefinite integral $\int \frac {1}{x\sqrt {9-x^2}} \; dx.$ Use a trigonometric substitution with
$x = \answer {3\cos (\theta )}$ The differential is
$dx = \answer {-3\sin (\theta )}\,d\theta$ Substituting these and simplifying the integrand gives
$\int \frac {1}{x\sqrt {9-x^2}} \; dx = -\frac 13\int \answer {\sec (\theta )} d\theta$ Computing this integral gives $\int \sec (\theta ) \; d\theta = \answer {\ln |\sec (\theta ) + \tan (\theta )|} + C.$ Based on the substitution $x = 3 \cos (\theta )$ ,
$\sec (\theta ) = \answer {\frac {3}{x}} \; \text { and } \; \tan (\theta ) = \answer {\sqrt {9-x^2}/(x)}$ The final answer is $\int \frac {1}{x\sqrt {9-x^2}} \; dx =$

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

(problem 3b)

Compute the indefinite integral $\int \frac {1}{x\sqrt {4-25x^2}} \; dx.$ Use a trigonometric substitution with
$x = \answer {\frac 25\cos (\theta )}$ The differential is
$dx = \answer {-\frac 25\sin (\theta )}\,d\theta$ Substituting these and simplifying the integrand gives
$\int \frac {1}{x\sqrt {4-25x^2}} \; dx = -\frac 12\int \answer {\sec (\theta )} d\theta$ Computing this integral gives $\int \sec (\theta ) \; d\theta = \answer {\ln |\sec (\theta ) + \tan (\theta )|} + C.$ Based on the substitution $x = \frac 25 \cos (\theta )$ ,
$\sec (\theta ) = \answer {\frac {2}{5x}} \; \text { and } \; \tan (\theta ) = \answer {\sqrt {4-25x^2}/(5x)}$ The final answer is $\int \frac {1}{x\sqrt {4-25x^2}} \; dx =$

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

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

example 4 Compute the indefinite integral $\int \frac {1}{\sqrt {x^2-1}} \; dx.$ We will use the substitution $x = \sec (\theta )$ , so that $dx = \sec (\theta )\tan (\theta ) \, d\theta$ .

The integral becomes

$\begin{align*} \int \frac{1}{\sqrt{x^2-1}}\; dx &= \int \frac{\sec(\theta)\tan(\theta)}{\sqrt{\sec^2(\theta)-1}}\; d\theta\\[6pt] &= \int \frac{\sec(\theta)\tan(\theta)}{\sqrt{\tan^2(\theta)}}\; d\theta\\[6pt] &= \int \frac{\sec(\theta)\tan(\theta)}{\tan(\theta)}\; d\theta\\[6pt] &= \int \sec(\theta)\; d\theta\\[6pt] &= \ln|\sec(\theta)+ \tan(\theta)| + C. \end{align*}$

From the triangle above, we can see that $\sec (\theta ) = x$ and $\tan (\theta ) = \sqrt {x^2-1}$ so that: $\int \frac {1}{\sqrt {x^2-1}} \; dx = \ln |\sec (\theta )+ \tan (\theta )| + C = \ln |x+\sqrt {x^2-1}| + C.$

Technically, $\sqrt {\tan ^2(\theta )} = |\tan (\theta )|$ . However, in the substitution $x = \sec (\theta )$ we choose the angle $\theta$ to be in the interval $[0, \pi /2)$ or $[\pi , 3\pi /2)$ . This way, $\tan (\theta ) \geq 0$ and $|\tan (\theta )| = \tan (\theta )$ .

(problem 4)

Compute the indefinite integral $\int \frac {1}{\sqrt {x^2 - 4}} \; dx$ Use a trigonometric substitution with
$x = \answer {2\sec (\theta )} \;\; dx = \answer {2\sec (\theta )\tan (\theta )} \; d\theta$ 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 $\int \sec (\theta ) \; d\theta = \answer {\ln |\sec (\theta ) + \tan (\theta )|} + C.$ Based on the substitution $x = 2\sec (\theta )$ ,
$\tan (\theta ) = \answer {\frac 12\sqrt {x^2-4}}$ The final answer is $\int \frac {1}{\sqrt {x^2 - 4}}\; dx =$

$\ln |x+\sqrt {x^2 - 4}| + C$ $\ln |\frac {x}{2}+ \frac {\sqrt {x^2 - 4}}{2}| + C$ $2\ln |x+\sqrt {x^2 - 4}| + C$

example 5 Compute the indefinite integral $\int x^3\sqrt {x^2-4} \; dx.$ We will use the substitution $x = 2\sec (\theta )$ , so that $dx = 2\sec (\theta )\tan (\theta ) \, d\theta$ .

