Question Number 145487 by Mrsof last updated on 05/Jul/21
Commented by Mrsof last updated on 05/Jul/21
$${help}\:{me}\:{sir} \\ $$
Answered by Ar Brandon last updated on 05/Jul/21
![I=∫_0 ^π tan^5 (x)sec^3 (x)dx =∫_0 ^π tan^4 (x)sec^2 (x)∙sec(x)tan(x)dx =∫_0 ^π (sec^2 (x)−1)^2 sec^2 (x)d(sec(x)) =∫_0 ^π (sec^6 (x)−2sec^4 (x)+sec^2 (x))d(sec(x)) =[((sec^7 (x))/7)−((2sec^5 (x))/5)+((sec^3 (x))/3)]_0 ^π =(−(1/7)+(2/5)−(1/3))−((1/7)−(2/5)+(1/3))=−((16)/(105)) ∣I∣=∣−((16)/(105))∣=((16)/(105))](https://www.tinkutara.com/question/Q145497.png)
$$\mathrm{I}=\int_{\mathrm{0}} ^{\pi} \mathrm{tan}^{\mathrm{5}} \left(\mathrm{x}\right)\mathrm{sec}^{\mathrm{3}} \left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\pi} \mathrm{tan}^{\mathrm{4}} \left(\mathrm{x}\right)\mathrm{sec}^{\mathrm{2}} \left(\mathrm{x}\right)\centerdot\mathrm{sec}\left(\mathrm{x}\right)\mathrm{tan}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\pi} \left(\mathrm{sec}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{1}\right)^{\mathrm{2}} \mathrm{sec}^{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{d}\left(\mathrm{sec}\left(\mathrm{x}\right)\right) \\ $$$$\:\:=\int_{\mathrm{0}} ^{\pi} \left(\mathrm{sec}^{\mathrm{6}} \left(\mathrm{x}\right)−\mathrm{2sec}^{\mathrm{4}} \left(\mathrm{x}\right)+\mathrm{sec}^{\mathrm{2}} \left(\mathrm{x}\right)\right)\mathrm{d}\left(\mathrm{sec}\left(\mathrm{x}\right)\right) \\ $$$$\:\:=\left[\frac{\mathrm{sec}^{\mathrm{7}} \left(\mathrm{x}\right)}{\mathrm{7}}−\frac{\mathrm{2sec}^{\mathrm{5}} \left(\mathrm{x}\right)}{\mathrm{5}}+\frac{\mathrm{sec}^{\mathrm{3}} \left(\mathrm{x}\right)}{\mathrm{3}}\right]_{\mathrm{0}} ^{\pi} \\ $$$$\:\:=\left(−\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{3}}\right)−\left(\frac{\mathrm{1}}{\mathrm{7}}−\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}}\right)=−\frac{\mathrm{16}}{\mathrm{105}} \\ $$$$\mid\mathrm{I}\mid=\mid−\frac{\mathrm{16}}{\mathrm{105}}\mid=\frac{\mathrm{16}}{\mathrm{105}} \\ $$
Commented by Ar Brandon last updated on 05/Jul/21
$$\mathrm{I}=\int_{\mathrm{0}} ^{\pi} \mathrm{tan}^{\mathrm{5}} \left(\mathrm{x}\right)\mathrm{sec}^{\mathrm{3}} \left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}^{\mathrm{5}} \left(\mathrm{x}\right)\mathrm{sec}^{\mathrm{3}} \left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\:\:=\beta\left(\mathrm{3},−\frac{\mathrm{7}}{\mathrm{2}}\right)=\frac{\mathrm{2}\Gamma\left(−\frac{\mathrm{7}}{\mathrm{2}}\right)}{\Gamma\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\:\:=\mathrm{2}\left(−\frac{\mathrm{2}}{\mathrm{7}}\right)\left(−\frac{\mathrm{2}}{\mathrm{5}}\right)\left(−\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$$\:\:=−\frac{\mathrm{16}}{\mathrm{105}}\:\Rightarrow\mid\mathrm{I}\mid=\mid−\frac{\mathrm{16}}{\mathrm{105}}\mid=\frac{\mathrm{16}}{\mathrm{105}} \\ $$
Answered by puissant last updated on 05/Jul/21
$$=\mid\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sin}^{\mathrm{5}} \left(\mathrm{x}\right)}{\mathrm{cos}^{\mathrm{8}} \left(\mathrm{x}\right)}\mathrm{dx}\mid\:=\:\mid\int_{\mathrm{0}} ^{\pi} \frac{\left(\mathrm{1}−\mathrm{2cos}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{cos}^{\mathrm{4}} \left(\mathrm{x}\right)\right)\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{cos}^{\mathrm{8}} \left(\mathrm{x}\right)}\mathrm{dx}\mid \\ $$$$\mathrm{u}=\mathrm{cos}\left(\mathrm{x}\right)\Rightarrow\mathrm{du}=−\mathrm{sin}\left(\mathrm{x}\right)\mathrm{dx}\Rightarrow\mathrm{dx}=−\frac{\mathrm{du}}{\mathrm{sin}\left(\mathrm{x}\right)} \\ $$$$\mid\int_{\mathrm{1}} ^{−\mathrm{1}} \frac{\left(\mathrm{1}−\mathrm{2u}^{\mathrm{2}} +\mathrm{u}^{\mathrm{4}} \right)\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{u}^{\mathrm{8}} }×\left(−\frac{\mathrm{du}}{\mathrm{sin}\left(\mathrm{x}\right)}\right)\mid \\ $$$$=\mid\int_{−\mathrm{1}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{8}} }−\frac{\mathrm{2}}{\mathrm{u}^{\mathrm{6}} }+\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{4}} }\right)\mathrm{du}\mid=…. \\ $$