integral de 0 a 1 de sqrt(1+(\sqrt{2-x^2))^2}

Solución

\int_{0}^{1} 1 + (2 - x^{2})^{2} d x

\frac{3 ( 2 arcsin ( \frac{1}{3} ) + sin ( 2 arcsin ( \frac{1}{3} ) ) )}{4}

Notación decimal

1.63032 \dots

Pasos de solución

\int_{0}^{1} 1 + (2 - x^{2})^{2} d x

Simplificar

1 + (2 - x^{2})^{2} : - x^{2} + 3

= \int_{0}^{1} - x^{2} + 3 d x

Aplicar integración por sustitución trigonométrica

: \int_{0}^{a r c s i n (\frac{1}{3})} 3 cos^{2} (u) d u

= \int_{0}^{a r c s i n (\frac{1}{3})} 3 cos^{2} (u) d u

Sacar la constante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int_{0}^{a r c s i n (\frac{1}{3})} cos^{2} (u) d u

Re-escribir usando identidades trigonométricas

= 3 \cdot \int_{0}^{a r c s i n (\frac{1}{3})} \frac{1 + cos ( 2 u )}{2} d u

Sacar la constante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \frac{1}{2} \cdot \int_{0}^{a r c s i n (\frac{1}{3})} 1 + cos (2 u) d u

Aplicar la regla de la suma:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 \cdot \frac{1}{2} (\int_{0}^{a r c s i n (\frac{1}{3})} 1 d u + \int_{0}^{a r c s i n (\frac{1}{3})} cos (2 u) d u)

\int_{0}^{a r c s i n (\frac{1}{3})} 1 d u = arcsin (\frac{1}{3})

\int_{0}^{a r c s i n (\frac{1}{3})} cos (2 u) d u = \frac{sin ( 2 arcsin ( \frac{1}{3} ) )}{2}

= 3 \cdot \frac{1}{2} arcsin (\frac{1}{3}) + \frac{sin ( 2 arcsin ( \frac{1}{3} ) )}{2}

Simplificar

3 \cdot \frac{1}{2} arcsin (\frac{1}{3}) + \frac{sin ( 2 arcsin ( \frac{1}{3} ) )}{2} : \frac{3 ( 2 arcsin ( \frac{1}{3} ) + sin ( 2 arcsin ( \frac{1}{3} ) ) )}{4}

= \frac{3 ( 2 arcsin ( \frac{1}{3} ) + sin ( 2 arcsin ( \frac{1}{3} ) ) )}{4}

\int arcsin (9 x) d x

y^{^{'}} + 8 y - 6 x = 0

y^{^{'}} = \frac{y}{y - x}

n = 1 \sum \infty n^{- 2.4}

\int \frac{1}{1 + y ^{2}} d y