The integral becomes

$\begin{align*} \int x^3\sqrt{x^2-4}\; dx &= \int 8\sec^3(\theta)\sqrt{4\sec^2(\theta)-4}\cdot 2\sec(\theta)\tan(\theta)\, d\theta\\[6pt] &= 16\int \sec^4(\theta)\sqrt{4\tan^2(\theta)}\cdot \tan(\theta)\, d\theta\\[6pt] &= 16\int \sec^4(\theta)\cdot2\tan(\theta)\cdot \tan(\theta)\, d\theta\\[6pt] &= 32\int \sec^4(\theta)\tan^2(\theta)\, d\theta \;\; \text{(even power of secant)} \\[6pt] &= 32\int \sec^2(\theta) \cdot \tan^2(\theta)\cdot\sec^2(\theta)\, d\theta\\[6pt] &= 32\int [1 + \tan^2(\theta)] \cdot \tan^2(\theta)\cdot\sec^2(\theta)\, d\theta\\[6pt] \text{Let } \; & u = \tan(\theta), \; du = \sec^2(\theta) \, d\theta.\\[6pt] &= 32\int \left(1 + u^2\right)u^2\, du\\[6pt] &= 32\int \left(u^2 + u^4\right) \, du\\[6pt] &= \frac{32}{3} u^3 + \frac{32}{5} u^5 + C\\[6pt] &= \frac{32}{3} \tan^3(\theta) + \frac{32}{5} \tan^5(\theta) + C \end{align*}$

From the triangle above, we can see that $\tan (\theta ) = \frac {\sqrt {x^2-4}}{2}$ so that: $\int x^3\sqrt {x^2-4} \; dx = \frac {32}{3} \tan ^3(\theta ) + \frac {32}{5} \tan ^5(\theta ) + C = \frac 43 (x^2 - 4)^{3/2} + \frac {1}{5} (x^2 - 4)^{5/2} + C.$

(problem 5) Compute the indefinite integral $\int x^3 \sqrt {x^2 - 16} \; dx$ Use a trigonometric substitution with
$x = \answer {4\sec (\theta )} \;\; dx = \answer {4\sec (\theta )\tan (\theta )} \; d\theta$ 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 $\int \sec ^4(\theta )\tan ^2(\theta ) \; d\theta = \answer {(1/3)\tan ^3(\theta )+ (1/5)\tan ^5(\theta ) } + C.$ Based on the substitution $x = 4\sec (\theta )$ ,
$\tan (\theta ) = \answer {\sqrt {x^2 -16}/4}$ The final answer is $\int x^3\sqrt {x^2 -16}\; dx =$

$\frac {1}{5}(x^2 -16)^{5/2} + \frac {1}{3}(x^2 -16)^{3/2} + C$ $\frac {16}{5}(x^2 -16)^{5/2} + \frac {16}{3}(x^2 -16)^{3/2} + C$ $\frac {16}{5}(x^2 -16)^{5/2} + \frac {1}{3}(x^2 -16)^{3/2} + C$ $\frac {1}{5}(x^2-16)^{5/2} + \frac {16}{3}(x^2-16)^{3/2} + C$

example 6 Compute the indefinite integral $\int \frac {1}{x\sqrt {x^2-16}} \; dx.$ We will use the substitution $x = 4\sec (\theta )$ , so that $dx = 4\sec (\theta )\tan (\theta ) \, d\theta$ .

The integral becomes

$\begin{align*} \int \frac{1}{x\sqrt{x^2-16}}\; dx &= \int \frac{4\sec(\theta)\tan(\theta)}{4\sec(\theta)\sqrt{16\sec^2(\theta)-16}}\; d\theta\\[6pt] &= \int \frac{\tan(\theta)}{\sqrt{16\tan^2(\theta)}}\; d\theta\\[6pt] &= \int \frac{\tan(\theta)}{4\tan(\theta)}\; d\theta\\[6pt] &= \int \frac14 \; d\theta\\[6pt] &= \frac14 \theta + C\\[6pt] &= \frac14 \sec^{-1}\left(\frac{x}{4}\right)+C. \end{align*}$

