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Question Number 210517 by depressiveshrek last updated on 11/Aug/24

∫(1/(sinx−cos2x))dx

$$\int\frac{\mathrm{1}}{\mathrm{sin}{x}−\mathrm{cos2}{x}}{dx} \\ $$

Commented by Frix last updated on 11/Aug/24

Simply use t=tan (x/2) to get (2/(3(1+tan (x/2))))+((2(√3))/9)ln ∣((2−(√3)−tan (x/2))/(2+(√3)−tan (x/3)))∣ +C

$$\mathrm{Simply}\:\mathrm{use}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\mathrm{to}\:\mathrm{get} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}\left(\mathrm{1}+\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)}+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{ln}\:\mid\frac{\mathrm{2}−\sqrt{\mathrm{3}}−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{3}}−\mathrm{tan}\:\frac{{x}}{\mathrm{3}}}\mid\:+{C} \\ $$

Commented by depressiveshrek last updated on 11/Aug/24

This answer is unacceptable. Please show your work.

$$\mathrm{This}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{unacceptable}.\:\mathrm{Please} \\ $$$$\mathrm{show}\:\mathrm{your}\:\mathrm{work}. \\ $$

Answered by Frix last updated on 12/Aug/24

∫(dx/(sin x −cos 2x))= =−(1/3)∫((2/(1−2sin x))+(1/(1+sin x)))dx =^(t=((cos x)/(1+sin x))) =(1/3)∫((4/(3t^2 −1))+(1/3))dt= =((2(√3))/9)ln (((√3)t−1)/( (√3)t+1)) +(t/3)= =((cos x)/(3(1+sin x)))+((2(√3))/9)ln ∣((1−(√3)cos x +sin x)/(1+(√3)cos x +sin x))∣ +C

$$\int\frac{{dx}}{\mathrm{sin}\:{x}\:−\mathrm{cos}\:\mathrm{2}{x}}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\int\left(\frac{\mathrm{2}}{\mathrm{1}−\mathrm{2sin}\:{x}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}\:{x}}\right){dx}\:\overset{{t}=\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}} {=} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\left(\frac{\mathrm{4}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{3}}\right){dt}= \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{ln}\:\frac{\sqrt{\mathrm{3}}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}{t}+\mathrm{1}}\:+\frac{{t}}{\mathrm{3}}= \\ $$$$=\frac{\mathrm{cos}\:{x}}{\mathrm{3}\left(\mathrm{1}+\mathrm{sin}\:{x}\right)}+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{ln}\:\mid\frac{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}\mid\:+{C} \\ $$

Answered by Frix last updated on 12/Aug/24

∫(dx/(sin x −cos 2x)) =^(t=(√3)cot ((x/2)+(π/4))) =((√3)/9)∫((4/((t−1)(t+1)))+1)dt= =((2(√3))/9)ln ((t−1)/(t+1)) +(((√3)t)/9)= =(1/3)cot ((x/2)+(π/4)) +((2(√3))/9)ln ∣((cos ((x/2)+((5π)/(12))))/(cos ((x/2)+(π/(12)))))∣ +C

$$\int\frac{{dx}}{\mathrm{sin}\:{x}\:−\mathrm{cos}\:\mathrm{2}{x}}\:\overset{{t}=\sqrt{\mathrm{3}}\mathrm{cot}\:\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)} {=} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\int\left(\frac{\mathrm{4}}{\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)}+\mathrm{1}\right){dt}= \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{ln}\:\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\:+\frac{\sqrt{\mathrm{3}}{t}}{\mathrm{9}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cot}\:\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\:+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{ln}\:\mid\frac{\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}+\frac{\mathrm{5}\pi}{\mathrm{12}}\right)}{\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{12}}\right)}\mid\:+{C} \\ $$

Answered by mr W last updated on 12/Aug/24

=∫(1/(2 sin^2 x+sin x−1)) dx =∫(1/((2 sin x−1)(sin x+1))) dx =(1/3)∫((2/(2 sin x−1))−(1/(sin x+1)))dx =(1/3)∫((1/(sin x−(1/2)))−(1/(sin x+1)))dx =....

$$=\int\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}+\mathrm{sin}\:{x}−\mathrm{1}}\:{dx} \\ $$$$=\int\frac{\mathrm{1}}{\left(\mathrm{2}\:\mathrm{sin}\:{x}−\mathrm{1}\right)\left(\mathrm{sin}\:{x}+\mathrm{1}\right)}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\left(\frac{\mathrm{2}}{\mathrm{2}\:\mathrm{sin}\:{x}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{sin}\:{x}+\mathrm{1}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\left(\frac{\mathrm{1}}{\mathrm{sin}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{sin}\:{x}+\mathrm{1}}\right){dx} \\ $$$$=.... \\ $$

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