Ii
CAL1 • Integrales
CAL1_INT_239
Guía de Cálculo II
Enunciado
Evaluar:
$$ \int \frac{dx}{(1+x^{2})\sqrt{1-x^{2}}} $$
$$ \int \frac{dx}{(1+x^{2})\sqrt{1-x^{2}}} $$
Solución Paso a Paso
1. Sustitución: Sea $x = \sin(\theta)$, entonces $dx = \cos(\theta)d\theta$.
El radical se convierte en $\sqrt{1-\sin^{2}(\theta)} = \cos(\theta)$.
2. Transformación:
$$ \int \frac{\cos(\theta)d\theta}{(1+\sin^{2}(\theta))\cos(\theta)} = \int \frac{d\theta}{1+\sin^{2}(\theta)} $$
Dividimos numerador y denominador por $\cos^{2}(\theta)$:
$$ \int \frac{\sec^{2}(\theta)d\theta}{\sec^{2}(\theta) + \tan^{2}(\theta)} = \int \frac{\sec^{2}(\theta)d\theta}{1+2\tan^{2}(\theta)} $$
3. Integración: Sea $u = \sqrt{2}\tan(\theta)$, $du = \sqrt{2}\sec^{2}(\theta)d\theta$:
$$ \frac{1}{\sqrt{2}} \int \frac{du}{1+u^{2}} = \frac{1}{\sqrt{2}} \arctan(u) + C $$
Regresando variables:
$$ \boxed{\frac{1}{\sqrt{2}} \arctan\left(\frac{\sqrt{2}x}{\sqrt{1-x^{2}}}\right) + C} $$
El radical se convierte en $\sqrt{1-\sin^{2}(\theta)} = \cos(\theta)$.
2. Transformación:
$$ \int \frac{\cos(\theta)d\theta}{(1+\sin^{2}(\theta))\cos(\theta)} = \int \frac{d\theta}{1+\sin^{2}(\theta)} $$
Dividimos numerador y denominador por $\cos^{2}(\theta)$:
$$ \int \frac{\sec^{2}(\theta)d\theta}{\sec^{2}(\theta) + \tan^{2}(\theta)} = \int \frac{\sec^{2}(\theta)d\theta}{1+2\tan^{2}(\theta)} $$
3. Integración: Sea $u = \sqrt{2}\tan(\theta)$, $du = \sqrt{2}\sec^{2}(\theta)d\theta$:
$$ \frac{1}{\sqrt{2}} \int \frac{du}{1+u^{2}} = \frac{1}{\sqrt{2}} \arctan(u) + C $$
Regresando variables:
$$ \boxed{\frac{1}{\sqrt{2}} \arctan\left(\frac{\sqrt{2}x}{\sqrt{1-x^{2}}}\right) + C} $$