calculate-0-3pi-4-dt-1-sin-2-t-2-

Question Number 36941 by maxmathsup by imad last updated on 07/Jun/18

c a l c u l a t e \int_{0}^{\frac{3 π}{4}} \frac{d t}{{(1 + {s i n}^{2} t)}^{2}}

Commented by math khazana by abdo last updated on 08/Jun/18

w e h a v e {c o s}^{2} t = \frac{1}{1 + {t a n}^{2} t} \Rightarrow {s i n}^{2} t = 1 - \frac{1}{1 + {t a n}^{2} t}

= \frac{{t a n}^{2} t}{1 + {t a n}^{2} t} \Rightarrow I = \int_{0}^{\frac{3 π}{4}} \frac{d t}{{(1 + \frac{{t a n}^{2} t}{1 + {t a n}^{2} t})}^{2}}

=_{t a n t = - x} \int_{0}^{1} \frac{1}{{(1 + \frac{x^{2}}{1 + x^{2}})}^{2}} . \frac{- d x}{1 + x^{2}}

= - \int_{0}^{1} \frac{{(1 + x^{2})}^{2}}{(1 + x^{2}) {(1 + 2 x^{2})}^{2}} d x

= - \int_{0}^{1} \frac{1 + x^{2}}{{(1 + 2 x^{2})}^{2}} d x c h a n g e m e n t x \sqrt{2} = s h (u)

g i v e I = - \int_{0}^{a r g s h (\sqrt{2})} \frac{1 + \frac{1}{2} {s h}^{2} u}{({c h}^{4} u)} \frac{1}{\sqrt{2}} c h (u) d u

= - \frac{1}{2 \sqrt{2}} \int_{0}^{l n (\sqrt{2} + \sqrt{3})} \frac{2 + {s h}^{2} u}{{c h}^{3} u} d u

= - \frac{1}{2 \sqrt{2}} \int_{0}^{l n (\sqrt{2} + \sqrt{3})} \frac{2 + \frac{c h (2 u) - 1}{2}}{\frac{1 + c h (2 u)}{2} c h (u)} d u

= \frac{1}{2 \sqrt{2}} \int_{0}^{l n (\sqrt{2} + \sqrt{3})} \frac{3 + c h (2 u)}{c h u + c h (u) c h (2 u)} d u \dots

b e c o n t i n u e d \dots