Question Number 45669 by arvinddayama01@gmail.com last updated on 15/Oct/18
![∫tan^(−1) (√((1−sinx)/(1+sinx))) dx=?](https://www.tinkutara.com/question/Q45669.png)
$$\int{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}−{sinx}}{\mathrm{1}+{sinx}}}\:{dx}=? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18
![t=tan(x/2) now tan^(−1) (√((1−((2t)/(1+t^2 )))/(1+((2t)/(1+t^2 ))))) =(√(((1−t)^2 )/((1+t)^2 ))) =((1−t)/(1+t))=((1−tan(x/2))/(1+tan(x/2)))=tan((π/4)−(x/2)) so tan^(−1) {tan((π/4)−(x/2))}=(π/4)−(x/2) ∫(π/4)−(x/2) dx =((πx)/4)−(x^2 /4)+c](https://www.tinkutara.com/question/Q45676.png)
$${t}={tan}\frac{{x}}{\mathrm{2}}\:\:\: \\ $$$$ \\ $$$${now}\:{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}−\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}}\: \\ $$$$=\sqrt{\frac{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }}\:=\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}=\frac{\mathrm{1}−{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}\frac{{x}}{\mathrm{2}}}={tan}\left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right) \\ $$$${so}\:{tan}^{−\mathrm{1}} \left\{{tan}\left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)\right\}=\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}} \\ $$$$\int\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\:\:{dx} \\ $$$$=\frac{\pi{x}}{\mathrm{4}}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+{c} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18
![most welcome...](https://www.tinkutara.com/question/Q45679.png)
$${most}\:{welcome}… \\ $$
Commented by arvinddayama01@gmail.com last updated on 15/Oct/18
![Thanku very much](https://www.tinkutara.com/question/Q45678.png)
$${Thanku}\:{very}\:{much} \\ $$
Answered by vermasir last updated on 16/Oct/18
![∫tan^(−1) (√((1−cos((π/2)−x))/(1+cos((π/2)+x)))) ∫tan^(−1) (√((2sin^2 ((π/4)−(x/2)))/(2cos^2 ((π/4)−(x/2))))) ∫tan^(−1) ∣tan((π/4)−(x/2))∣dx ∫((π/4)−(x/2))dx (π/4)x−(x^2 /4)+c](https://www.tinkutara.com/question/Q45757.png)
$$\int\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{\mathrm{1}+{cos}\left(\frac{\pi}{\mathrm{2}}+{x}\right)}} \\ $$$$\int\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)}} \\ $$$$\int\mathrm{tan}^{−\mathrm{1}} \mid{tan}\left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)\mid{dx} \\ $$$$\int\left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$\frac{\pi}{\mathrm{4}}{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+{c} \\ $$