VideoBlog Matemático: Integral Sustitución Trigonométrica 02

Ejemplos Resueltos:
Ejemplo 01

1.	$\displaystyle {\int \frac{dx}{\sqrt{4+x^{2}}}}$

Sea $\displaystyle {x = 2\;tan\;\theta, \theta \varepsilon \left]\frac{-\Pi}{2}, \frac{\Pi}{2}\right[}$
$\displaystyle {dx = 2\;sec^{2}\theta\;d\theta}$
Luego: $\displaystyle {4+x^{2} = 4+4\;tan^{2}\theta = 4(1 + tan^{2}\theta)}$
$\displaystyle {4+x^{2} = 4\;sec^{2}\theta }$
$\displaystyle {\sqrt{4+x^{2}} = \sqrt{4\;sec^{2}\theta} = \vert 2\;sec\;\theta\vert = 2\;sec\;\Theta}$
Sustituyendo
$\displaystyle {\int \frac{dx}{\sqrt{4+x^{2}}} = \int \frac{2\;sec^{2}\theta\;d\theta}{2\;sec\;\theta} = \int sec\;\theta\;d\theta}$
$\displaystyle {ln\;\vert sec\;\theta+ tan\;\theta\vert + C}$
$\displaystyle {\int \frac{dx}{\sqrt{4+x^{2}}} = ln\left\vert \frac{\sqrt{4+x^{2}}}{2} + \frac{x}{2}\right\vert + C}$

Ejemplo 02

2.	$\displaystyle {\int \frac{x^{2}\;dx}{\sqrt{x^{2}+6}}}$

Sea $\displaystyle {x = \sqrt{6}\;tan\;\theta, \theta \varepsilon \left]\frac{-\Pi}{2}, \frac{\Pi}{2}\right[}$

$\displaystyle {dx = \sqrt{6}\;sec^{2}\theta\;d\theta}$

Luego: $\displaystyle {x^{2} + 6 = 6\;tan^{2}\theta + 6 = 6(tan^{2}\theta + 1) = 6\;sec^{2}\theta}$

$\displaystyle {\sqrt{x^{2}+6} = \sqrt{6\;sec^{2}\theta} = \sqrt{6}\;sec\;\theta... ...heta>0 si \theta \varepsilon \left]\frac{-\Pi}{2}, \frac{\Pi}{2}\right[\right)}$

Sustituyendo

$\displaystyle {\int \frac{x^{2}}{\sqrt{^{2}+6}}\;dx = \int \frac{6\;tan^{2}\the... ...a\;d\theta}{\sqrt{6}\;sec\;\theta} = 6\int tan^{2}\theta\;sec\;\theta\;d\theta}$

$\displaystyle {= 6 \int (sec^{2}\theta - 1)\;sec\;\theta\;d\theta = 6\int(sec^{3} - sec\;\theta)\;d\theta}$

$\displaystyle {= 6 \left[\frac{1}{2}(sec\;\theta\;tan\;\theta) + ln\;\vert sec\... ...eta + tan\;\theta\vert\right] -6\;ln\;\vert sec\;\theta + tan\;\theta\vert + C}$

$\displaystyle {= 3 sec\;\theta\;tan\;\theta - 3\;ln\;\vert sec\;\theta + tan\;\theta\vert + C}$

$\displaystyle {= 3 \cdot \frac{\sqrt{x^{2}+6}}{\sqrt{6}}\cdot \frac{x}{\sqrt{6}... ...;\left\vert\frac{\sqrt{x^{2}+6}}{\sqrt{6}} + \frac{x}{\sqrt{6}}\right\vert + C}$

$\displaystyle {= \frac{x\sqrt{x^{2}+6}}{2} - 3\;ln\;\left\vert\frac{\sqrt{x^{2}+6} + x}{\sqrt{6}}\right\vert + C}$

Ejemplo 03

3.	$\displaystyle {\int \frac{x\;dx}{(9+4x^{2})^{\frac{3}{2}}}}$

Sea $\displaystyle {x = \frac{3}{2}\;tan\;\theta, \theta \varepsilon \left]\frac{-\Pi}{2}, \frac{\Pi}{2}\right[}$

