let-f-x-0-x-t-1-sint-dt-1-find-a-explicit-form-of-f-x-2-calculate-0-t-1-sint-dt-

Question Number 43337 by math khazana by abdo last updated on 09/Sep/18

l e t f (x) = \int_{0}^{x} \frac{t}{1 + s i n t} d t

1) f i n d a e x p l i c i t f o r m o f f (x)

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

Commented by maxmathsup by imad last updated on 11/Sep/18

1) c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e f (x) = \int_{0}^{t a n (\frac{x}{2})} \frac{2 a r c t a n u}{1 + \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= \int_{0}^{t a n (\frac{x}{2})} \frac{4 a r c t a n u}{1 + u^{2} + 2 u} d u = 4 \int_{0}^{t a n (\frac{x}{2})} \frac{a r c t a n u}{{(u + 1)}^{2}} d u b y p a r t s w e g e t

f (x) = 4 {{[- \frac{1}{1 + u} a r c t a n u]}_{0}^{t a n (\frac{x}{2})} + \int_{0}^{t a n (\frac{x}{2})} \frac{d u}{(u + 1) (1 + u^{2}}}

= - 4 \frac{x}{2 (1 + t a n (\frac{x}{2}))} + 4 \int_{0}^{t a n (\frac{x}{2})} \frac{d u}{(u + 1) (u^{2} + 1)} b u t

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

= \frac{1}{2} {[l n ∣ u + 1 ∣]}_{0}^{t a n (\frac{x}{2})} - \frac{1}{4} {[l n (u^{2} + 1)]}_{0}^{t a n (\frac{x}{2})} + \frac{1}{2} {[a r c t a n u]}_{0}^{t a n (\frac{x}{2})}

= \frac{1}{2} l n ∣ 1 + t a n (\frac{x}{2}) ∣ - \frac{1}{4} l n ∣ 1 + {t a n}^{2} (\frac{x}{2}) ∣ + \frac{x}{4} \Rightarrow

f (x) = \frac{- 2 x}{1 + t a n (\frac{x}{2})} + \frac{1}{2} l n ∣ 1 + t a n (\frac{x}{2}) ∣ - \frac{1}{4} l n ∣ 1 + {t a n}^{2} (\frac{x}{2}) ∣ + \frac{x}{4} .

Commented by maxmathsup by imad last updated on 11/Sep/18

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

Commented by maxmathsup by imad last updated on 11/Sep/18

\int_{0}^{\frac{π}{2}} \frac{t}{1 + s i n t} d t = f (\frac{π}{2}) = \frac{- π}{1 + 1} + \frac{1}{2} l n (2) - \frac{1}{4} l n (2) + \frac{π}{4}

= - \frac{π}{4} + \frac{l n (2)}{2} .