Question Number 179799 by yaslm last updated on 02/Nov/22
Answered by MJS_new last updated on 02/Nov/22
![∫(dx/( (√(cos^3 x))(√(1+sin x))))= [t=tan x +(1/(cos x)) ⇔ x=arctan ((t^2 −1)/(2t)) → dx=((2dt)/(t^2 +1))] =(1/2)∫((t^2 +1)/t^(5/2) )dt=∫((1/(2t^(1/2) ))+(1/(2t^(5/2) )))dt= =(√t)−(1/(3t^(3/2) ))= =((2(1+2sin x))/(3(√((1+sin x)cos x))))+C](https://www.tinkutara.com/question/Q179804.png)
$$\int\frac{{dx}}{\:\sqrt{\mathrm{cos}^{\mathrm{3}} \:{x}}\sqrt{\mathrm{1}+\mathrm{sin}\:{x}}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:+\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\:\Leftrightarrow\:{x}=\mathrm{arctan}\:\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{t}}\:\rightarrow\:{dx}=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{5}/\mathrm{2}} }{dt}=\int\left(\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{1}/\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{5}/\mathrm{2}} }\right){dt}= \\ $$$$=\sqrt{{t}}−\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{3}/\mathrm{2}} }= \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}+\mathrm{2sin}\:{x}\right)}{\mathrm{3}\sqrt{\left(\mathrm{1}+\mathrm{sin}\:{x}\right)\mathrm{cos}\:{x}}}+{C} \\ $$
Answered by Frix last updated on 02/Nov/22
$$\int\frac{{dx}}{\:\sqrt{\mathrm{cos}^{\mathrm{3}} \:{x}}\sqrt{\mathrm{sin}\:{x}\:+\mathrm{1}}}\overset{{t}=\mathrm{tan}\:{x}} {=} \\ $$$$=\int\frac{{dt}}{\:\sqrt{{t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}}\overset{{u}=\frac{\mathrm{1}}{\:\sqrt{{t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}}} {=} \\ $$$$=−\int\frac{{u}^{\mathrm{4}} +\mathrm{1}}{{u}^{\mathrm{2}} }{du}=−\frac{{u}^{\mathrm{4}} −\mathrm{3}}{\mathrm{3}{u}}=… \\ $$