integral de (sec^3(x))/(tan^3(x))

Solución

\int \frac{sec ^{3} ( x )}{tan ^{3} ( x )} d x

\frac{1}{8} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2})) + C

Pasos de solución

\int \frac{sec ^{3} ( x )}{tan ^{3} ( x )} d x

Re-escribir usando identidades trigonométricas

= \int \frac{1}{sin ^{3} ( x )} d x

Aplicar integración por sustitución

: \int \frac{( 1 + u ^{2} ) ^{2}}{4 u ^{3}} d u

= \int \frac{( 1 + u ^{2} ) ^{2}}{4 u ^{3}} d u

Sacar la constante:

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

= \frac{1}{4} \cdot \int \frac{( 1 + u ^{2} ) ^{2}}{u ^{3}} d u

Expandir

\frac{( 1 + u ^{2} ) ^{2}}{u ^{3}} : \frac{1}{u ^{3}} + \frac{2}{u} + u

= \frac{1}{4} \cdot \int \frac{1}{u ^{3}} + \frac{2}{u} + 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

= \frac{1}{4} (\int \frac{1}{u ^{3}} d u + \int \frac{2}{u} d u + \int u d u)

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int u d u = \frac{u ^{2}}{2}

= \frac{1}{4} (- \frac{1}{2 u ^{2}} + 2 ln ∣ u ∣ + \frac{u ^{2}}{2})

Sustituir en la ecuación

u = tan (\frac{x}{2})

= \frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{x}{2} )} + 2 ln tan (\frac{x}{2}) + \frac{tan ^{2} ( \frac{x}{2} )}{2})

Simplificar

\frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{x}{2} )} + 2 ln tan (\frac{x}{2}) + \frac{tan ^{2} ( \frac{x}{2} )}{2}) : \frac{1}{8} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2}))

= \frac{1}{8} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2}))

Agregar una constante a la solución

= \frac{1}{8} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2})) + C

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