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Question Number 209687 by uuuuu last updated on 18/Jul/24

	Answered by lepuissantcedricjunior last updated on 18/Jul/24

	$k = \int \frac{s i n 2 x}{s i \overset{4}{n x} + c o \overset{2}{s x}} d x$ $= \int \frac{2 s i n x c o s x}{c o \overset{4}{s x} (1 + t a \overset{4}{n x})} d x$ $= \int \frac{2 s i n x}{c o \overset{3}{s x}} \times \frac{1}{1 + t a \overset{4}{n x}} d x$ $=> k = a r c t a n (t a \overset{2}{n x}) + k$
	Answered by mr W last updated on 19/Jul/24

	$= \int \frac{\sin 2 x}{{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x} d x$ $= \int \frac{\sin 2 x}{1 - \frac{\sin^{2} 2 x}{2}} d x$ $= \int \frac{2 \sin 2 x}{2 - \sin^{2} 2 x} d x$ $= - \int \frac{d (\cos 2 x)}{1 + \cos^{2} 2 x} = - \int \frac{d u}{1 + u^{2}}$ $= - \tan^{- 1} u + C$ $= - \tan^{- 1} (\cos 2 x) + C$ $= \tan^{- 1} (\tan^{2} x) + C$
	Answered by mathmax last updated on 19/Jul/24
	$I=∫ ((sin(2x))/(sin^4 x+cos^4 x))dx ⇒I=∫((sin(2x))/((sin^2 x+cos^2 x)^2 −2sin^2 x cos^2 x))dx =∫ ((sin(2x))/(1−(1/2)sin^2 (2x)))dx =∫ ((2sin(2x))/(2−sin^2 (2x)))dx (2x=t) =∫ ((sint)/(2−sin^2 t))dt =∫((sint)/(((√2)−sint)((√2)+sint)))dt =(1/(2(√2)))∫((1/( (√2)−sint))+(1/( (√2)+sint)))dt =(1/(2(√2))){ I_1 −I_2 } I_1 =_(tan((t/2))=y) ∫ ((2dy)/((1+y^2 )((√2)−((2y)/(1+y^2 ))))) =∫ ((2dy)/( (√2)+(√2)y^2 −2y))=(√2)∫(dy/(y^2 −(√2)y+1)) Δ=2−1=1 ⇒y_1 =(((√2)+1)/2) et y_2 =(((√2)−1)/2) ⇒I_1 =(√2)∫((1/(y−y_1 ))−(1/(y−y_2 )))dy =(√2)ln∣((y−y_1 )/(y−y_2 ))∣+c_1 =(√2)ln∣((tanx−(((√2)+1)/2))/(tanx−(((√2)−1)/2)))∣+c_1 =(√2)ln∣((2tanx−(√2)−1)/(2tanx−(√2)+1))∣+c_1 I_2 =∫ (dt/( (√2)+sint)) =_(tan((t/2))=y) ∫ ((2dy)/((1+y^2 )((√2)+((2y)/(1+y^2 ))))) =∫ ((2dy)/( (√2)+(√2)y^2 +2y))=(√2)∫(dy/(y^2 +(√2)y+1)) y_1 =((−(√2)+1)/2) and y_2 =((−(√2)−1)/2) I_2 =(√2)∫((1/(y−y_1 ))−(1/(y−y_2 )))dy =(√2)ln∣((y−y_1 )/(y−y_2 ))∣ +c_2 =(√2)ln∣((tanx+(((√2)−1)/2))/(tanx+(((√2)+1)/2)))∣+c_2 =(√2)ln∣((2tanx+(√2)−1)/(2tanx+(√2)+1))∣ +c_2 ⇒ I=(1/(2(√2))){I_1 −I_2 }=(1/2){ln∣((2tanx−(√2)−1)/(2tanx−(√2)+1))∣ −ln∣((2tanx+(√2)−1)/(2tanx+(√2)+1))∣ +c$
