calculate ∫_0 ^(π/2) ((ln(1+sinx))/(sinx))dx


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Question Number 57325 by turbo msup by abdo last updated on 02/Apr/19

$c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{l n (1 + s i n x)}{s i n x} d x$
Commented by Abdo msup. last updated on 05/Apr/19
$let f(t) =∫_0 ^(π/2) ((ln(1+t sinx))/(sinx))dx with 0≤t≤1 we have f^′ (t) =∫_0 ^(π/2) ((sinx)/(sinx(1+tsinx)))dx =∫_0 ^(π/2) (dx/(1+tsinx)) =_(tan((x/2))=u) ∫_0 ^1 ((2du)/((1+u^2 )(1+t ((2u)/(1+u^2 ))))) =∫_0 ^1 ((2du)/(1+u^2 +2tu)) =∫_0 ^1 ((2du)/(u^2 +2tu +1)) =∫_0 ^1 ((2du)/(u^2 +2tu +u^2 +1−u^2 )) =∫_0 ^1 ((2du)/((u+t)^2 +1−t^2 )) =_(u+t =(√(1−t^2 ))α) ∫_(t/(√(1−t2))) ^((1+t)/(√(1−t^2 ))) ((2(√(1−t^2 ))dα)/((1−t^2 )(1+α^2 ))) =(2/(√(1−t^2 ))) [ arctan(α)]_(t/(√(1−t^2 ))) ^((1+t)/(√(1−t^2 ))) =(2/(√(1−t^2 ))){ arctan(((1+t)/(√(1−t^2 ))))−arctan((t/(√(1−t^2 ))))} ⇒ f(t) = ∫ (2/(√(1−t^2 ))) arctan(((1+t)/(√(1−t^2 ))))dt −∫ (2/(√(1−t^2 ))) arctan((t/(√(1−t^2 ))))dt +c ....be continued..=$
$l e t f (t) = \int_{0}^{\frac{π}{2}} \frac{l n (1 + t s i n x)}{s i n x} d x w i t h 0 ⩽ t ⩽ 1$ $w e h a v e f^{^{'}} (t) = \int_{0}^{\frac{π}{2}} \frac{s i n x}{s i n x (1 + t s i n x)} d x$ $= \int_{0}^{\frac{π}{2}} \frac{d x}{1 + t s i n x} =_{t a n (\frac{x}{2}) = u} \int_{0}^{1} \frac{2 d u}{(1 + u^{2}) (1 + t \frac{2 u}{1 + u^{2}})}$ $= \int_{0}^{1} \frac{2 d u}{1 + u^{2} + 2 t u} = \int_{0}^{1} \frac{2 d u}{u^{2} + 2 t u + 1}$ $= \int_{0}^{1} \frac{2 d u}{u^{2} + 2 t u + u^{2} + 1 - u^{2}} = \int_{0}^{1} \frac{2 d u}{{(u + t)}^{2} + 1 - t^{2}}$ $=_{u + t = \sqrt{1 - t^{2}} α} \int_{\frac{t}{\sqrt{1 - t 2}}}^{\frac{1 + t}{\sqrt{1 - t^{2}}}} \frac{2 \sqrt{1 - t^{2}} d α}{(1 - t^{2}) (1 + α^{2})}$ $= \frac{2}{\sqrt{1 - t^{2}}} {[a r c t a n (α)]}_{\frac{t}{\sqrt{1 - t^{2}}}}^{\frac{1 + t}{\sqrt{1 - t^{2}}}}$ $= \frac{2}{\sqrt{1 - t^{2}}} {a r c t a n (\frac{1 + t}{\sqrt{1 - t^{2}}}) - a r c t a n (\frac{t}{\sqrt{1 - t^{2}}})} \Rightarrow$ $f (t) = \int \frac{2}{\sqrt{1 - t^{2}}} a r c t a n (\frac{1 + t}{\sqrt{1 - t^{2}}}) d t$ $- \int \frac{2}{\sqrt{1 - t^{2}}} a r c t a n (\frac{t}{\sqrt{1 - t^{2}}}) d t + c . . . . b e c o n t i n u e d . . =$
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