calculate-0-pi-4-1-2tanx-dx-

Question Number 83441 by mathmax by abdo last updated on 02/Mar/20

c a l c u l a t e \int_{0}^{\frac{π}{4}} \sqrt{1 + 2 t a n x} d x

Answered by M±th+et£s last updated on 06/Mar/20

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

= \int_{0}^{\frac{π}{4}} \frac{1 + 2 t a n (x)}{1 + {t a n}^{2} (x)} \frac{{s e c}^{2} (x)}{\sqrt{1 + 2 t a n (x)}} d x

y = \sqrt{1 + 2 t a n x} d y = \frac{{s e c}^{2} (x)}{\sqrt{1 + 2 t a n (x)}}

\frac{y^{2} - 1}{2} = t a n (x)

i f x = 0 y = 1

i f x = \frac{π}{4} y = \sqrt{3}

\int_{1}^{\sqrt{3}} \frac{y^{2}}{1 + {(\frac{y^{2} - 1}{2})}^{2}} d y

4 \int_{1}^{\sqrt{3}} \frac{y^{2}}{y^{4} - 2 y^{2} + 5} d y = 2 \int_{1}^{\sqrt{3}} \frac{y^{2} - \sqrt{5} + y^{2} + \sqrt{5}}{y^{4} - 2 y^{2} + 5} d y

= 2 \int_{1}^{\sqrt{3}} \frac{1 - \frac{\sqrt{5}}{y^{2}}}{y^{2} + \frac{5}{y^{2}} - 2} d y + 2 \int_{1}^{\sqrt{3}} \frac{1 + \frac{5}{y^{2}}}{y^{2} + \frac{5}{y^{2}} - 2} d y

= 2 \int_{1}^{\sqrt{3}} \frac{d (y + \frac{\sqrt{5}}{y})}{{(y + \frac{\sqrt{5}}{y})}^{2} - 2 - 2 \sqrt{5}} + 2 \int_{1}^{\sqrt{3}} \frac{d (y - \frac{\sqrt{5}}{y})}{{(y - \frac{\sqrt{5}}{y})}^{2} + 2 \sqrt{5} - 2}

= \frac{- 2}{\sqrt{2 + 2 \sqrt{5}}} {[{t a n h}^{- 1} (\frac{y + \frac{\sqrt{5}}{y}}{\sqrt{2 + 2 \sqrt{5}}})]}_{1}^{\sqrt{3}} + \frac{2}{\sqrt{2 \sqrt{5} - 2}} {[{t a n}^{- 1} (\frac{y - \frac{\sqrt{5}}{y}}{\sqrt{2 \sqrt{5} - 2}})]}_{1}^{\sqrt{3}}

\frac{- 2}{\sqrt{2 \sqrt{5} + 2}} [{t a n h}^{- 1} (\frac{\sqrt{3} + \frac{\sqrt{5}}{\sqrt{3}}}{\sqrt{2 + 2 \sqrt{5}}}) - {t a n h}^{- 1} (\frac{1 + \sqrt{5}}{\sqrt{2 + 2 \sqrt{5}}})] + \frac{2}{\sqrt{2 \sqrt{5} - 2}} [{t a n}^{- 1} (\frac{\sqrt{3} - \frac{\sqrt{5}}{\sqrt{3}}}{\sqrt{2 \sqrt{5} - 2}}) - {t a n}^{- 1} (\frac{1 - \sqrt{5}}{\sqrt{2 \sqrt{5} - 2}})]