Integration by Trigonometric Substitution

Key Questions

When integrating by trigonometric substitution, what are some useful identities to know?

Useful Trigonometric Identities

${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2theta+sin^2theta=1$

$1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2theta=sec^2theta$

$sin 2 θ = 2 sin θ cos θ$ $sin2theta=2sin theta cos theta$

$cos 2 θ = {cos}^{2} θ - {sin}^{2} θ = 2 {cos}^{2} θ - 1 = 1 - 2 {sin}^{2} θ$ $cos2theta=cos^2theta-sin^2theta=2cos^2theta-1=1-2sin^2theta$

${cos}^{2} θ = \frac{1}{2} (1 + cos 2 θ)$ $cos^2theta=1/2(1+cos2theta)$

${sin}^{2} θ = \frac{1}{2} (1 - cos 2 θ)$ $sin^2theta=1/2(1-cos2theta)$

I hope that this was helpful.

Wataru · · Oct 18 2014
How do you find the integral $\int \frac{d t}{\sqrt{t^{2} - 6 t + 13}}$ $intdt/(sqrt(t^2-6t+13))$ ?

By completing the square,
$t^{2} - 6 t + 13 = (t^{2} - 6 t + 9) + 4 = {(t - 3)}^{2} + 2^{2}$ $t^2-6t+13=(t^2-6t+9)+4=(t-3)^2+2^2$

So, we can rewrite the integral as
$\int \frac{d t}{\sqrt{{(t - 3)}^{2} + 2^{2}}}$ $int{dt}/{sqrt{(t-3)^2+2^2}}$

Let $t - 3 = 2 tan θ$ $t-3=2tan theta$ .
$\Rightarrow \frac{d t}{d θ} = 2 {sec}^{2} θ \Rightarrow d t = 2 {sec}^{2} θ d θ$ $Rightarrow {dt}/{d theta}=2sec^2 theta Rightarrow dt=2sec^2 theta d theta$

by the above substitution,
$= \int \frac{2 {sec}^{2} θ d θ}{\sqrt{{(2 tan θ)}^{2} + 2^{2}}} = \int \frac{{sec}^{2} θ}{\sqrt{{tan}^{2} θ + 1}} d θ$ $=int{2sec^2theta d theta}/{sqrt{(2tan theta)^2+2^2)} =int{sec^2theta}/{sqrt{tan^2theta+1}}d theta$
by the identity ${tan}^{2} θ + 1 = {sec}^{2} θ$ $tan^2theta+1=sec^2theta$ ,
$= \int \frac{{sec}^{2} θ}{\sqrt{{sec}^{2} θ}} d θ = \int sec θ d θ = ln | sec θ + tan θ | + C_{1}$ $=int{sec^2theta}/{sqrt{sec^2theta}}d theta=int sec theta d theta=ln|sec theta + tan theta|+C_1$
by using
$tan θ = \frac{t - 3}{2}$ $tan theta={t-3}/2$ and $sec θ = \sqrt{{tan}^{2} θ + 1} = \frac{\sqrt{t^{2} - 6 + 13}}{2}$ $sec theta=sqrt{tan^2theta+1}={sqrt{t^2-6+13}}/2$ ,
we have
$= ln ∣ ∣ ∣ \frac{\sqrt{t^{2} - 6 + 13}}{2} + \frac{t - 3}{2} ∣ ∣ ∣ + C_{1}$ $=ln|{sqrt{t^2-6+13}}/2+{t-3}/2|+C_1$
$= ln ∣ ∣ ∣ \frac{\sqrt{t^{2} - 6 t + 13} + t - 3}{2} ∣ ∣ ∣ + C_{1}$ $=ln|{sqrt{t^2-6t+13}+t-3}/2|+C_1$
by log property,
$= ln ∣ ∣ \sqrt{t^{2} - 6 t + 13} + t - 3 ∣ ∣ - ln 2 + C_{1}$ $=ln|sqrt{t^2-6t+13}+t-3|-ln2+C_1$
by setting $C = ln 2 + C_{1}$ $C=ln2+C_1$ ,
$= ln ∣ ∣ \sqrt{t^{2} - 6 t + 13} + t - 3 ∣ ∣ + C$ $=ln|sqrt{t^2-6t+13}+t-3|+C$

Wataru · 2 · Sep 10 2014
How do we solve an integral using trigonometric substitution?

