integral de 5/(16+9cos^2(x))

Solución

\int \frac{5}{16 + 9 cos ^{2} ( x )} d x

\frac{1}{4} arctan (\frac{4}{5} tan (x)) + C

Pasos de solución

\int \frac{5}{16 + 9 cos ^{2} ( x )} d x

Sacar la constante:

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

= 5 \cdot \int \frac{1}{16 + 9 cos ^{2} ( x )} d x

Aplicar integración por sustitución trigonométrica

: \int \frac{1}{16 u ^{2} + 25} d u

= 5 \cdot \int \frac{1}{16 u ^{2} + 25} d u

Aplicar integración por sustitución

= 5 \cdot \int \frac{1}{20 ( v ^{2} + 1 )} d v

Sacar la constante:

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

= 5 \cdot \frac{1}{20} \cdot \int \frac{1}{v ^{2} + 1} d v

Aplicar la regla de integración:

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

= 5 \cdot \frac{1}{20} arctan (v)

Sustitución hacia atrás

= 5 \cdot \frac{1}{20} arctan (\frac{4}{5} tan (x))

Simplificar

5 \cdot \frac{1}{20} arctan (\frac{4}{5} tan (x)) : \frac{1}{4} arctan (\frac{4}{5} tan (x))

= \frac{1}{4} arctan (\frac{4}{5} tan (x))

Agregar una constante a la solución

= \frac{1}{4} arctan (\frac{4}{5} tan (x)) + C

\int \frac{tan ^{3} ( x )}{cos ^{4} ( x )} d x

\frac{\partial}{\partial x} (ln ((\frac{y}{x})^{2}))

\int \frac{x ^{2} - 8 x - 17}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

\frac{\partial}{\partial x} (41 - x^{2} - y^{2})

\frac{\partial}{\partial x} (sin (2 y))