0-pi-2-ln-2-sinx-dx-

Question Number 113821 by 675480065 last updated on 15/Sep/20

\int_{0}^{\frac{π}{2}} \ln (2 - sinx) dx

Commented by Dwaipayan Shikari last updated on 15/Sep/20

I (a) = \int_{0}^{\frac{π}{2}} l o g (2 + a s i n x) d x

I^{'} (a) = \int_{0}^{\frac{π}{2}} \frac{s i n x}{2 + a s i n x} d x

I^{'} (a) = \frac{1}{a} \int_{0}^{\frac{π}{2}} 1 - \frac{2}{2 + a s i n x}

I^{'} (a) = \frac{π}{2 a} - 2 \int_{0}^{\frac{π}{2}} \frac{1}{2 + a s i n x} d x

I^{'} (a) = \frac{π}{2 a} - 4 \int_{0}^{\frac{π}{2}} \frac{1}{2 + \frac{2 a t}{1 + t^{2}}} . \frac{1}{1 + t^{2}} d t (t a n \frac{x}{2} = t)

I^{'} (a) = \frac{π}{2 a} - 4 \int_{0}^{1} \frac{1}{2 + 2 t^{2} + 2 a t} d t

I^{'} (a) = \frac{π}{2 a} - 2 \int_{0}^{1} \frac{1}{{(t + \frac{a}{2})}^{2} + 1 - \frac{a^{2}}{4}} d t

I^{'} (a) = \frac{π}{2 a} - 2 \frac{1}{\sqrt{1 - \frac{a^{2}}{4}}} . {[{t a n}^{- 1} \frac{2 t + a}{\sqrt{4 - a^{2}}}]}_{0}^{1}

I (a) = \frac{π}{2} l o g (a) - \int \frac{4}{\sqrt{4 - a^{2}}} ({t a n}^{- 1} \sqrt{\frac{2 + a}{2 - a}} - {t a n}^{- 1} \frac{a}{\sqrt{4 - a^{2}}}) \dots .

\dots . .

Answered by mathdave last updated on 15/Sep/20

s o l u t i o n

l e t

A = \int_{0}^{\frac{π}{2}} \ln [2 (1 - \frac{1}{2} \sin x)] d x = \ln 2 \int^{\frac{π}{2}} d x + \int^{\frac{π}{2}} (1 - \frac{1}{2} \sin x) d x

A = \frac{π}{2} \ln 2 + \int_{0}^{\frac{π}{2}} (1 - \frac{1}{2} \sin x) d x

l e t I = \int_{0}^{\frac{π}{2}} (1 - \frac{1}{2} \sin x) d x \dots \dots \dots (1)

Commented by mathdave last updated on 15/Sep/20

Commented by mathdave last updated on 15/Sep/20

Commented by mathdave last updated on 15/Sep/20

Commented by Tawa11 last updated on 06/Sep/21

great sir

Answered by Olaf last updated on 15/Sep/20

I = \int_{0}^{\frac{π}{2}} \ln (2 - \sin x) d x

\frac{1}{1 - \frac{1}{2} \sin x} = \underset{n = 0}{\sum^{\infty}} \frac{\sin^{n} x}{2^{n}}

\frac{- \frac{1}{2} \cos x}{1 - \frac{1}{2} \sin x} = - \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \cos x \frac{\sin^{n} x}{2^{n}}

\ln (1 - \frac{1}{2} \sin x) = - \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{\sin^{n + 1} x}{2^{n} (n + 1)}

I = \frac{π}{2} \ln 2 + \int_{0}^{\frac{π}{2}} \ln (1 - \frac{1}{2} \sin x) d x

I = \frac{π}{2} \ln 2 - \int_{0}^{\frac{π}{2}} \underset{n = 0}{\sum^{\infty}} \frac{\sin^{n + 1} x}{2^{n + 1} (n + 1)} d x

I = \frac{π}{2} \ln 2 - \int_{0}^{\frac{π}{2}} \underset{n = 1}{\sum^{\infty}} \frac{\sin^{n} x}{n 2^{n}} d x

I = \frac{π}{2} \ln 2 - \underset{n = 1}{\sum^{\infty}} \frac{1}{n 2^{n}} \int_{0}^{\frac{π}{2}} \sin^{n} x d x

W_{n} = \int_{0}^{\frac{π}{2}} \sin^{n} x d x (Wallis)

I = \frac{π}{2} \ln 2 - \underset{n = 1}{\sum^{\infty}} \frac{W_{n}}{n 2^{n}} \dots

Answered by mathmax by abdo last updated on 15/Sep/20

let I = \int_{0}^{\frac{π}{2}} \ln (2 - sinx) dx \Rightarrow I = \int_{0}^{\frac{π}{2}} \ln (2) + \ln (1 - \frac{1}{2} sinx) dx