	$I = \int \frac{s i n (2 x)}{{s i n}^{4} x + {c o s}^{4} x} d x \Rightarrow I = \int \frac{s i n (2 x)}{{({s i n}^{2} x + {c o s}^{2} x)}^{2} - 2 {s i n}^{2} x {c o s}^{2} x} d x$ $= \int \frac{s i n (2 x)}{1 - \frac{1}{2} {s i n}^{2} (2 x)} d x$ $= \int \frac{2 s i n (2 x)}{2 - {s i n}^{2} (2 x)} d x (2 x = t)$ $= \int \frac{s i n t}{2 - {s i n}^{2} t} d t = \int \frac{s i n t}{(\sqrt{2} - s i n t) (\sqrt{2} + s i n t)} d t$ $= \frac{1}{2 \sqrt{2}} \int (\frac{1}{\sqrt{2} - s i n t} + \frac{1}{\sqrt{2} + s i n t}) d t$ $= \frac{1}{2 \sqrt{2}} {I_{1} - I_{2}}$ $I_{1} =_{t a n (\frac{t}{2}) = y} \int \frac{2 d y}{(1 + y^{2}) (\sqrt{2} - \frac{2 y}{1 + y^{2}})}$ $= \int \frac{2 d y}{\sqrt{2} + \sqrt{2} y^{2} - 2 y} = \sqrt{2} \int \frac{d y}{y^{2} - \sqrt{2} y + 1}$ $Δ = 2 - 1 = 1 \Rightarrow y_{1} = \frac{\sqrt{2} + 1}{2}$ $e t y_{2} = \frac{\sqrt{2} - 1}{2} \Rightarrow I_{1} = \sqrt{2} \int (\frac{1}{y - y_{1}} - \frac{1}{y - y_{2}}) d y$ $= \sqrt{2} l n ∣ \frac{y - y_{1}}{y - y_{2}} ∣ + c_{1} = \sqrt{2} l n ∣ \frac{t a n x - \frac{\sqrt{2} + 1}{2}}{t a n x - \frac{\sqrt{2} - 1}{2}} ∣ + c_{1}$ $= \sqrt{2} l n ∣ \frac{2 t a n x - \sqrt{2} - 1}{2 t a n x - \sqrt{2} + 1} ∣ + c_{1}$ $I_{2} = \int \frac{d t}{\sqrt{2} + s i n t} =_{t a n (\frac{t}{2}) = y} \int \frac{2 d y}{(1 + y^{2}) (\sqrt{2} + \frac{2 y}{1 + y^{2}})}$ $= \int \frac{2 d y}{\sqrt{2} + \sqrt{2} y^{2} + 2 y} = \sqrt{2} \int \frac{d y}{y^{2} + \sqrt{2} y + 1}$ $y_{1} = \frac{- \sqrt{2} + 1}{2} a n d y_{2} = \frac{- \sqrt{2} - 1}{2}$ $I_{2} = \sqrt{2} \int (\frac{1}{y - y_{1}} - \frac{1}{y - y_{2}}) d y$ $= \sqrt{2} l n ∣ \frac{y - y_{1}}{y - y_{2}} ∣ + c_{2} = \sqrt{2} l n ∣ \frac{t a n x + \frac{\sqrt{2} - 1}{2}}{t a n x + \frac{\sqrt{2} + 1}{2}} ∣ + c_{2}$ $= \sqrt{2} l n ∣ \frac{2 t a n x + \sqrt{2} - 1}{2 t a n x + \sqrt{2} + 1} ∣ + c_{2} \Rightarrow$ $I = \frac{1}{2 \sqrt{2}} {I_{1} - I_{2}} = \frac{1}{2} {l n ∣ \frac{2 t a n x - \sqrt{2} - 1}{2 t a n x - \sqrt{2} + 1} ∣$ $- l n ∣ \frac{2 t a n x + \sqrt{2} - 1}{2 t a n x + \sqrt{2} + 1} ∣ + c$
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