(problem 6) Compute the indefinite integral $\int \frac {1}{x\sqrt {4x^2 - 1}}\; dx$ Use a trigonometric substitution with
$x = \answer {(1/2)\sec (\theta )} \;\; dx = \answer {(1/2)\sec (\theta )\tan (\theta )} \; d\theta$ Substituting these and simplifying the integrand gives
$\int \frac {1}{x\sqrt {4x^2-1}} \; dx = \int \answer {1} \; d\theta$

Computing this integral gives $\int 1 \; d\theta = \answer {\theta } + C.$ Based on the substitution $x = \frac 12\sec (\theta )$ , the final answer is

$\int \frac {1}{x\sqrt {4x^2 - 1}}\; dx = \answer {\sec ^{-1}(2x)} + C$

example 7 Compute the indefinite integral $\int \frac {1}{x^2\sqrt {9x^2-4}} \; dx.$ We will use the substitution $x = \frac 23\sec (\theta )$ , so that $dx = \frac 23\sec (\theta )\tan (\theta ) \, d\theta$ .

The integral becomes

$\begin{align*} \int \frac{1}{x^2\sqrt{9x^2-4}}\; dx &= \int \frac{(2/3)\sec(\theta)\tan(\theta)}{(2/3)^2\sec^2(\theta)\sqrt{4\sec^2(\theta)-4}}\; d\theta\\[6pt] &= \frac32\int \frac{\tan(\theta)}{\sec(\theta)\sqrt{4\tan^2(\theta)}}\; d\theta\\[6pt] &= \frac32\int \frac{\tan(\theta)}{2\sec(\theta)\tan(\theta)}\; d\theta\\[6pt] &= \frac34 \int \frac{1}{\sec(\theta)} \; d\theta\\[6pt] &= \frac34 \int \cos(\theta) \; d\theta\\[6pt] &= \frac34 \sin(\theta) + C. \end{align*}$

From the triangle above, we can see that $\sin (\theta ) = \frac {\sqrt {9x^2-4}}{3x}$ so that: $\int \frac {1}{x^2\sqrt {9x^2-4}} \; dx = \frac 34 \sin (\theta ) + C = \frac {\sqrt {9x^2 - 4}}{4x} + C.$

(problem 7) Compute the indefinite integral $\int \frac {1}{x^2\sqrt {25x^2-16}} \; dx$

Use a trigonometric substitution with
$x = \answer {4/5 \sec (\theta )} \;\; dx = \answer {4/5 \sec (\theta )\tan (\theta )} \; d\theta$ 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 $\int \cos (\theta ) \; d\theta = \answer {\sin (\theta ) } + C.$ Based on the substitution $x = \frac 45\sec (\theta )$ ,
$\sin (\theta ) = \answer {\sqrt {25x^2 -16}/5x}$ The final answer is $\int \frac {1}{x^2\sqrt {25x^2-16}} \; dx =$

$\frac {16\sqrt {25x^2 - 16}}{5x} + C$ $\frac {5\sqrt {25x^2 - 16}}{16x} + C$ $\frac {\sqrt {25x^2 - 16}}{5x} + C$ $\frac {\sqrt {25x^2 - 16}}{16x} + C$

example 8 Compute the indefinite integral $\int \frac {1}{x^3\sqrt {x^2-25}} \; dx.$ We will use the substitution $x = 5\sec (\theta )$ , so that $dx = 5\sec (\theta )\tan (\theta ) \, d\theta$ .

The integral becomes

$\begin{align*} \int \frac{1}{x^3\sqrt{x^2-25}}\; dx &= \int \frac{5\sec(\theta)\tan(\theta)}{5^3\sec^3(\theta)\sqrt{25\sec^2(\theta)-25}}\; d\theta\\[6pt] &= \frac{1}{25}\int \frac{\tan(\theta)}{\sec^2(\theta)\sqrt{25\tan^2(\theta)}}\; d\theta\\[6pt] &= \frac{1}{25}\int \frac{\tan(\theta)}{5\sec^2(\theta)\tan(\theta)}\; d\theta\\[6pt] &= \frac{1}{125} \int \frac{1}{\sec^2(\theta)} \; d\theta\\[6pt] &= \frac{1}{125} \int \cos^2(\theta) \; d\theta\\[6pt] &= \frac{1}{125} \int \frac{1+\cos(2\theta)}{2} \; d\theta\\[6pt] &= \frac{1}{250}\left[\theta + \frac12 \sin(2\theta)\right] + C\\[6pt] &= \frac{1}{250} \theta + \frac{1}{500} \cdot 2\sin(\theta)\cos(\theta) + C. \end{align*}$

