let-f-x-0-pi-2-cos-1-xsin-d-1-determine-a-explicit-form-of-f-x-2-calculate-0-pi-2-sin-2-1-xsin-2-d-3-find-the-values-of-0-pi-2-cos-1-2cos-d-

Question Number 43100 by maxmathsup by imad last updated on 07/Sep/18

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

1) d e t e r m i n e 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}^{\frac{π}{2}} \frac{s i n (2 θ)}{{(1 + x s i n θ)}^{2}} d θ

3) f i n d t h e v a l u e s o f \int_{0}^{\frac{π}{2}} \frac{c o s θ}{1 + 2 c o s θ} d θ a n d \int_{0}^{\frac{π}{2}} \frac{s i n (2 θ)}{{(1 + 3 s i n θ)}^{2}} d θ .

Commented by maxmathsup by imad last updated on 08/Sep/18

1) x = 0 \Rightarrow f (x) = \int_{0}^{\frac{π}{2}} c o s θ d θ = 1

x \neq 0 \Rightarrow f (x) = \frac{1}{x} \int_{0}^{\frac{π}{2}} \frac{x c o s θ}{1 + x s i n θ} d θ = \frac{1}{x} {[l n ∣ 1 + x s i n θ ∣]}_{0}^{\frac{π}{2}} = \frac{l n ∣ 1 + x ∣}{x} w i t h

x \neq - 1 a n d w e m u s t s t u d y t h e c a s e x = - 1

2) w e h a v e f^{^{'}} (x) = - \int_{0}^{\frac{π}{2}} \frac{s i n θ c o s θ}{{(1 + x s i n θ)}^{2}} d θ = - \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{s i n (2 θ)}{{(1 + x s i n θ)}^{2}} d θ \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{s i n (2 θ)}{{(1 + x s i n θ)}^{2}} d θ = - 2 f^{^{'}} (x) b u t f^{^{'}} (x) = - \frac{1}{x^{2}} l n ∣ 1 + x ∣ + \frac{1}{x (1 + x)} \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{s i n (2 θ)}{{(1 + x s i n θ)}^{2}} d θ = \frac{2 l n ∣ 1 + x ∣}{x^{2}} - \frac{2}{x (x + 1)} .

Commented by maxmathsup by imad last updated on 08/Sep/18

3) \int_{0}^{\frac{π}{2}} \frac{c o s θ}{1 + 2 s i n θ} d θ = f (2) = \frac{l n (3)}{2} a l s o

\int_{0}^{\frac{π}{2}} \frac{s i n (2 θ)}{{(1 + 3 s i n θ)}^{2}} = - 2 f^{^{'}} (3) = \frac{2 l n (4)}{9} - \frac{2}{12} = \frac{4 l n (2)}{9} - \frac{1}{6} .

Answered by alex041103 last updated on 08/Sep/18

l e t u = 1 + x s i n θ

d u / x = c o s θ d θ

\Rightarrow f (x) = \frac{1}{x} \int_{1}^{1 + x} \frac{d u}{u} = \frac{l n (1 + x)}{x} = f (x)