(1) ∫ (dx/(1−2cos x)) (2) ∫ ((sin 2x)/(sin x−sin^2 2x)) dx (3) ∫ (dx/(cos 2x−sin x))


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Question Number 161285 by cortano last updated on 15/Dec/21

$(1) \int \frac{d x}{1 - 2 \cos x}$ $(2) \int \frac{\sin 2 x}{\sin x - \sin^{2} 2 x} d x$ $(3) \int \frac{d x}{\cos 2 x - \sin x}$
	Answered by bobhans last updated on 15/Dec/21
	$(1) ∫ (dx/(1−2cos x)) = ∫ (dx/(1−2(2cos^2 ((x/2))−1))) = ∫ (dx/(3−4cos^2 ((x/2)))) ; [tan (x/2)=t → { ((cos (x/2)=(1/( (√(1+t^2 )))))),((dx=(2/(1+t^2 )) dt)) :}] = ∫ (2/(1+t^2 )) ((1/(3−(4/(1+t^2 )))))dt = ∫ (2/(3t^2 −1)) dt = ∫ (2/((t(√3)−1)(t(√3)+1))) dt = ∫ ((1/(t(√3)−1)) −(1/(t(√3)+1)))dt = (1/( (√3))) ln ∣t(√3)−1∣−(1/( (√3))) ln ∣t(√3)+1∣ + c = (1/( (√3))) ln ∣(((√3) tan ((x/2))−1)/( (√3) tan ((x/2))+1))∣ + c$
	$(1) \int \frac{dx}{1 - 2 \cos x} = \int \frac{dx}{1 - 2 (2 \cos^{2} (\frac{x}{2}) - 1)}$ $= \int \frac{dx}{3 - 4 \cos^{2} (\frac{x}{2})}; [\tan \frac{x}{2} = t \to {\begin{cases} \cos \frac{x}{2} = \frac{1}{\sqrt{1 + t^{2}}} \\ dx = \frac{2}{1 + t^{2}} dt \end{cases}]$ $= \int \frac{2}{1 + t^{2}} (\frac{1}{3 - \frac{4}{1 + t^{2}}}) dt$ $= \int \frac{2}{{3 t}^{2} - 1} dt = \int \frac{2}{(t \sqrt{3} - 1) (t \sqrt{3} + 1)} dt$ $= \int (\frac{1}{t \sqrt{3} - 1} - \frac{1}{t \sqrt{3} + 1}) dt$ $= \frac{1}{\sqrt{3}} \ln ∣ t \sqrt{3} - 1 ∣ - \frac{1}{\sqrt{3}} \ln ∣ t \sqrt{3} + 1 ∣ + c$ $= \frac{1}{\sqrt{3}} \ln ∣ \frac{\sqrt{3} \tan (\frac{x}{2}) - 1}{\sqrt{3} \tan (\frac{x}{2}) + 1} ∣ + c$
	Answered by bobhans last updated on 17/Dec/21
	$(Q) ∫ (dx/(cos 2x−sin x)) =? (⇒ ) ∫ (dx/(1−2sin^2 x−sin x)) = ∫ (dx/(−2sin^2 x−sin x+1)) =−∫ (dx/(2sin^2 x+sin x−1)) = −∫ (dx/((2sin x−1)(sin x+1))) = −(1/3)∫ ((2/(2sin x−1))−(1/(sin x+1)))dx [ tan (x/2)=u → { ((sin x = ((2u)/(1+u^2 )))),((dx=(2/(1+u^2 )))) :} ] I_1 =−(2/3) ∫ (2/(1+u^2 )) .(1/(((4u)/(1+u^2 ))−1)) du I_1 = −(4/3)∫ (du/(4u−u^2 −1))= (4/3)∫ (du/((u−2)^2 −5)) [ let u−2 =(√5) sec t ] I_1 = (4/3) ∫ (((√5) sec t tan t dt)/(5 tan^2 t))=(4/(3(√5)))∫csc t dt I_1 = (4/(3(√5))) ln ∣ csc t−cot t ∣ + c_(1 ) I_2 = (1/3)∫ (dx/(sin x+1)) = (1/3)∫ (2/(1+u^2 )).(1/(((2u)/(1+u^2 ))+1)) du I_2 = (2/3)∫ (du/((u+1)^2 )) = (2/3)∫ (u+1)^(−2) du I_2 = −(2/(3(u+1))) + c_2 =−(2/(3(tan ((x/2))+1))) + c_2 ∴ I = I_1 +I_2$
	$(Q) \int \frac{dx}{\cos 2 x - \sin x} = ?$ $(\Rightarrow) \int \frac{dx}{1 - 2 \sin^{2} x - \sin x} = \int \frac{dx}{- {2 \sin}^{2} x - \sin x + 1}$ $= - \int \frac{dx}{2 \sin^{2} x + \sin x - 1} = - \int \frac{dx}{(2 \sin x - 1) (\sin x + 1)}$ $= - \frac{1}{3} \int (\frac{2}{2 \sin x - 1} - \frac{1}{\sin x + 1}) dx$ $[\tan \frac{x}{2} = u \to {\begin{cases} \sin x = \frac{2 u}{1 + u^{2}} \\ dx = \frac{2}{1 + u^{2}} \end{cases}]$ $I_{1} = - \frac{2}{3} \int \frac{2}{1 + u^{2}} . \frac{1}{\frac{4 u}{1 + u^{2}} - 1} du$ $I_{1} = - \frac{4}{3} \int \frac{du}{4 u - u^{2} - 1} = \frac{4}{3} \int \frac{du}{{(u - 2)}^{2} - 5}$ $[let u - 2 = \sqrt{5} \sec t]$ $I_{1} = \frac{4}{3} \int \frac{\sqrt{5} \sec t \tan t dt}{5 \tan^{2} t} = \frac{4}{3 \sqrt{5}} \int \csc t dt$ $I_{1} = \frac{4}{3 \sqrt{5}} \ln ∣ \csc t - \cot t ∣ + c_{1}$ $I_{2} = \frac{1}{3} \int \frac{dx}{\sin x + 1} = \frac{1}{3} \int \frac{2}{1 + u^{2}} . \frac{1}{\frac{2 u}{1 + u^{2}} + 1} du$ $I_{2} = \frac{2}{3} \int \frac{du}{{(u + 1)}^{2}} = \frac{2}{3} \int {(u + 1)}^{- 2} du$ $I_{2} = - \frac{2}{3 (u + 1)} + c_{2} = - \frac{2}{3 (\tan (\frac{x}{2}) + 1)} + c_{2}$ $∴ I = I_{1} + I_{2}$
	Answered by Ar Brandon last updated on 15/Dec/21

