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integral de (tan(x))/(1+cos^4(x))

Solución

\int \frac{tan ( x )}{1 + cos ^{4} ( x )} d x

- \frac{1}{4} (ln (cos^{4} (x)) - ln 1 + cos^{4} (x)) + C

Pasos de solución

\int \frac{tan ( x )}{1 + cos ^{4} ( x )} d x

Aplicar integración por sustitución

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

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

Sacar la constante:

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

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

Aplicar integración por sustitución

: \int \frac{1}{4 ( v - 1 ) v} d v

= - \int \frac{1}{4 ( v - 1 ) v} d v

Sacar la constante:

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

= - \frac{1}{4} \cdot \int \frac{1}{( v - 1 ) v} d v

Tomar la fracción parcial de

\frac{1}{( v - 1 ) v} : \frac{1}{v - 1} - \frac{1}{v}

= - \frac{1}{4} \cdot \int \frac{1}{v - 1} - \frac{1}{v} d v

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}{v - 1} d v - \int \frac{1}{v} d v)

\int \frac{1}{v - 1} d v = ln ∣ v - 1 ∣

\int \frac{1}{v} d v = ln ∣ v ∣

= - \frac{1}{4} (ln ∣ v - 1 ∣ - ln ∣ v ∣)

Sustitución hacia atrás

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

Simplificar

- \frac{1}{4} (ln 1 + cos^{4} (x) - 1 - ln 1 + cos^{4} (x)) : - \frac{1}{4} (ln (cos^{4} (x)) - ln 1 + cos^{4} (x))

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

Agregar una constante a la solución

= - \frac{1}{4} (ln (cos^{4} (x)) - ln 1 + cos^{4} (x)) + C

\frac{1}{x - 2}

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