Question Number 211344 by Nadirhashim last updated on 06/Sep/24
$$\:\:\:{find}\:\int\frac{\boldsymbol{{dx}}}{\boldsymbol{{sin}}^{\mathrm{3}} \left(\boldsymbol{{x}}\right)\:\boldsymbol{{cos}}^{\mathrm{5}} \left(\boldsymbol{{x}}\right)}\:.\boldsymbol{{dx}}\: \\ $$
Answered by Frix last updated on 06/Sep/24
![∫(dx/(cos^5 x sin^3 x))=∫(((1+tan^2 x)^4 )/(tan^3 x))dx =^([t=tan x]) =∫(((t^2 +1)^3 )/t^3 )dt=∫(t^3 +3t+(3/t)+(1/t^3 ))dt= =(t^4 /4)+((3t^2 )/2)+3ln t −(1/(2t^2 ))=((t^6 +6t^4 −2)/(4t^2 ))+3ln t = =((tan^6 x +6tan^4 x −2)/(4tan^2 x))+3ln ∣tan x∣ +C](https://www.tinkutara.com/question/Q211356.png)
$$\int\frac{{dx}}{\mathrm{cos}^{\mathrm{5}} \:{x}\:\mathrm{sin}^{\mathrm{3}} \:{x}}=\int\frac{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}\right)^{\mathrm{4}} }{\mathrm{tan}^{\mathrm{3}} \:{x}}{dx}\:\overset{\left[{t}=\mathrm{tan}\:{x}\right]} {=} \\ $$$$=\int\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{{t}^{\mathrm{3}} }{dt}=\int\left({t}^{\mathrm{3}} +\mathrm{3}{t}+\frac{\mathrm{3}}{{t}}+\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\right){dt}= \\ $$$$=\frac{{t}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{3}{t}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3ln}\:{t}\:−\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }=\frac{{t}^{\mathrm{6}} +\mathrm{6}{t}^{\mathrm{4}} −\mathrm{2}}{\mathrm{4}{t}^{\mathrm{2}} }+\mathrm{3ln}\:{t}\:= \\ $$$$=\frac{\mathrm{tan}^{\mathrm{6}} \:{x}\:+\mathrm{6tan}^{\mathrm{4}} \:{x}\:−\mathrm{2}}{\mathrm{4tan}^{\mathrm{2}} \:{x}}+\mathrm{3ln}\:\mid\mathrm{tan}\:{x}\mid\:+{C} \\ $$