$\displaystyle {dx = \frac{3}{2}\;sec^{2}\theta\;d\theta}$

Luego $\displaystyle {9 + 4x^{2} = 9 + 4\cdot \frac{9}{4}\;tan^{2}\theta = 9 + 9\;tan^{2}\theta = 9(1 + tan^{2}\theta)}$

$\displaystyle {9 + 4x^{2} = 9\;sec^{2}\theta}$

$\displaystyle {(9 + 4x^{2})^{\frac{3}{2}} = (9\;sec^{2}\theta)^{\frac{3}{2}} = (9\;sec^{2}\theta)^{3}}$

$\displaystyle {(9 + 4x^{2})^{\frac{3}{2}} = (3\;sec\;\theta)^{3} = 27\;sec^{3}\theta}$

Sustituyendo

$\displaystyle {\int \frac{x\;dx}{(9+4x^{2})^{\frac{3}{2}}} = \int \frac{\frac{3... ...}\theta}\;d\theta = \frac{1}{12} \int \frac{tan\;\theta\;d\theta}{sec\;\theta}}$

$\displaystyle {= \frac{1}{12} \int \frac{\frac{sen\;\theta}{cos\;\theta}}{\frac... ...heta = \frac{1}{12} \int sen\;\theta\;d\theta = \frac{1}{12}(-cos\;\theta) + C}$

Como

$\displaystyle {tan\;\theta = \frac{2x}{3}}$ de la sustitución inicial

Por tanto:

$\displaystyle {\int \frac{x\;dx}{(9+4x^{2})^{\frac{3}{2}}} = \frac{-1}{12} \cdot\frac{3}{\sqrt{9+4x^{2}}} + C}$

$\displaystyle {= \frac{-1}{4\sqrt{9+4x^{2}}} + C }$

Ejemplo 04

4.	$\displaystyle {\int \frac{dx}{x^{4}\sqrt{x^{2}+3}}}$

Sea $\displaystyle {x = \sqrt{3}\;tan\;\theta, \theta \varepsilon \left]\frac{-\Pi}{2}, \frac{\Pi}{2}\right[}$

$\displaystyle {dx = \sqrt{3}\;sec^{2}\theta\;d\theta}$

Luego $\displaystyle {x^{2} + 3 = 3\;tan^{2}\theta + 3 = 3(tan^{2}\theta + 1) = 3\;sec^{2}\theta}$

$\displaystyle {\sqrt{x^{2}+3} = \sqrt{3\;sec^{2}\theta} = \sqrt{3}\;sec\;\theta}$

Sustituyendo

$\displaystyle {\int \frac{dx}{x^{4}\sqrt{x^{2}+3}} = \int \frac{\sqrt{3}\;sec^{... ...{3}\;sec\;\theta} = \frac{1}{9}\int \frac{sec\;\theta\;d\theta}{tan^{4}\theta}}$

$\displaystyle {= \frac{1}{9} \int \frac{cos^{4}\theta}{cos\;\theta\cdot sen^{4}\theta}\;d\theta = \frac{1}{9}\int \frac{cos^{3}\theta}{sen^{4}\theta}\;d\theta}$

$\displaystyle {= \frac{1}{9} \int \frac{(1-sen^{2}\theta)cos\;\theta}{sen^{4}\t... ...n^{4}\theta}- \frac{sen^{2}\theta\;cos\;\theta}{sen^{4}\theta}\right)\;d\theta}$

$\displaystyle {= \frac{1}{9} \int cos\;\theta(sen\;\theta)^{-4}\;d\theta - \frac{1}{9}\int cos\;\theta(sen\;\theta)^{-2}\;d\theta}$

$\displaystyle {= \frac{1}{9}\;\frac{(sen\;\theta)^{-3}}{-3} - \frac{1}{9}\;\frac{(sen\;\theta)^{-1}}{-1} + C}$

$\displaystyle {= \frac{-1}{27\;sen^{3}\theta} + \frac{csc\;\theta}{9} + C}$

Como $\displaystyle {x = \sqrt{3}\;tan\;\theta}$ entonces $\displaystyle {tan\;\theta = \frac{x}{3}}$

Por lo que: