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Question Number 201351 by sonukgindia last updated on 05/Dec/23

	Answered by Sutrisno last updated on 05/Dec/23

	$= \int_{0}^{π} \frac{x}{1 + s i n x} . \frac{1 - s i n x}{1 - s i n x} d x$ $= \int_{0}^{π} \frac{x - x s i n x}{1 - {s i n}^{2} x} d x$ $= \int_{0}^{π} \frac{x - x s i n x}{{c o s}^{2} x} d x$ $= \int_{0}^{π} {x s e c}^{2} x - x s i n x . {c o s}^{- 2} x d x (u s e i n t e g r a l b y p a r t s)$ $= (x t a n x - l n ∣ s e c x ∣) - (x s e c x - l n ∣ s e c x + t a n x) \underset{0}{\overset{π}{∣}}$ $= π$
	Answered by Calculusboy last updated on 05/Dec/23
	$Solution: let y=𝛑−x dy=−dx when x=𝛑 y=0 and when x=0 y=𝛑 I=∫_𝛑 ^0 (((𝛑−y))/(1+sin(𝛑−y)))−dy ⇔ I=∫_0 ^𝛑 (((𝛑−y))/(1+sin(𝛑−y)))dy NB: sin(𝛑−y)=sin𝛑cosy−cos𝛑siny=siny I=∫_0 ^𝛑 (𝛑/(1+siny))dy−∫_0 ^𝛑 (y/(1+siny))dy I=𝛑∫_0 ^𝛑 (1/(1+siny))dy−I 2I=𝛑∫_0 ^𝛑 (1/(1+siny))dy (changing of variable) I=(𝛑/2)∫_0 ^𝛑 (1/(1+sinx))dx ⇔ I=(𝛑/2)∫_0 ^𝛑 (1/(1+sinx))∙((1−sinx)/(1−sinx))dx NB: (a+b)(a−b)=a^2 −b^2 ∴ (1+sinx)(1−sinx)=1−sin^2 x=cos^2 x I=(𝛑/2)∫_0 ^𝛑 ((1−sinx)/(1−sin^2 x))dx ⇔ I=(𝛑/2)∫_0 ^𝛑 ((1−sinx)/(cos^2 x))dx I=(𝛑/2)∫_0 ^𝛑 [(1/(cos^2 x))−((sinx)/(cos^2 x))]dx ⇔ I=(𝛑/2)∫_0 ^𝛑 [sec^2 x−secxtanx]dx I=(𝛑/2)[∫_0 ^𝛑 sec^2 xdx−∫_0 ^𝛑 secxtanxdx] I=(𝛑/2)[∣tanx∣_0 ^𝛑 −∣secx∣_0 ^𝛑 ] I=(𝛑/2){[tan(𝛑)−tan(0)]−[sec(𝛑)−sec(0)] I=(𝛑/2){(0−0)−(−1−1)} I=(𝛑/2){−(−2)} I=(𝛑/2)×2 I=𝛑$
	$S o l u t i o n : l e t y = π - x d y = - d x$ $w h e n x = π y = 0 a n d w h e n x = 0 y = π$ $I = \int_{π}^{0} \frac{(π - y)}{1 + s i n (π - y)} - d y \Leftrightarrow I = \int_{0}^{π} \frac{(π - y)}{1 + s i n (π - y)} d y$ $N B : s i n (π - y) = s i n π c o s y - c o s π s i n y = s i n y$ $I = \int_{0}^{π} \frac{π}{1 + s i n y} d y - \int_{0}^{π} \frac{y}{1 + s i n y} d y$ $I = π \int_{0}^{π} \frac{1}{1 + s i n y} d y - I$ $2 I = π \int_{0}^{π} \frac{1}{1 + s i n y} d y (c h a n g i n g o f v a r i a b l e)$ $I = \frac{π}{2} \int_{0}^{π} \frac{1}{1 + s i n x} d x \Leftrightarrow I = \frac{π}{2} \int_{0}^{π} \frac{1}{1 + s i n x} \cdot \frac{1 - s i n x}{1 - s i n x} d x$ $N B : (a + b) (a - b) = a^{2} - b^{2} ∴ (1 + s i n x) (1 - s i n x) = 1 - {s i n}^{2} x = {c o s}^{2} x$ $I = \frac{π}{2} \int_{0}^{π} \frac{1 - s i n x}{1 - {s i n}^{2} x} d x \Leftrightarrow I = \frac{π}{2} \int_{0}^{π} \frac{1 - s i n x}{{c o s}^{2} x} d x$ $I = \frac{π}{2} \int_{0}^{π} [\frac{1}{{c o s}^{2} x} - \frac{s i n x}{{c o s}^{2} x}] d x \Leftrightarrow I = \frac{π}{2} \int_{0}^{π} [{s e c}^{2} x - s e c x t a n x] d x$ $I = \frac{π}{2} [\int_{0}^{π} {s e c}^{2} x d x - \int_{0}^{π} s e c x t a n x d x]$ $I = \frac{π}{2} [∣ t a n x ∣_{0}^{π} - ∣ s e c x ∣_{0}^{π}]$ $I = \frac{π}{2} {[t a n (π) - t a n (0)] - [s e c (π) - s e c (0)]$ $I = \frac{π}{2} {(0 - 0) - (- 1 - 1)}$ $I = \frac{π}{2} {- (- 2)}$ $I = \frac{π}{2} \times 2$ $I = π$
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