calculate ∫_0 ^π ((2xsinx)/(3 +cos(2x)))dx .


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Question Number 57825 by maxmathsup by imad last updated on 13/Apr/19

$c a l c u l a t e \int_{0}^{π} \frac{2 x s i n x}{3 + c o s (2 x)} d x .$
Commented by maxmathsup by imad last updated on 13/Apr/19

$l e t I = \int_{0}^{π} \frac{2 x s i n x}{3 + c o s (2 x)} d x c h a n g e m e n t x = π - t g i v e$ $I = \int_{0}^{π} \frac{2 (π - t) s i n t}{3 + c o s (2 t)} d t = 2 π \int_{0}^{π} \frac{s i n t}{3 + c o s (2 t)} - I \Rightarrow 2 I = 2 π \int_{0}^{π} \frac{s i n t}{3 + c o s (2 t)} d t$ $\Rightarrow I = π \int_{0}^{π} \frac{s i n t}{3 + c o s (2 t)} d t b u t$ $\int_{0}^{π} \frac{s i n t}{3 + c o s (2 t)} d t = \int_{0}^{π} \frac{s i n t}{3 + 2 {c o s}^{2} t - 1} d t = \int_{0}^{π} \frac{s i n t d t}{2 (1 + {c o s}^{2} t)}$ $=_{u = c o s t} - \int_{- 1}^{1} \frac{- d u}{2 (1 + u^{2})} = \frac{1}{2} \int_{- 1}^{1} \frac{d u}{1 + u^{2}} = \frac{1}{2} {[a r c t a n u]}_{- 1}^{1} = \frac{1}{2} (\frac{π}{2}) = \frac{π}{4} \Rightarrow$ $★ I = \frac{π^{2}}{4} ★$
	Answered by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19
	$∫_0 ^π ((2(π−x)sin(π−x))/(3+cos2(π−x)))dx ∫_0 ^π ((2πsinx−2xsinx)/(3+cos2x))dx 2I=∫_0 ^π ((2πsinx)/(3+cos2x))dx (I/π)=∫_0 ^π ((−d(cosx))/(3+2cos^2 x−1)) ((−I)/π)=(1/2)∫_0 ^π ((d(cosx))/(1+cos^2 x)) ((−I)/π)=(1/2)×∣tan^(−1) (cosx)∣_0 ^π ((−I)/π)=(1/2)×{tan^(−1) (−1)−tan^(−1) (1)} ((−I)/π)=(1/2)×−2×(π/4) I=(π^2 /4)$
	$\int_{0}^{π} \frac{2 (π - x) s i n (π - x)}{3 + c o s 2 (π - x)} d x$ $\int_{0}^{π} \frac{2 π s i n x - 2 x s i n x}{3 + c o s 2 x} d x$ $2 I = \int_{0}^{π} \frac{2 π s i n x}{3 + c o s 2 x} d x$ $\frac{I}{π} = \int_{0}^{π} \frac{- d (c o s x)}{3 + 2 {c o s}^{2} x - 1}$ $\frac{- I}{π} = \frac{1}{2} \int_{0}^{π} \frac{d (c o s x)}{1 + {c o s}^{2} x}$ $\frac{- I}{π} = \frac{1}{2} \times ∣ {t a n}^{- 1} (c o s x) ∣_{0}^{π}$ $\frac{- I}{π} = \frac{1}{2} \times {{t a n}^{- 1} (- 1) - {t a n}^{- 1} (1)}$ $\frac{- I}{π} = \frac{1}{2} \times - 2 \times \frac{π}{4}$ $I = \frac{π^{2}}{4}$
	Commented by maxmathsup by imad last updated on 13/Apr/19

	$s i r t a n m a y y o u r a n s w e r i s c o r r e c t t h a n k s .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19

	$m o s t w e l c o m e s i r$
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