Integral de $$$\frac{9}{\sqrt{1 - x^{2}}}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \frac{9}{\sqrt{1 - x^{2}}}\, dx$$$.
Solución
Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=9$$$ y $$$f{\left(x \right)} = \frac{1}{\sqrt{1 - x^{2}}}$$$:
$${\color{red}{\int{\frac{9}{\sqrt{1 - x^{2}}} d x}}} = {\color{red}{\left(9 \int{\frac{1}{\sqrt{1 - x^{2}}} d x}\right)}}$$
La integral de $$$\frac{1}{\sqrt{1 - x^{2}}}$$$ es $$$\int{\frac{1}{\sqrt{1 - x^{2}}} d x} = \operatorname{asin}{\left(x \right)}$$$:
$$9 {\color{red}{\int{\frac{1}{\sqrt{1 - x^{2}}} d x}}} = 9 {\color{red}{\operatorname{asin}{\left(x \right)}}}$$
Por lo tanto,
$$\int{\frac{9}{\sqrt{1 - x^{2}}} d x} = 9 \operatorname{asin}{\left(x \right)}$$
Añade la constante de integración:
$$\int{\frac{9}{\sqrt{1 - x^{2}}} d x} = 9 \operatorname{asin}{\left(x \right)}+C$$
Respuesta
$$$\int \frac{9}{\sqrt{1 - x^{2}}}\, dx = 9 \operatorname{asin}{\left(x \right)} + C$$$A