From the triangle above, we can see that $\sin (\theta ) = \frac {\sqrt {x^2-25}}{x}$ so that:

$\begin{align*} \int \frac{1}{x^3\sqrt{x^2-25}} \; dx &= \frac{\theta}{250} + \frac{\sin(\theta)\cos(\theta)}{250} + C \\ &= \frac{1}{250} \sec^{-1}\left(\frac{x}{5}\right) + \frac{1}{250} \cdot \frac{\sqrt{x^2 - 25}}{x}\cdot \frac{5}{x} + C\\ &= \frac{1}{250} \sec^{-1}\left(\frac{x}{5}\right) + \frac{\sqrt{x^2 - 25}}{50x^2}+ C. \end{align*}$

(problem 8) Compute the indefinite integral $\int \frac {1}{x^3\sqrt {x^2-1}} \; dx$ Use a trigonometric substitution with
$x = \answer {\sec (\theta )} \;\; dx = \answer { \sec (\theta )\tan (\theta )} \; d\theta$ 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 $\cos ^2(\theta ) = \frac 12 + \frac 12 \cos (2\theta )$

$\int \cos ^2(\theta ) \; d\theta = \answer {\frac 12 \theta + \frac 14 \sin (2\theta ) } + C.$

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

$\frac 12 \theta + \frac 12 \answer {\sin (\theta ) \cos (\theta )} + C.$

Based on the substitution $x = \sec (\theta )$ ,
$\sin (\theta ) = \answer {\sqrt {x^2 -1}/x} \;\; \text {and} \;\; \cos (\theta ) = \answer {1/x}$ The final answer is $\int \frac {1}{x^3\sqrt {x^2-1}} \; dx =$

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

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

example 9 Compute the indefinite integral: $\int x^3\sqrt {1+x^2} \; dx.$ We will use the substitution $x = \tan (\theta )$ . Then the differential, $dx = \sec ^2(\theta ) \, d\theta$ . Substituting these gives $\begin{align*} \int x^3\sqrt{1+x^2} \; dx &= \int \tan^3(\theta) \sqrt{1+\tan^2(\theta)} \cdot \sec^2(\theta) \; d\theta\\[6pt] &= \int \tan^3(\theta) \sqrt{\sec^2(\theta)} \cdot \sec^2(\theta) \; d\theta\\[6pt] &= \int \tan^3(\theta) \cdot \sec(\theta) \cdot \sec^2(\theta) \; d\theta\\[6pt] &= \int \tan^3(\theta) \sec^3(\theta) \; d\theta\\[6pt] \text{u-sub with } \; & u=\sec(\theta), \, du = \sec(\theta) \tan(\theta) d\theta\\[6pt] &= \int \tan^2(\theta) \sec^2(\theta) \cdot \sec(\theta) \tan(\theta)\; d\theta\\[6pt] &= \int \left(\sec^2(\theta)-1 \right) \sec^2(\theta) \cdot \sec(\theta) \tan(\theta)\; d\theta\\[6pt] &= \int (u^2 -1)u^2 \; du\\[6pt] &= \int (u^4 -u^2) \; du\\[6pt] &\text{now we can integrate}\\[6pt] &= \frac{u^5}{5} - \frac{u^3}{3} + C\\[6pt] &= \frac15 \sec^5(\theta) - \frac13 \sec^3(\theta) + C. \end{align*}$

From the triangle above, we see that $\sec (\theta )= \sqrt {1+x^2}$ . Back-substituting gives $\int x^3\sqrt {1+x^2} \; dx = \frac 15 \sec ^5(\theta ) - \frac 13 \sec ^3(\theta ) + C = \frac 15 (1+x^2)^{5/2} - \frac 13 (1+x^2)^{3/2} + C$

Technically, $\sqrt {\sec ^2(\theta )} = |\sec (\theta )|$ . However, in our substitution, $x = \tan (\theta )$ , we can specify that the angle $\theta$ be in the interval $(-\pi /2, \pi /2)$ . This way, $\sec (\theta ) > 0$ and $\sqrt {\sec ^2(\theta )} = |\sec (\theta )| = \sec (\theta )$ .