= \frac{π}{2} \ln (2) + \int_{0}^{\frac{π}{2}} \ln (1 - \frac{1}{2} sinx) dx let f (a) = \int_{0}^{\frac{π}{2}} \ln (1 - asinx) dx

with 0 < a < 1

f^{^{'}} (a) = \int_{0}^{\frac{π}{2}} \frac{- sinx}{1 - asinx} dx =_{\tan (\frac{x}{2}) = t} - \int_{0}^{1} \frac{\frac{2 t}{1 + t^{2}}}{1 - a \frac{2 t}{1 + t^{2}}} \times \frac{2 dt}{1 + t^{2}}

= - \int_{0}^{1} \frac{4 t}{(1 + t^{2}) (1 + t^{2} - 2 at)} dt = - 4 \int_{0}^{1} \frac{tdt}{(t^{2} + 1) (t^{2} - 2 at + 1)}

let decomoose F (t) = \frac{t}{(t^{2} - 2 at + 1) (t^{2} + 1)}

t^{2} - 2 at + 1 = 0 \to Δ^{^{'}} = a^{2} - 1 < 0 \Rightarrow F (t) = \frac{α t + β}{t^{2} - 2 at + 1} + \frac{mt + n}{t^{2} + 1}

\Rightarrow (α t + β) (t^{2} + 1) + (mt + n) (t^{2} - 2 at + 1) = t \Rightarrow

α t^{3} + α t + β t^{2} + β + {mt}^{3} - {2 amt}^{2} + mt + {nt}^{2} - 2 ant + n = t \Rightarrow

(α + m) t^{3} + (β - 2 am + n) t^{2} + (α + m - 2 an) t + β + n = t \Rightarrow

{\begin{cases} α + m = 0 \\ β - 2 am + n = 0 and {\begin{cases} α + m - 2 an = 1 \\ β + n = 0 \Rightarrow \end{cases} \end{cases}

{\begin{cases} m = - α and {\begin{cases} - 2 an = 1 \\ β = - n \Rightarrow n = - \frac{1}{2 a} \end{cases} \\ - 2 a α = 0 \end{cases}

and α = 0 \Rightarrow m = 0 \Rightarrow F (t) = \frac{\frac{1}{2 a}}{t^{2} - 2 at + 1} + \frac{- \frac{1}{2 a}}{t^{2} + 1}

= \frac{1}{2 a} {\frac{1}{t^{2} - 2 at + 1} - \frac{1}{t^{2} + 1}} \Rightarrow

f^{^{'}} (a) = - \frac{2}{a} \int_{0}^{1} \frac{dt}{t^{2} - 2 at + 1} + \frac{2}{a} \int_{0}^{1} \frac{dt}{t^{2} + 1}

= \frac{2}{a} \times \frac{π}{4} - \frac{2}{a} \int_{0}^{1} \frac{dt}{t^{2} - 2 at + 1} = \frac{π}{2 a} - \frac{2}{a} \int_{0}^{1} \frac{dt}{t^{2} - 2 at + 1}

but \int_{0}^{1} \frac{dt}{t^{2} - 2 at + 1} = \int_{0}^{1} \frac{dt}{t^{2} - 2 at + a^{2} + 1 - a^{2}} = \int_{0}^{1} \frac{dt}{{(t - a)}^{2} + 1 - a^{2}}

=_{t - a = \sqrt{1 - a^{2}} u} \int_{\frac{- a}{\sqrt{1 - a^{2}}}}^{\frac{1 - a}{\sqrt{1 - a^{2}}}} \frac{\sqrt{1 - a^{2}}}{(1 - a^{2}) (1 + u^{2})} du

= \frac{1}{\sqrt{1 - a^{2}}} {\arctan (\frac{1 - a}{\sqrt{1 - a^{2}}}) + \arctan (\frac{a}{\sqrt{1 - a^{2}}})} \Rightarrow

f (a) = \frac{π}{2} lna - 2 \int \frac{1}{a \sqrt{1 - a^{2}}} \arctan (\frac{1 - a}{\sqrt{1 - a^{2}}}) da - 2 \int \frac{1}{a \sqrt{1 - a^{2}}} \arctan (\frac{a}{\sqrt{1 - a^{2}}}) da

\dots . be continued \dots .

Commented by mathmax by abdo last updated on 16/Sep/20

we have \int \frac{1}{a \sqrt{1 - a^{2}}} \arctan (\frac{1 - a}{\sqrt{1 - a^{2}}}) da

=_{a = cost} \int \frac{1}{cost sint} \arctan (\frac{1 - cost}{sint}) (- sint) dt

= - \int \frac{1}{cost} \arctan (\frac{{2 \sin}^{2} (\frac{t}{2})}{2 \sin (\frac{t}{2}) \cos (\frac{t}{2})}) dt

= - \int \frac{1}{cost} \arctan (\tan (\frac{t}{2}) dt = - \frac{1}{2} \int \frac{t}{cost} dt

=_{\tan (\frac{t}{2}) = z} - \frac{1}{2} \int \frac{2 \arctan (z)}{\frac{1 - z^{2}}{1 + z^{2}}} \frac{2 dz}{1 + z^{2}}

= - 2 \int \frac{\arctan (z)}{1 - z^{2}} dz \dots . . be continued \dots