In general trigonometric substitutions are useful to solve the integrals of algebraic functions containing radicals in the form $\sqrt{x^{2} \pm a^{2}}$ $sqrt(x^2+-a^2)$ or $\sqrt{a^{2} \pm x^{2}}$ $sqrt(a^2+-x^2)$ . Consider the different cases:

A. Let $f (x)$ $f(x)$ be a rational function of $x$ $x$ and $\sqrt{x^{2} + a^{2}}$ $sqrt(x^2+a^2)$ :

$\int f (x) d x = \int R (x, \sqrt{x^{2} + a^{2}}) d x$ $int f(x)dx = int R(x, sqrt(x^2+a^2))dx$

Substitute: $x = a tan t$ $x = atant$ , $d x = a {sec}^{2} t d t$ $dx = asec^2tdt$ with $t \in (- \frac{π}{2}, \frac{π}{2})$ $t in (-pi/2,pi/2)$ and use the trigonometric identity:

$1 + {tan}^{2} t = {sec}^{2} t$ $1+tan^2t = sec^2t$

Considering that for $t \in (- \frac{π}{2}, \frac{π}{2})$ $t in (-pi/2,pi/2)$ the secant is positive:

$\sqrt{x^{2} + a^{2}} = \sqrt{a^{2} {tan}^{2} t + a^{2}}$ $sqrt(x^2+a^2)= sqrt(a^2tan^2t+a^2)$

$\sqrt{x^{2} + a^{2}} = a \sqrt{{tan}^{2} t + 1} = a sec t$ $sqrt(x^2+a^2)=asqrt(tan^2t+1) = asect$

Then:

$\int f (x) d x = \int R (a tan t, a sec t) {sec}^{2} t d t$ $int f(x)dx = int R(atant, asect)sec^2t dt$

B. Let $f (x)$ $f(x)$ be a rational function of $x$ $x$ and $\sqrt{x^{2} - a^{2}}$ $sqrt(x^2-a^2)$ :

$\int f (x) d x = \int R (x, \sqrt{x^{2} - a^{2}}) d x$ $int f(x)dx = int R(x, sqrt(x^2-a^2))dx$

Restrict the function to $x \in (a, + \infty)$ $x in (a,+oo)$ and substitute: $x = a sec t$ $x = asect$ , $d x = a sec t tan t d t$ $dx = asect tantdt$ with $t \in (0, \frac{π}{2})$ $t in (0,pi/2)$ and use the trigonometric identity:

${sec}^{2} t - 1 = {tan}^{2} t$ $sec^2t-1 = tan^2t$

Considering that for $t \in (0, \frac{π}{2})$ $t in (0,pi/2)$ the tangent is positive:

$\sqrt{x^{2} - a^{2}} = \sqrt{a^{2} {sec}^{2} t - a^{2}}$ $sqrt(x^2-a^2)= sqrt(a^2sec^2t-a^2)$

$\sqrt{x^{2} - a^{2}} = a \sqrt{{sec}^{2} t - 1} = a tan t$ $sqrt(x^2-a^2)=asqrt(sec^2t-1) = atant$

Then:

$\int f (x) d x = \int R (a sec t, a tan t) sec t tan t d t$ $int f(x)dx = int R(asect, atant)sect tant dt$

Normally you can see by differentiation that the solution that is found is valid also for $x \in (- \infty, - a)$ $x in (-oo, -a)$

C. Let $f (x)$ $f(x)$ be a rational function of $x$ $x$ and $\sqrt{a^{2} - x^{2}}$ $sqrt(a^2-x^2)$ :

$\int f (x) d x = \int R (x, \sqrt{a^{2} - x^{2}}) d x$ $int f(x)dx = int R(x, sqrt(a^2-x^2))dx$

Substitute: $x = a sin t$ $x = a sint$ , $d x = a cos t$ $dx = a cost$ with $t \in (- \frac{π}{2}, \frac{π}{2})$ $t in (-pi/2,pi/2)$ and use the trigonometric identity:

$1 - {sin}^{2} t = {cos}^{2} t$ $1-sin^2t = cos^2t$

Considering that for $t \in (- \frac{π}{2}, \frac{π}{2})$ $t in (-pi/2,pi/2)$ the cosine is positive:

$\sqrt{a^{2} - x^{2}} = \sqrt{a^{2} - a^{2} {sin}^{2} t}$ $sqrt(a^2-x^2)= sqrt(a^2-a^2sin^2t)$

$\sqrt{a^{2} - x^{2}} = a \sqrt{1 - {sin}^{2} t} = a cos t$ $sqrt(a^2-x^2)=a sqrt(1-sin^2t) = acost$

Then:

$\int f (x) d x = \int R (a sin t, a cos t) cos t d t$ $int f(x)dx = int R(asint, acost)cost dt$

Andrea S. · 1 · May 25 2018

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Integration by Trigonometric Substitution

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