(problem 9) Compute the indefinite integral $\int x^3\sqrt {9 + x^2}\; dx.$ Use a trigonometric substitution with
$x = \answer {3\tan (\theta )}.$ The differential is
$dx = \answer {3\sec ^2(\theta )} \; d\theta$ Substituting these and simplifying the integrand gives
$\int x^3\sqrt {9 + x^2}\; dx = 243\int \answer {\tan ^3(\theta )\sec ^3(\theta )} d\theta$ Computing this integral gives $\int \tan ^3(\theta )\sec ^3(\theta ) \; d\theta = \answer {(1/5)\sec ^5(\theta ) - (1/3)\sec ^3(\theta )} + C.$ Based on the substitution $x = 3\tan (\theta )$ ,
$\sec (\theta ) = \answer {\sqrt {9+x^2}/3}$ The final answer is $\int x^3\sqrt {9 + x^2}\; dx =$

$\frac {1}{5}(9+x^2)^{5/2} - 3(9+x^2)^{3/2} + C$ $\frac {1}{5}(9+x^2)^{5/2} + 3(9+x^2)^{3/2} + C$ $\frac {1}{3}(9+x^2)^{3/2} - (9+x^2)^{5/2} + C$ $\frac {1}{3}(9+x^2)^{3/2} + (9+x^2)^{5/2} + C$

example 10 Compute the indefinite integral $\int \frac {x^3}{\sqrt {9 + x^2}} \; dx.$ We will let $x = 3\tan (\theta )$ . Then the differential $dx = 3\sec ^2(\theta ) \, d\theta$ . Substituting these, we get $\begin{align*} \int \frac{x^3}{\sqrt{9 + x^2}} \; dx &= \int \frac{27\tan^3(\theta)}{\sqrt{9 + 9\tan^2(\theta)}} \cdot 3\sec^2(\theta) \; d\theta\\[6pt] &= \int \frac{27\tan^3(\theta)}{\sqrt{9\sec^2(\theta)}} \cdot 3\sec^2(\theta) \; d\theta\\[6pt] &= \int \frac{27\tan^3(\theta)}{3\sec(\theta)} \cdot 3\sec^2(\theta) \; d\theta\\[6pt] &= 27\int \tan^3(\theta) \sec(\theta) \; d\theta\\[6pt] & \text{let} \; u = \sec(\theta), \, du = \sec(\theta) \tan(\theta) \, d\theta\\[6pt] &= 27\int \tan^2(\theta) \cdot \sec(\theta) \tan(\theta) \; d\theta\\[6pt] &= 27\int \left(\sec^2(\theta)-1\right) \cdot \sec(\theta) \tan(\theta) \; d\theta\\[6pt] &= 27\int (u^2-1) \; du\\[6pt] &= 9u^3 - \frac{u}{27} + C\\[6pt] &= 9\sec^3(\theta) - 27 \sec(\theta) + C. \end{align*}$

From the triangle above, we can see that $\sec (\theta ) = \frac {\sqrt {9+x^2}}{3}$ and so

$\int \frac {x^3}{\sqrt {9 + x^2}} \; dx = 9\sec ^3(\theta ) - 27 \sec (\theta ) + C = \frac 13 (9+x^2)^{3/2} - 9(9+x^2)^{1/2} + C.$

(problem 10) Compute the indefinite integral $\int \frac {x^3}{\sqrt {4 + x^2}} \; dx.$ Use a trigonometric substitution with
$x = \answer {2\tan (\theta )}$ The differential is
$dx = \answer {2\sec ^2(\theta )} \; d\theta$ Substituting these and simplifying the integrand gives
$\int \frac {x^3}{\sqrt {4 + x^2}} \; dx = 8\int \answer {\tan ^3(\theta )\sec (\theta )} d\theta$ Computing this integral gives $\int \tan ^3(\theta )\sec (\theta ) \; d\theta = \answer {(1/3)\sec ^3(\theta ) - \sec (\theta )} + C.$ Based on the substitution $x = 2\tan (\theta )$ ,
$\sec (\theta ) = \answer {\sqrt {4+x^2}/2}$ The final answer is $\int \frac {x^3}{\sqrt {4 + x^2}} \; dx =$

