Question Number 124334 by Ar Brandon last updated on 02/Dec/20
$$\int\sqrt{\frac{\mathrm{cos}{x}−\mathrm{cos}^{\mathrm{3}} {x}}{\left(\mathrm{1}−\mathrm{cos}^{\mathrm{3}} {x}\right)}}\mathrm{d}{x} \\ $$
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](https://www.tinkutara.com/question/Q124335.png)
$$=\int\frac{\mathrm{sin}\:{x}\:\sqrt{\mathrm{cos}\:{x}}}{\:\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{3}} \:{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{cos}\:{x}\:\rightarrow\:{dx}=−\frac{{dt}}{\mathrm{sin}\:{x}}\right] \\ $$$$=−\int\frac{\sqrt{{t}}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{3}} }}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{arcsin}\:{t}^{\mathrm{3}/\mathrm{2}} \:\rightarrow\:{dt}=\frac{\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{3}} }}{\mathrm{3}\sqrt{{t}}}{du}\right] \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}\int{du}=−\frac{\mathrm{2}{u}}{\mathrm{3}}=−\frac{\mathrm{2}}{\mathrm{3}}\mathrm{arcsin}\:{t}^{\mathrm{3}/\mathrm{2}} \:= \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}\mathrm{arcsin}\:\mathrm{cos}^{\mathrm{3}/\mathrm{2}} \:{x}\:+{C} \\ $$
Commented by mr W last updated on 02/Dec/20
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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
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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..