Question Number 215496 by MATHEMATICSAM last updated on 08/Jan/25
$$\int_{\mathrm{0}} ^{\pi} \frac{{x}\mathrm{tan}{x}}{\mathrm{sec}{x}\:+\:\mathrm{tan}{x}}\:{dx} \\ $$$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{integral}. \\ $$
Commented by mr W last updated on 09/Jan/25
$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\pi \\ $$
Answered by Frix last updated on 10/Jan/25
![∫((xtan x)/(sec x +tan x))dx=∫((xsin x)/(1+sin x))dx =^([by parts]) =((x(1+xcos x −sin x))/(cos x))−∫(x+(1/(cos x))−tan x)dx Should be easy from here...](https://www.tinkutara.com/question/Q215568.png)
$$\int\frac{{x}\mathrm{tan}\:{x}}{\mathrm{sec}\:{x}\:+\mathrm{tan}\:{x}}{dx}=\int\frac{{x}\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}{dx}\:\overset{\left[\mathrm{by}\:\mathrm{parts}\right]} {=}\: \\ $$$$=\frac{{x}\left(\mathrm{1}+{x}\mathrm{cos}\:{x}\:−\mathrm{sin}\:{x}\right)}{\mathrm{cos}\:{x}}−\int\left({x}+\frac{\mathrm{1}}{\mathrm{cos}\:{x}}−\mathrm{tan}\:{x}\right){dx} \\ $$$$\mathrm{Should}\:\mathrm{be}\:\mathrm{easy}\:\mathrm{from}\:\mathrm{here}… \\ $$