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Question Number 44573 by Raj Singh last updated on 01/Oct/18

Commented by maxmathsup by imad last updated on 01/Oct/18
$let A =∫_0 ^(π/4) ln(sin(2θ))dθ ⇒ A =_(2θ=t) ∫_0 ^(π/2) ln(sin(t))(dt/2) =(1/2) ∫_0 ^(π/2) ln(sint)dt let I =∫_0 ^(π/2) ln(sin(t))dt and J =∫_0 ^(π/2) ln(cost)dt changement t =(π/2)−u show that I =J ⇒2I =∫_0 ^(π/2) (ln(sint) +ln(cost))dt =∫_0 ^(π/2) ln((1/2)sin(2t))dt =−(π/2)ln(2) + ∫_0 ^(π/2) ln(sin(2t))dt but ∫_0 ^(π/2) ln(sin(2t)dt =_(2t =u) ∫_0 ^π ln(sin(u))(du/2) =(1/2){ ∫_0 ^(π/2) ln(sin(u))du + ∫_(π/2) ^π ln(sinu)du} =(1/2) I +(1/2) ∫_(π/2) ^π ln(sinu)du but ∫_(π/2) ^π ln(sin(u))du =_(u =(π/2) +θ) ∫_0 ^(π/2) ln(cosθ)dθ =I ⇒ 2I =−(π/2)ln(2) +I ⇒ I =−(π/2)ln(2) ⇒ A =−(π/4)ln(2) .$
$l e t A = \int_{0}^{\frac{π}{4}} l n (s i n (2 θ)) d θ \Rightarrow A =_{2 θ = t} \int_{0}^{\frac{π}{2}} l n (s i n (t)) \frac{d t}{2}$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} l n (s i n t) d t l e t I = \int_{0}^{\frac{π}{2}} l n (s i n (t)) d t a n d J = \int_{0}^{\frac{π}{2}} l n (c o s t) d t$ $c h a n g e m e n t t = \frac{π}{2} - u s h o w t h a t I = J \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} (l n (s i n t) + l n (c o s t)) d t$ $= \int_{0}^{\frac{π}{2}} l n (\frac{1}{2} s i n (2 t)) d t = - \frac{π}{2} l n (2) + \int_{0}^{\frac{π}{2}} l n (s i n (2 t)) d t b u t$ $\int_{0}^{\frac{π}{2}} l n (s i n (2 t) d t =_{2 t = u} \int_{0}^{π} l n (s i n (u)) \frac{d u}{2}$ $= \frac{1}{2} {\int_{0}^{\frac{π}{2}} l n (s i n (u)) d u + \int_{\frac{π}{2}}^{π} l n (s i n u) d u} = \frac{1}{2} I + \frac{1}{2} \int_{\frac{π}{2}}^{π} l n (s i n u) d u b u t$ $\int_{\frac{π}{2}}^{π} l n (s i n (u)) d u =_{u = \frac{π}{2} + θ} \int_{0}^{\frac{π}{2}} l n (c o s θ) d θ = I \Rightarrow$ $2 I = - \frac{π}{2} l n (2) + I \Rightarrow I = - \frac{π}{2} l n (2) \Rightarrow A = - \frac{π}{4} l n (2) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 01/Oct/18

	$I = \int_{0}^{\frac{π}{4}} l n s i n 2 θ d θ$ $I = \int_{0}^{\frac{π}{4}} l n s i n 2 (\frac{π}{4} - θ) d θ$ $= \int_{0}^{\frac{π}{2}} l n c o s 2 θ d θ$ $2 I = \int_{0}^{\frac{π}{4}} l n (s i n 2 θ) + l n c o s 2 θ d θ$ $2 I = \int_{0}^{\frac{π}{4}} l n (\frac{s i n 4 θ}{2}) d θ$ $2 I = \int_{0}^{\frac{π}{4}} l n s i n 4 θ d θ - l n 2 \int_{0}^{\frac{π}{4}} d θ$ $2 θ = t s o d θ = \frac{d t}{2}$ $2 I = \frac{1}{2} \int_{0}^{\frac{π}{2}} l n s i n 2 t d t - l n 2 \times \frac{π}{4}$ $2 I = \frac{1}{2} \times 2 \int_{0}^{\frac{π}{4}} l n s i n 2 t d t - l n 2 \times \frac{π}{4}$ $2 I = I - l n 2 \times \frac{π}{4}$ $I = - l n 2 \times \frac{π}{4}$
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