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)
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)
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18
![most welcome...](https://www.tinkutara.com/question/Q45679.png)
Commented by arvinddayama01@gmail.com last updated on 15/Oct/18
![Thanku very much](https://www.tinkutara.com/question/Q45678.png)
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)