integral de cos^7(5x)\sqrt[5]{sin(5x)}

Solución

\frac{1}{6} sin^{\frac{6}{5}} (5 x) - \frac{3}{16} sin^{\frac{16}{5}} (5 x) + \frac{3}{26} sin^{\frac{26}{5}} (5 x) - \frac{1}{36} sin^{\frac{36}{5}} (5 x) + C

Pasos de solución

Aplicar las leyes de los exponentes:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{5} = 5^{- 1}

= \int cos^{7} (5 x) sin^{5^{- 1}} (5 x) d x

Aplicar integración por sustitución

Sacar la constante:

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

Re-escribir usando identidades trigonométricas

Aplicar integración por sustitución

Expandir

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

\int 3 v^{\frac{11}{5}} d v = \frac{15}{16} v^{\frac{16}{5}}

\int 3 v^{\frac{21}{5}} d v = \frac{15}{26} v^{\frac{26}{5}}

\int v^{\frac{31}{5}} d v = \frac{5}{36} v^{\frac{36}{5}}

= \frac{1}{5} (\frac{5}{6} v^{\frac{6}{5}} - \frac{15}{16} v^{\frac{16}{5}} + \frac{15}{26} v^{\frac{26}{5}} - \frac{5}{36} v^{\frac{36}{5}})

Sustitución hacia atrás

= \frac{1}{5} (\frac{5}{6} sin^{\frac{6}{5}} (5 x) - \frac{15}{16} sin^{\frac{16}{5}} (5 x) + \frac{15}{26} sin^{\frac{26}{5}} (5 x) - \frac{5}{36} sin^{\frac{36}{5}} (5 x))

Simplificar

\frac{1}{5} (\frac{5}{6} sin^{\frac{6}{5}} (5 x) - \frac{15}{16} sin^{\frac{16}{5}} (5 x) + \frac{15}{26} sin^{\frac{26}{5}} (5 x) - \frac{5}{36} sin^{\frac{36}{5}} (5 x)) : \frac{1}{6} sin^{\frac{6}{5}} (5 x) - \frac{3}{16} sin^{\frac{16}{5}} (5 x) + \frac{3}{26} sin^{\frac{26}{5}} (5 x) - \frac{1}{36} sin^{\frac{36}{5}} (5 x)

= \frac{1}{6} sin^{\frac{6}{5}} (5 x) - \frac{3}{16} sin^{\frac{16}{5}} (5 x) + \frac{3}{26} sin^{\frac{26}{5}} (5 x) - \frac{1}{36} sin^{\frac{36}{5}} (5 x)

Agregar una constante a la solución

= \frac{1}{6} sin^{\frac{6}{5}} (5 x) - \frac{3}{16} sin^{\frac{16}{5}} (5 x) + \frac{3}{26} sin^{\frac{26}{5}} (5 x) - \frac{1}{36} sin^{\frac{36}{5}} (5 x) + C

\int_{- 2}^{2} (x^{3} - 1) d x

\frac{\partial}{\partial x} (x^{3} + x^{2} y^{3} - 2 y^{2})

f (x) = (6 x^{2} - 9 x) e^{x}

\frac{d}{d x} (\frac{( x ^{2} )}{x ^{2} + 1})

n = 1 \sum \infty (\frac{1}{6})^{n}