$\frac {1}{3}(4+x^2)^{3/2} + 4(4+x^2)^{1/2} + C$ $\frac {1}{3}(4+x^2)^{3/2} - 4(4+x^2)^{1/2} + C$ $\frac {1}{3}(4+x^2)^{1/2} + 4(4+x^2)^{3/2} + C$ $\frac {1}{3}(4+x^2)^{1/2} - 4(4+x^2)^{3/2} + C$

example 11 Compute the indefinite integral $\int \sqrt {16 + 9x^2} \; dx.$ Let $x = \frac 43 \tan (\theta )$ . Then $dx = \frac 43 \sec ^2(\theta ) \, d\theta$ . Substituting these, the integral becomes, $\begin{align*} \int \sqrt{16 + 9x^2} \; dx &= \int \sqrt{16+16\tan^2(\theta)} \cdot \frac43\sec^2(\theta) \; d\theta\\[6pt] &= \frac43\int \sqrt{16\sec^2(\theta)} \cdot \sec^2(\theta) \; d\theta\\[6pt] &= \frac43\int 4\sec(\theta) \cdot \sec^2(\theta) \; d\theta\\[6pt] &= \frac{16}{3}\int \sec^3(\theta) \; d\theta\\[6pt] &\;\;\;\;\;\text{(see the IBP section for this integral).}\\[6pt] &= \frac{8}{3}\sec(\theta) \tan(\theta) + \frac{8}{3} \ln|\sec(\theta) + \tan(\theta)| + C \end{align*}$

From the triangle above, we can see that $\sec (\theta ) = \frac {\sqrt {16+9x^2}}{4 }$ . Back substitution gives

$\begin{align*} \int \sqrt{16 + 9x^2} \; dx &= \frac{8}{3} \sec(x) \tan(x) + \frac{8}{3} \ln|\sec(x) + \tan(x)| + C\\ &= \frac{8}{3} \cdot\frac{\sqrt{16+9x^2}}{4 } \cdot \frac{3x}{4} + \frac{8}{3} \ln\bigg|\frac{\sqrt{16+9x^2}}{4 } + \frac{3x}{4}\bigg| + C\\ &= \frac12 x\sqrt{16+9x^2} + \frac83 \ln\left(3x + \sqrt{16+9x^2}\right) + C\\ \end{align*}$ Note that the absolute value bars were removed because $3x + \sqrt {16+9x^2} > 0$ for any $x$ . Also note that the $1/4$ was removed from inside the log because $\ln (u/v) = \ln (u) - \ln (v)$ and $-\frac 83\ln (4)$ can be combined with the constant of integration.

(problem 11) Compute the indefinite integral $\int \sqrt {9 + 49x^2} \; dx.$ Use a trigonometric substitution with
$x = \answer {(3/7)\tan (\theta )}$ The differential is
$dx = \answer {(3/7)\sec ^2(\theta )} \; d\theta$ Substituting these and simplifying the integrand gives
$\int \sqrt {9 + 49x^2} \; dx = \frac 97 \int \answer {\sec ^3(\theta )} d\theta$ Computing this integral gives $\int \sec ^3(\theta ) \; d\theta = \answer {(1/2)\sec (\theta ) \tan (\theta ) + (1/2)\ln |\sec (\theta )+\tan (\theta )|} + C.$ Based on the substitution $x = \frac 37 \tan (\theta )$ ,
$\sec (\theta ) = \answer {\sqrt {9+49x^2}/3}$ The final answer is $\int \frac {x^3}{\sqrt {4 + x^2}} \; dx =$

$\frac 12 x\sqrt {9+49x^2} + \frac {3}{14}\ln \left (7x + \sqrt {9+49x^2}\right ) + C$ $\frac 12 x\sqrt {9+49x^2} - \frac {9}{14}\ln \left (7x + \sqrt {9+49x^2}\right ) + C$ $\frac 12 x\sqrt {9+49x^2} + \frac {9}{14}\ln \left (7x+\sqrt {9+49x^2}\right ) + C$ $\frac 12 x\sqrt {9+49x^2} - \frac {3}{14}\ln |7x - \sqrt {9+49x^2}| + C$

Here is a detailed, lecture style video on trigonometric substitution:

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 $x = \answer {2\sec \theta }$
The resulting trig integral is

$32\dis \int \sec ^4\theta \tan ^2\theta \; d\theta$ $16\dis \int \sec ^4\theta \tan ^2\theta \; d\theta$
$32\dis \int \sec ^3\theta \tan \theta \; d\theta$ $16\dis \int \sec ^3\theta \tan \theta \; d\theta$

Trigonometric identities

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

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

Integrands involving \sqrt {a^2 + u^2}

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

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

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