calculate ∫_0 ^(π/3) ((sinx cos(cosx))/(1+2sin(cosx)))dx


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Question Number 35588 by abdo mathsup 649 cc last updated on 20/May/18

$c a l c u l a t e \int_{0}^{\frac{π}{3}} \frac{s i n x c o s (c o s x)}{1 + 2 s i n (c o s x)} d x$
Commented by abdo imad last updated on 21/May/18
$we remark that (sin(cosx))^′ =−sinx cos(cosx) ⇒ I =−(1/2) ∫_0 ^(π/3) (((1+2sin(cosx))^′ )/(1+2 sin(cosx)))dx =−(1/2)[ln∣ 1+2sin(cosx)∣]_0 ^(π/3) = =−(1/2){ln(1+2sin((1/2)) −ln(1 +2sin(1)}$
$w e r e m a r k t h a t {(s i n (c o s x))}^{^{'}} = - s i n x c o s (c o s x) \Rightarrow$ $I = - \frac{1}{2} \int_{0}^{\frac{π}{3}} \frac{{(1 + 2 s i n (c o s x))}^{^{'}}}{1 + 2 s i n (c o s x)} d x$ $= - \frac{1}{2} {[l n ∣ 1 + 2 s i n (c o s x) ∣]}_{0}^{\frac{π}{3}} =$ $= - \frac{1}{2} {l n (1 + 2 s i n (\frac{1}{2}) - l n (1 + 2 s i n (1)}$
	Answered by RAMANUJAN last updated on 20/May/18

	$[\ln {(2 Π + 3) / 3}] / 2 - 0$ $is the correct answer$
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