	$I = \int \frac{d x}{1 - 2 \cos x}, t = \tan \frac{x}{2}$ $= \int \frac{2}{1 - 2 \frac{1 - t^{2}}{1 + t^{2}}} \cdot \frac{d t}{1 + t^{2}} = 2 \int \frac{d t}{3 t^{2} - 1}$ $= - \frac{2}{\sqrt{3}} argth (\sqrt{3} t) + C = - \frac{1}{\sqrt{3}} \ln ∣ \frac{1 + \sqrt{3} t}{1 - \sqrt{3} t} ∣ + C$ $= \frac{\sqrt{3}}{3} \ln ∣ \frac{1 - \sqrt{3} \tan \frac{x}{2}}{1 + \sqrt{3} \tan \frac{x}{2}} ∣ + C$
	Answered by Ar Brandon last updated on 15/Dec/21

	$J = \int \frac{\sin 2 x}{\sin x - \sin^{2} 2 x} d x = \int \frac{2 \sin x \cos x}{\sin x - {4 \sin}^{2} x \cos^{2} x} d x$ $= 2 \int \frac{\cos x}{1 - 4 \sin x (1 - \sin^{2} x)} d x = 2 \int \frac{d (\sin x)}{{4 \sin}^{3} x - 4 \sin x + 1}$
	Answered by Tyller last updated on 19/Dec/21

	$1) \int \frac{d x}{1 - 2 c o s x} .$ $f a z e n t o : c o s (x) = \frac{1 - u^{2}}{1 + u^{2}} .$ $u = t g (\frac{x}{2}) \Rightarrow d x = \frac{2 d u}{u^{2} + 1} \Rightarrow$ $\int \frac{2 d u}{3 u^{2} - 1} = I .$ $∴ I = \frac{2}{\sqrt{3}} l n ∣ \frac{u + \sqrt{3}}{\sqrt{u^{2} - 3}} ∣= \frac{2}{\sqrt{3}} l n ∣ \frac{t g (\frac{x}{2}) + \sqrt{3}}{\sqrt{{t g}^{2} (\frac{x}{2}) - 3}} ∣ + c$
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