Question Number 214618 by kuldeep52 last updated on 13/Dec/24
$$\int_{\mathrm{0}} ^{\Pi/\mathrm{2}} \frac{\mathrm{3}\sqrt{\mathrm{tan}\:{x}}}{\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} \:}{dx} \\ $$
Answered by Frix last updated on 14/Dec/24
![3∫((√(tan x))/((cos x +sin x)^2 ))dx =^([t=(√(tan x))]) =6∫_ (t^2 /((t^2 +1)^2 ))dt=3∫(dt/(t^2 +1))−3∫((1−t^2 )/((t^2 +1)^2 ))dt= =3arctan t −((3t)/(t^2 +1))= =3arctan (√(tan x)) −((3(√(cos x sin x)))/(cos x +sin x))+C ⇒ Answer is ((3π)/2)](https://www.tinkutara.com/question/Q214627.png)
$$\mathrm{3}\int\frac{\sqrt{\mathrm{tan}\:{x}}}{\left(\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }{dx}\:\overset{\left[{t}=\sqrt{\mathrm{tan}\:{x}}\right]} {=} \\ $$$$=\mathrm{6}\underset{} {\int}\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}=\mathrm{3}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}−\mathrm{3}\int\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}= \\ $$$$=\mathrm{3arctan}\:{t}\:−\frac{\mathrm{3}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\mathrm{3arctan}\:\sqrt{\mathrm{tan}\:{x}}\:−\frac{\mathrm{3}\sqrt{\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}+{C} \\ $$$$\Rightarrow \\ $$$$\mathrm{Answer}\:\mathrm{is}\:\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$