∫_( 0) ^π ((x tan x)/(sec x+cos x)) dx =


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Question Number 52947 by gunawan last updated on 15/Jan/19

$\underset{0}{\int^{π}} \frac{x \tan x}{\sec x + \cos x} d x =$
	Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19
	$∫_0 ^π ((x×((sinx)/(cosx)))/((1/(cosx))+cosx))dx ∫_0 ^π ((xsinx)/(1+cos^2 x))dx=∫_0 ^π (((π−x)sin(π−x))/(1+cos^2 (π−x)))dx 2I=∫_0 ^π ((xsinx+(π−x)sinx)/(1+cos^2 x))dx 2I=π∫_0 ^π ((sinx)/(1+cos^2 x))dx 2I=(−π)∫_0 ^π ((d(cosx))/(1+cos^2 x)) −2I=π×∣tan^(−1) (cosx)∣_0 ^π I=(((−π)/2) )×{tan^(−1) (−1)−tan^(−1) (1)} =(((−π))/2)×{−2tan^(−1) (1)} =(π/2)×2×(π/4)=(π^2 /4) pls check...$
	$\int_{0}^{π} \frac{x \times \frac{s i n x}{c o s x}}{\frac{1}{c o s x} + c o s x} d x$ $\int_{0}^{π} \frac{x s i n x}{1 + {c o s}^{2} x} d x = \int_{0}^{π} \frac{(π - x) s i n (π - x)}{1 + {c o s}^{2} (π - x)} d x$ $2 I = \int_{0}^{π} \frac{x s i n x + (π - x) s i n x}{1 + {c o s}^{2} x} d x$ $2 I = π \int_{0}^{π} \frac{s i n x}{1 + {c o s}^{2} x} d x$ $2 I = (- π) \int_{0}^{π} \frac{d (c o s x)}{1 + {c o s}^{2} x}$ $- 2 I = π \times ∣ {t a n}^{- 1} (c o s x) ∣_{0}^{π}$ $I = (\frac{- π}{2}) \times {{t a n}^{- 1} (- 1) - {t a n}^{- 1} (1)}$ $= \frac{(- π)}{2} \times {- 2 {t a n}^{- 1} (1)}$ $= \frac{π}{2} \times 2 \times \frac{π}{4} = \frac{π^{2}}{4} p l s c h e c k . . .$
	Commented by gunawan last updated on 16/Jan/19

	$Good bless you Sir$
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