solve-pi-2-pi-2-sin-R-2-r-2-2rR-cos-d-

Question Number 33759 by 33 last updated on 24/Apr/18

s o l v e :

\underset{- π / 2}{\int^{π / 2}} \frac{s i n θ}{\sqrt{R^{2} + r^{2} - 2 r R c o s θ}} d θ

Commented by MJS last updated on 24/Apr/18

\int \frac{\cos θ}{\sqrt{R^{2} + r^{2} - 2 r R \cos θ}} d θ =

R^{2} + r^{2} = p

- 2 r R = q

= \int \frac{\cos θ}{\sqrt{p + q \cos θ}} d θ =

thanks to my friend Prof . H . E .

\cos θ = \frac{1}{q} (p + q \cos θ) - \frac{p}{q}

= \int \frac{\frac{1}{q} (p + q \cos θ) - \frac{p}{q}}{\sqrt{p + q \cos θ}} d θ =

= \frac{1}{q} \int \frac{p + q \cos θ}{\sqrt{p + q \cos θ}} d θ - \frac{p}{q} \int \frac{1}{\sqrt{p + q \cos θ}} d θ =

= \frac{1}{q} \int \sqrt{p + q \cos θ} d θ - \frac{p}{q} \int \frac{1}{\sqrt{p + q \cos θ}} d θ =

\sin^{2} \frac{θ}{2} = \frac{1 - \cos θ}{2}

\cos θ = 1 - {2 \sin}^{2} \frac{θ}{2}

p + q \cos θ = p + q - 2 q \sin^{2} \frac{θ}{2} =

= (p + q) (1 - \frac{2 q}{p + q} \sin^{2} \frac{θ}{2})

= \frac{\sqrt{p + q}}{q} \int \sqrt{(1 - \frac{2 q}{p + q} \sin^{2} \frac{θ}{2})} d θ - \frac{p \sqrt{p + q}}{q (p + q)} \int \frac{1}{\sqrt{(1 - \frac{2 q}{p + q} \sin^{2} \frac{θ}{2})}} d θ

u = \frac{x}{2} \to d x = 2 d u

= \frac{2 \sqrt{p + q}}{q} \int \sqrt{(1 - \frac{2 q}{p + q} \sin^{2} u)} d u - \frac{2 p \sqrt{p + q}}{q (p + q)} \int \frac{1}{\sqrt{(1 - \frac{2 q}{p + q} \sin^{2} u)}} d u =

the first is an elliptic integral of the

2^{nd} kind, the second one of the 1^{st} kind

the notation follows, but {it}^{'} s beyond

my understanding \dots

= \frac{2 \sqrt{p + q}}{q} E (u ∣ \frac{2 q}{p + q}) - \frac{2 p \sqrt{p + q}}{q (p + q)} F (u ∣ \frac{2 q}{p + q}) =

u = \frac{x}{2}

= \frac{2 \sqrt{p + q}}{q} (E (\frac{x}{2} ∣ \frac{2 q}{p + q}) - \frac{p}{p + q} F (\frac{x}{2} ∣ \frac{2 q}{p + q}))

Commented by prof Abdo imad last updated on 24/Apr/18

l e t p u t I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{c o s θ}{\sqrt{R^{2} + r^{2} - 2 r R c o s θ}} d θ

w e h a v e R^{2} + r^{2} - 2 r R c o s θ

= R^{2} (1 + {(\frac{r}{R})}^{2} - 2 \frac{r}{R} c o s θ l e t p u t λ = \frac{r}{R}

I = \frac{1}{R} \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{c o s θ}{\sqrt{1 + λ^{2} - 2 λ c o s θ}} d θ

t a n (\frac{θ}{2}) = t \Rightarrow I = \frac{1}{R} \int_{- 1}^{1} \frac{\frac{1 - t^{2}}{1 + t^{2}}}{\sqrt{1 + λ^{2} - 2 λ \frac{1 - t^{2}}{1 + t^{2}}}} \frac{2 d t}{1 + t^{2}}

I = \frac{2}{R} \int_{- 1}^{1} \frac{1 - t^{2}}{{(1 + t^{2})}^{2}} \sqrt{1 + t^{2}} \frac{d t}{\sqrt{(1 + λ^{2}) (1 + t^{2}) - 2 λ (1 - t^{2})}}

