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integral de sqrt(pi/(12)) a sqrt(pi/9 de)5xtan^4(3x^2)sec^2(3x^2)

Solución

\int_{\frac{π}{12}}^{\frac{π}{9}} 5 x tan^{4} (3 x^{2}) sec^{2} (3 x^{2}) d x

\frac{9 3 - 1}{6}

Decimal

2.43140 \dots

Pasos de solución

\int_{\frac{π}{12}}^{\frac{π}{9}} 5 x tan^{4} (3 x^{2}) sec^{2} (3 x^{2}) d x

Sacar la constante:

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

= 5 \cdot \int_{\frac{π}{12}}^{\frac{π}{9}} x tan^{4} (3 x^{2}) sec^{2} (3 x^{2}) d x

Aplicar integración por sustitución

: \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{tan ^{4} ( u ) sec ^{2} ( u )}{6} d u

= 5 \cdot \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{tan ^{4} ( u ) sec ^{2} ( u )}{6} d u

Sacar la constante:

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

= 5 \cdot \frac{1}{6} \cdot \int_{\frac{π}{4}}^{\frac{π}{3}} tan^{4} (u) sec^{2} (u) d u

Re-escribir usando identidades trigonométricas

= 5 \cdot \frac{1}{6} \cdot \int_{\frac{π}{4}}^{\frac{π}{3}} (1 + tan^{2} (u)) tan^{4} (u) d u

Aplicar integración por sustitución

: \int_{1}^{3} v^{4} d v

= 5 \cdot \frac{1}{6} \cdot \int_{1}^{3} v^{4} d v

Aplicar la regla de la potencia

: [\frac{v ^{5}}{5}]_{1}^{3}

= 5 \cdot \frac{1}{6} [\frac{v ^{5}}{5}]_{1}^{3}

Simplificar

5 \cdot \frac{1}{6} [\frac{v ^{5}}{5}]_{1}^{3} : \frac{5}{6} [\frac{v ^{5}}{5}]_{1}^{3}

= \frac{5}{6} [\frac{v ^{5}}{5}]_{1}^{3}

Calcular los limites:

\frac{9 3}{5} - \frac{1}{5}

= \frac{5}{6} (\frac{9 3}{5} - \frac{1}{5})

Simplificar

= \frac{9 3 - 1}{6}

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