Question Number 226290 by Spillover last updated on 25/Nov/25
Answered by Frix last updated on 25/Nov/25
![=∫(√((cos x (1−cos^2 x))/(1−cos^3 x))) dx= =∫sin x(√((cos x)/(1−cos^3 x))) dx =^([t=cos^(3/2) x]) =−(2/3)∫(dt/( (√(1−t^2 ))))= =−(2/3)sin^(−1) t = =−(2/3)sin^(−1) (cos^(3/2) x) +C](https://www.tinkutara.com/question/Q226345.png)
$$=\int\sqrt{\frac{\mathrm{cos}\:{x}\:\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{x}\right)}{\mathrm{1}−\mathrm{cos}^{\mathrm{3}} \:{x}}}\:{dx}= \\ $$$$=\int\mathrm{sin}\:{x}\sqrt{\frac{\mathrm{cos}\:{x}}{\mathrm{1}−\mathrm{cos}^{\mathrm{3}} \:{x}}}\:{dx}\:\overset{\left[{t}=\mathrm{cos}^{\frac{\mathrm{3}}{\mathrm{2}}} \:{x}\right]} {=} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}= \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \:{t}\:= \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \:\left(\mathrm{cos}^{\frac{\mathrm{3}}{\mathrm{2}}} \:{x}\right)\:+{C} \\ $$
Commented by Spillover last updated on 26/Nov/25
$${great} \\ $$