integral de 1/(4sec(x)+5)

Solución

\int \frac{1}{4 sec ( x ) + 5} d x

2 (- \frac{2}{15} ln tan (\frac{x}{2}) + 3 + \frac{2}{15} ln tan (\frac{x}{2}) - 3 + \frac{1}{5} arctan (tan (\frac{x}{2}))) + C

Pasos de solución

\int \frac{1}{4 sec ( x ) + 5} d x

Re-escribir usando identidades trigonométricas

= \int \frac{cos ( x )}{4 + 5 cos ( x )} d x

Aplicar integración por sustitución

: \int \frac{2 ( 1 - u ^{2} )}{( - u ^{2} + 9 ) ( 1 + u ^{2} )} d u

= \int \frac{2 ( 1 - u ^{2} )}{( - u ^{2} + 9 ) ( 1 + u ^{2} )} d u

Sacar la constante:

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

= 2 \cdot \int \frac{1 - u ^{2}}{( - u ^{2} + 9 ) ( 1 + u ^{2} )} d u

Tomar la fracción parcial de

\frac{1 - u ^{2}}{( - u ^{2} + 9 ) ( 1 + u ^{2} )} : - \frac{2}{15 ( u + 3 )} + \frac{2}{15 ( u - 3 )} + \frac{1}{5 ( u ^{2} + 1 )}

= 2 \cdot \int - \frac{2}{15 ( u + 3 )} + \frac{2}{15 ( u - 3 )} + \frac{1}{5 ( u ^{2} + 1 )} 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

= 2 (- \int \frac{2}{15 ( u + 3 )} d u + \int \frac{2}{15 ( u - 3 )} d u + \int \frac{1}{5 ( u ^{2} + 1 )} d u)

\int \frac{2}{15 ( u + 3 )} d u = \frac{2}{15} ln ∣ u + 3 ∣

\int \frac{2}{15 ( u - 3 )} d u = \frac{2}{15} ln ∣ u - 3 ∣

\int \frac{1}{5 ( u ^{2} + 1 )} d u = \frac{1}{5} arctan (u)

= 2 (- \frac{2}{15} ln ∣ u + 3 ∣ + \frac{2}{15} ln ∣ u - 3 ∣ + \frac{1}{5} arctan (u))

Sustituir en la ecuación

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

= 2 (- \frac{2}{15} ln tan (\frac{x}{2}) + 3 + \frac{2}{15} ln tan (\frac{x}{2}) - 3 + \frac{1}{5} arctan (tan (\frac{x}{2})))

Agregar una constante a la solución

= 2 (- \frac{2}{15} ln tan (\frac{x}{2}) + 3 + \frac{2}{15} ln tan (\frac{x}{2}) - 3 + \frac{1}{5} arctan (tan (\frac{x}{2}))) + C

\int \frac{16 + x ^{2}}{x ^{2}} d x

\int \frac{x ln ( sin ( x ^{2} ) )}{tan ( x ^{2} )} d x

\int \frac{x + 5}{2 - x} d x

\int \frac{sec ( x ) csc ( x )}{ln ( tan ( x ) )} d x

\int \frac{24 x ^{3}}{( x ^{4} - 5 ) ^{5}} d x