How do we solve an integral using trigonometric substitution?

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How do we solve an integral using trigonometric substitution?

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Answer 1 · 2018-05-25T12:39:18

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:

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

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How do we solve an integral using trigonometric substitution?

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