integral de 0 a 1 de sqrt(tan(θ))

Solução

\int_{0}^{1} tan (θ) d θ

\frac{4 arctan ( 2 tan ( 1 ) + 1 ) + 4 arctan ( 2 tan ( 1 ) - 1 ) - 2 ln ( tan ( 1 ) + 2 tan ( 1 ) + 1 ) - 2 ln ( \frac{1}{t a n ( 1 ) - 2 t a n ( 1 ) + 1} )}{4 2}

Decimal

0.72729 \dots

Passos da solução

\int_{0}^{1} tan (θ) d θ

Aplicar integração por substituição trigonométrica

= \int_{0}^{t a n (1)} \frac{u}{1 + u ^{2}} d u

Aplicar integração por substituição

= \int_{0}^{2 t a n (1)} \frac{2 v}{4 + v ^{2}} d v

Remover a constante:

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

= 2 \cdot \int_{0}^{2 t a n (1)} \frac{v}{4 + v ^{2}} d v

Aplicar integração por substituição

= 2 \cdot \int_{0}^{2 t a n (1)} \frac{2 w ^{2}}{4 + w ^{4}} d w

Remover a constante:

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

= 2 \cdot 2 \cdot \int_{0}^{2 t a n (1)} \frac{w ^{2}}{4 + w ^{4}} d w

Tirar a fração parcial de

\frac{w ^{2}}{4 + w ^{4}} : - \frac{w}{4 ( w ^{2} + 2 w + 2 )} + \frac{w}{4 ( w ^{2} - 2 w + 2 )}

= 2 \cdot 2 \cdot \int_{0}^{2 t a n (1)} - \frac{w}{4 ( w ^{2} + 2 w + 2 )} + \frac{w}{4 ( w ^{2} - 2 w + 2 )} d w

Aplicar a regra da soma:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot 2 (- \int_{0}^{2 t a n (1)} \frac{w}{4 ( w ^{2} + 2 w + 2 )} d w + \int_{0}^{2 t a n (1)} \frac{w}{4 ( w ^{2} - 2 w + 2 )} d w)

\int_{0}^{2 t a n (1)} \frac{w}{4 ( w ^{2} + 2 w + 2 )} d w = \frac{1}{4} (\frac{1}{2} ln (tan (1) + 2 tan (1) + 1) - arctan (2 tan (1) + 1) + \frac{π}{4})

\int_{0}^{2 t a n (1)} \frac{w}{4 ( w ^{2} - 2 w + 2 )} d w = \frac{1}{4} (- \frac{1}{2} ln (\frac{1}{tan ( 1 ) - 2 tan ( 1 ) + 1}) + arctan (2 tan (1) - 1) + \frac{π}{4})

= 2 \cdot 2 (- \frac{1}{4} (\frac{1}{2} ln (tan (1) + 2 tan (1) + 1) - arctan (2 tan (1) + 1) + \frac{π}{4}) + \frac{1}{4} (- \frac{1}{2} ln (\frac{1}{tan ( 1 ) - 2 tan ( 1 ) + 1}) + arctan (2 tan (1) - 1) + \frac{π}{4}))

Simplificar

2 \cdot 2 (- \frac{1}{4} (\frac{1}{2} ln (tan (1) + 2 tan (1) + 1) - arctan (2 tan (1) + 1) + \frac{π}{4}) + \frac{1}{4} (- \frac{1}{2} ln (\frac{1}{tan ( 1 ) - 2 tan ( 1 ) + 1}) + arctan (2 tan (1) - 1) + \frac{π}{4})) : \frac{4 arctan ( 2 tan ( 1 ) + 1 ) + 4 arctan ( 2 tan ( 1 ) - 1 ) - 2 ln ( tan ( 1 ) + 2 tan ( 1 ) + 1 ) - 2 ln ( \frac{1}{t a n ( 1 ) - 2 t a n ( 1 ) + 1} )}{4 2}

= \frac{4 arctan ( 2 tan ( 1 ) + 1 ) + 4 arctan ( 2 tan ( 1 ) - 1 ) - 2 ln ( tan ( 1 ) + 2 tan ( 1 ) + 1 ) - 2 ln ( \frac{1}{t a n ( 1 ) - 2 t a n ( 1 ) + 1} )}{4 2}

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