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

Question Number 32351 by abdo imad last updated on 23/Mar/18

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

Answered by sma3l2996 last updated on 25/Mar/18

l e t x = t a n (t / 2) \Rightarrow d t = \frac{2 d x}{1 + x^{2}}

s i n t = \frac{2 x}{1 + x^{2}}

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

= \frac{2}{{s i n}^{2} θ} \int_{0}^{1} \frac{d x}{{(\frac{x + c o s θ}{s i n θ})}^{2} + 1}

l e t u = \frac{x + c o s θ}{s i n θ} \Rightarrow d x = s i n θ d u

\int_{0}^{π / 2} \frac{d t}{1 + c o s θ s i n t} = \frac{2}{s i n θ} \int_{c o t θ}^{\frac{1 + c o s θ}{s i n θ}} \frac{d u}{u^{2} + 1} = \frac{2}{s i n θ} {[a r c t a n (\frac{x + c o s θ}{s i n θ})]}_{0}^{1}

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