= \frac{2}{R} \int_{- 1}^{1} \frac{1 - t^{2}}{{(1 + t^{2})}^{\frac{3}{2}}} \frac{d t}{\sqrt{1 + t^{2} + λ^{2} + λ^{2} t^{2} - 2 λ + 2 λ t^{2}}}

= \frac{2}{R} \int_{- 1}^{1} \frac{(1 - t^{2}) d t}{{(1 + t^{2})}^{\frac{3}{2}} \sqrt{λ^{2} - 2 λ + 1 + (λ^{2} + 2 λ + 1) t^{2}}}

= \frac{2}{R} \int_{- 1}^{1} \frac{1 - t^{2}}{{(1 + t^{2})}^{\frac{3}{2}} \sqrt{{(λ - 1)}^{2} + {(λ + 1)}^{2} t^{2}}} d t

c h . (λ + 1) t = (λ - 1) u g i v e

I = \frac{2}{R} \int_{- 1}^{1} \frac{1 - {(\frac{λ - 1}{λ + 1} u)}^{2}}{{(1 + {(\frac{λ - 1}{λ + 1} u)}^{2})}^{\frac{3}{2}}} \frac{\frac{λ - 1}{λ + 1}}{∣ λ - 1 ∣ \sqrt{1 + u^{2}}} d u

a f t e r w e u s e t h e c h . u = t a n θ o r u = s h θ \dots b e

c o n t i n u e d \dots

Commented by 33 last updated on 24/Apr/18

T o s i r M J S, s i r w e c a n n o t

c o n s i d e r a, b a n d c a s

c o n s t a n t s a s t h e y a r e

d e p e n d e n t o n c o s θ .

a n d y o u d o n o t k n o w

a b c a s a f u n c t i o n o f α

a n d v i c e v e r s a .

Commented by MJS last updated on 24/Apr/18

{you}^{'} re right

Commented by MJS last updated on 24/Apr/18

\dots this leads to an elliptic integral and it

might be impossible to exactly solve it

Commented by 33 last updated on 24/Apr/18

w h a t i f t h e r e w e r e s i n e

i n s t e a d o f c o s i n n u m e r a t o r ?

Commented by MJS last updated on 24/Apr/18

then {it}^{'} s easy

\int \frac{\cos θ}{\sqrt{R^{2} + r^{2} - 2 R r \sin θ}} d θ =

u = R^{2} + r^{2} - 2 R r \sin θ

d x = - \frac{1}{2 r R \cos θ}

- \frac{1}{2 r R} \int \frac{1}{\sqrt{u}} d u = - \frac{1}{2 r R} \times 2 \sqrt{u} + C = - \frac{\sqrt{u}}{r R} + C =

= - \frac{\sqrt{R^{2} + r^{2} - 2 R r \sin θ}}{r R} + C

\underset{- \frac{π}{2}}{\int^{\frac{π}{2}}} \frac{\cos θ}{\sqrt{R^{2} + r^{2} - 2 R r \sin θ}} d θ =

= \frac{\sqrt{R^{2} + r^{2} + 2 r R} - \sqrt{R^{2} + r^{2} - 2 r R}}{r R} =

= \frac{\sqrt{{(R + r)}^{2}} - \sqrt{{(R - r)}^{2}}}{r R} =

= \frac{∣ R + r ∣ - ∣ R - r ∣}{r R} =

if R > r \land R > 0 \land r > 0

= \frac{R + r - (R - r)}{r R} = \frac{2}{R}

Commented by MJS last updated on 24/Apr/18

sorry, I misread \dots but {it}^{'} s the same way :

\int \frac{\sin θ}{\sqrt{R^{2} + r^{2} - 2 R r \cos θ}} d θ =

= \frac{\sqrt{R^{2} + r^{2} - 2 R r \cos θ}}{r R} + C

\underset{- \frac{π}{2}}{\int^{\frac{π}{2}}} \frac{\sin θ}{\sqrt{R^{2} + r^{2} - 2 R r \cos θ}} d θ = 0

Commented by prof Abdo imad last updated on 25/Apr/18

s o m e t h i n g w e n t w r o n g b e c s u s e θ \to \frac{s i n θ}{\sqrt{R^{2} + r^{2} - 2 r R c o s θ}}

i s o d d \Rightarrow \int_{- \frac{π}{2}}^{\frac{π}{2}} (\dots) d θ = 0