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Question Number 124334 by Ar Brandon last updated on 02/Dec/20

∫(√((cosx−cos^3 x)/((1−cos^3 x))))dx

$$\int\sqrt{\frac{\mathrm{cos}{x}−\mathrm{cos}^{\mathrm{3}} {x}}{\left(\mathrm{1}−\mathrm{cos}^{\mathrm{3}} {x}\right)}}\mathrm{d}{x} \\ $$

Answered by MJS_new last updated on 02/Dec/20

=∫((sin x (√(cos x)))/( (√(1−cos^3  x))))dx=       [t=cos x → dx=−(dt/(sin x))]  =−∫((√t)/( (√(1−t^3 ))))dt=       [u=arcsin t^(3/2)  → dt=((2(√(1−t^3 )))/(3(√t)))du]  =−(2/3)∫du=−((2u)/3)=−(2/3)arcsin t^(3/2)  =  =−(2/3)arcsin cos^(3/2)  x +C

$$=\int\frac{\mathrm{sin}\:{x}\:\sqrt{\mathrm{cos}\:{x}}}{\:\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{3}} \:{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{cos}\:{x}\:\rightarrow\:{dx}=−\frac{{dt}}{\mathrm{sin}\:{x}}\right] \\ $$$$=−\int\frac{\sqrt{{t}}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{3}} }}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{arcsin}\:{t}^{\mathrm{3}/\mathrm{2}} \:\rightarrow\:{dt}=\frac{\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{3}} }}{\mathrm{3}\sqrt{{t}}}{du}\right] \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}\int{du}=−\frac{\mathrm{2}{u}}{\mathrm{3}}=−\frac{\mathrm{2}}{\mathrm{3}}\mathrm{arcsin}\:{t}^{\mathrm{3}/\mathrm{2}} \:= \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}\mathrm{arcsin}\:\mathrm{cos}^{\mathrm{3}/\mathrm{2}} \:{x}\:+{C} \\ $$

Commented by mr W last updated on 02/Dec/20

To Mammadli sir:  you should have seen that all of  your posts are red flagged. it means,  people don′t like all of your posts.   why? because you posted your questions  in the threads of other people′s  questions. this disturbs and is also  not respectful to other people.   i think other people have told you  this for many times, but you just  ignored all. i think other people   have also told you how to open a new  thread for your own question, you  ignored also this. WHAT IS YOUR  PROBLEM? don′t you understand  English? which language do you  understand then?

$${To}\:{Mammadli}\:{sir}: \\ $$$${you}\:{should}\:{have}\:{seen}\:{that}\:{all}\:{of} \\ $$$${your}\:{posts}\:{are}\:{red}\:{flagged}.\:{it}\:{means}, \\ $$$${people}\:{don}'{t}\:{like}\:{all}\:{of}\:{your}\:{posts}.\: \\ $$$${why}?\:{because}\:{you}\:{posted}\:{your}\:{questions} \\ $$$${in}\:{the}\:{threads}\:{of}\:{other}\:{people}'{s} \\ $$$${questions}.\:{this}\:{disturbs}\:{and}\:{is}\:{also} \\ $$$${not}\:{respectful}\:{to}\:{other}\:{people}.\: \\ $$$${i}\:{think}\:{other}\:{people}\:{have}\:{told}\:{you} \\ $$$${this}\:{for}\:{many}\:{times},\:{but}\:{you}\:{just} \\ $$$${ignored}\:{all}.\:{i}\:{think}\:{other}\:{people}\: \\ $$$${have}\:{also}\:{told}\:{you}\:{how}\:{to}\:{open}\:{a}\:{new} \\ $$$${thread}\:{for}\:{your}\:{own}\:{question},\:{you} \\ $$$${ignored}\:{also}\:{this}.\:{WHAT}\:{IS}\:{YOUR} \\ $$$${PROBLEM}?\:{don}'{t}\:{you}\:{understand} \\ $$$${English}?\:{which}\:{language}\:{do}\:{you} \\ $$$${understand}\:{then}? \\ $$

Commented by Mammadli last updated on 02/Dec/20

Dear ser, I am asking questions that I do not know and I would like you to help

Commented by Mammadli last updated on 02/Dec/20

I do not share correctly or how dear ser

Commented by mr W last updated on 02/Dec/20

thanks for this clear answer!  you should not post your question   as comment to a question of other  people. you should open a new post  for your question in following way:    to type your question in text or  formula just use command (1).    to post your question as image  just use command (2).

$${thanks}\:{for}\:{this}\:{clear}\:{answer}! \\ $$$${you}\:{should}\:{not}\:{post}\:{your}\:{question}\: \\ $$$${as}\:{comment}\:{to}\:{a}\:{question}\:{of}\:{other} \\ $$$${people}.\:{you}\:{should}\:{open}\:{a}\:{new}\:{post} \\ $$$${for}\:{your}\:{question}\:{in}\:{following}\:{way}: \\ $$$$ \\ $$$${to}\:{type}\:{your}\:{question}\:{in}\:{text}\:{or} \\ $$$${formula}\:{just}\:{use}\:{command}\:\left(\mathrm{1}\right). \\ $$$$ \\ $$$${to}\:{post}\:{your}\:{question}\:{as}\:{image} \\ $$$${just}\:{use}\:{command}\:\left(\mathrm{2}\right). \\ $$

Commented by mr W last updated on 02/Dec/20

Commented by Mammadli last updated on 02/Dec/20

Thank you very much dear Ser..

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