Question Number 187135 by cortano12 last updated on 14/Feb/23
$$\:\underset{\:\:\mathrm{0}} {\overset{\:\:\pi/\mathrm{4}} {\int}}\:\frac{\mathrm{tan}\:^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{sin}\:{x}}\:{dx}\:=? \\ $$
Answered by horsebrand11 last updated on 14/Feb/23
![I=∫_( 0) ^( π/4) ((tan^2 x)/(1+sin x)) dx Let tan ((x/2))=((1−t)/(1+t)) { ((tan x=((1−t^2 )/(2t)))),((dx=−(1+sin x)dt)) :} λ =∫ (((1−t^2 )/(2t)))^2 dt=∫(((t^4 −2t^2 +1)/(4t^2 )))dt λ= ∫ ((1/4)t^2 −(1/2)+(1/4)t^(−2) )dt λ=(1/(12))t^3 −(1/2)t−(1/(4t)) +c I= [ (1/(12))t^3 −(1/2)t−(1/(4t)) ]_((√2)−1) ^1 I=((1/(12))−(1/2)−(1/4))−(((((√2)−1)^3 )/(12))−((((√2)−1))/2)−(1/(4((√2)−1)))) I= (((√2)−1)/3)](Q187136.png)
$$\:{I}=\underset{\:\mathrm{0}} {\overset{\:\pi/\mathrm{4}} {\int}}\:\frac{\mathrm{tan}\:^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{sin}\:{x}}\:{dx}\: \\ $$$$\:{Let}\:\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)=\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}} \\ $$$$\:\begin{cases}{\mathrm{tan}\:{x}=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}}\\{{dx}=−\left(\mathrm{1}+\mathrm{sin}\:{x}\right){dt}}\end{cases} \\ $$$$\:\lambda\:=\int\:\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}\right)^{\mathrm{2}} {dt}=\int\left(\frac{{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4}{t}^{\mathrm{2}} }\right){dt} \\ $$$$\:\lambda=\:\int\:\left(\frac{\mathrm{1}}{\mathrm{4}}{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}{t}^{−\mathrm{2}} \right){dt} \\ $$$$\:\lambda=\frac{\mathrm{1}}{\mathrm{12}}{t}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\mathrm{1}}{\mathrm{4}{t}}\:+{c}\: \\ $$$${I}=\:\left[\:\frac{\mathrm{1}}{\mathrm{12}}{t}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\mathrm{1}}{\mathrm{4}{t}}\:\right]_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\mathrm{1}} \\ $$$${I}=\left(\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\right)−\left(\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{12}}−\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}\right) \\ $$$${I}=\:\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{3}} \\ $$