Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 166488 by bobhans last updated on 21/Feb/22

B=∫ (√((sin 2x−1)/(cos 2x−1))) dx =?

$$\:\:\:\mathrm{B}=\int\:\sqrt{\frac{\mathrm{sin}\:\mathrm{2x}−\mathrm{1}}{\mathrm{cos}\:\mathrm{2x}−\mathrm{1}}}\:\mathrm{dx}\:=? \\ $$

Answered by greogoury55 last updated on 21/Feb/22

B=∫ (√((1−sin 2x)/(1−cos 2x))) dx B=∫ (√((sin^2 x+cos^2 x−2sin x cos x)/(2sin^2 x))) dx B=(1/( (√2))) ∫ ((∣sin x−cos x∣)/(∣sin x∣)) dx → { ((∣sin x−cos x∣= { ((sin x−cos x)),((cos x−sin x)) :})),((∣sin x∣= { ((sin x)),((−sin x)) :})) :} B_1 =(1/( (√2)))∫(1−cot x)dx=(1/( (√2)))(x−ln ∣sin x∣)+c B_2 =(1/( (√2)))∫(−1+cot x)dx=(1/( (√2)))(−x+ln ∣sin x∣)+c

$$\:\:\:{B}=\int\:\sqrt{\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}}\:{dx} \\ $$$$\:\:\:{B}=\int\:\sqrt{\frac{\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}}{\mathrm{2sin}\:^{\mathrm{2}} {x}}}\:{dx} \\ $$$$\:\:\:{B}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\frac{\mid\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\mid}{\mid\mathrm{sin}\:{x}\mid}\:{dx} \\ $$$$\:\rightarrow\begin{cases}{\mid\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\mid=\begin{cases}{\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}\\{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}\end{cases}}\\{\mid\mathrm{sin}\:{x}\mid=\begin{cases}{\mathrm{sin}\:{x}}\\{−\mathrm{sin}\:{x}}\end{cases}}\end{cases} \\ $$$$\:\:\:{B}_{\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\left(\mathrm{1}−\mathrm{cot}\:{x}\right){dx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left({x}−\mathrm{ln}\:\mid\mathrm{sin}\:{x}\mid\right)+{c} \\ $$$$\:\:{B}_{\mathrm{2}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\left(−\mathrm{1}+\mathrm{cot}\:{x}\right){dx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(−{x}+\mathrm{ln}\:\mid\mathrm{sin}\:{x}\mid\right)+{c} \\ $$$$\:\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com