calculate-0-pi-2-dt-1-cos-cost-

Question Number 50415 by Abdo msup. last updated on 16/Dec/18

c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{d t}{1 + c o s θ c o s t}

Commented by Abdo msup. last updated on 18/Dec/18

l e t p u t c o s θ = λ a n d f i n d A (λ) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + λ c o s t}

w i t h ∣ λ ∣⩽ 1 c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e

A (λ) = \int_{0}^{1} \frac{1}{1 + λ \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

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

= \frac{2}{1 - λ} \int_{0}^{1} \frac{d u}{u^{2} + \frac{1 + λ}{1 - λ}}

=_{u = \sqrt{\frac{1 + λ}{1 - λ}} t} \frac{2}{1 - λ} \int_{0}^{\sqrt{\frac{1 - λ}{1 + λ}}} \frac{\sqrt{\frac{1 + λ}{1 - λ}} d t}{\frac{1 + λ}{1 - λ} (1 + t^{2})}

= \frac{2}{1 + λ} \frac{\sqrt{1 + λ}}{\sqrt{1 - λ}} a r c t a n (\sqrt{\frac{1 - λ}{1 + λ}})

= \frac{2}{\sqrt{1 - λ^{2}}} a r c t a n (\sqrt{\frac{1 - λ}{1 + λ}}) \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{d t}{1 + c o θ c o s t} = A (c o s θ)

= \frac{2}{\sqrt{1 - {c o s}^{2} θ}} a r c t a n (\sqrt{\frac{1 - c o s θ}{1 + c o s θ}})

= \frac{2}{∣ s i n θ ∣} a r c t a n ∣ t a n (\frac{θ}{2}) ∣ .

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

\int \frac{d t}{1 + c o s θ c o s t}

\frac{1}{c o s θ} \int \frac{d t}{s e c θ + c o s t}

\frac{1}{c o s θ} \int \frac{d t}{s e c θ + \frac{1 - {t a n}^{2} \frac{t}{2}}{1 + {t a n}^{2} \frac{t}{2}}}

\frac{1}{c o s θ} \int \frac{{s e c}^{2} \frac{t}{2} d t}{s e c θ + s e c θ {t a n}^{2} \frac{t}{2} + 1 - {t a n}^{2} \frac{t}{2}}

= s e c θ \int \frac{{s e c}^{2} \frac{t}{2} \times d t}{(s e c θ + 1) + {t a n}^{2} \frac{t}{2} (s e c θ - 1)}

= s e c θ \int \frac{{s e c}^{2} \frac{t}{2}}{(s e c θ - 1) {\frac{s e c θ + 1}{s e c θ - 1} + {t a n}^{2} \frac{t}{2}}} d t

= \frac{s e c θ}{s e c θ - 1} \int \frac{{s e c}^{2} \frac{t}{2}}{\frac{1 + c o s θ}{1 - c o s θ} + {t a n}^{2} \frac{t}{2}} d t

= \frac{1}{1 - c o s θ} \int \frac{{s e c}^{2} \frac{t}{2} d t}{{c o t}^{2} \frac{θ}{2} + {t a n}^{2} \frac{t}{2}}

k = t a n \frac{t}{2} d k = \frac{1}{2} {s e c}^{2} \frac{t}{2} d t

= \frac{1}{2 {s i n}^{2} \frac{θ}{2}} \int \frac{2 d k}{{c o t}^{2} \frac{θ}{2} + k^{2}}

= \frac{1}{{s i n}^{2} \frac{θ}{2}} \int \frac{d k}{{c o t}^{2} \frac{θ}{2} + k^{2}}

= \frac{1}{{s i n}^{2} \frac{θ}{2}} \times \frac{1}{c o t \frac{θ}{2}} {t a n}^{- 1} (\frac{k}{c o t \frac{θ}{2}})

= \frac{2}{s i n θ} {t a n}^{- 1} (\frac{t a n \frac{t}{2}}{c o t \frac{θ}{2}}) + c

r e q u i r e d a n s w e r i z

\frac{2}{s i n θ} ∣ {t a n}^{- 1} (\frac{t a n \frac{t}{2}}{c o t \frac{θ}{2}}) ∣_{0}^{\frac{π}{2}}

= \frac{2}{s i n θ} [{t a n}^{- 1} (\frac{t a n \frac{π}{4}}{c o t \frac{θ}{2}}) - {t a n}^{- 1} (\frac{t a n 0}{c o t \frac{θ}{2}})]

= \frac{2}{s i n θ} [{t a n}^{- 1} (t a n \frac{θ}{2})]

= \frac{2}{s i n θ} \times \frac{θ}{2} = \frac{θ}{s i n θ}