Question-156739

Question Number 156739 by cortano last updated on 15/Oct/21

Commented by john_santu last updated on 15/Oct/21

B = \int \sqrt{\frac{\sin x + \cos x}{\sin x - \cos x}} d x

= \int \sqrt{\frac{\sin (x + \frac{π}{4})}{- \cos (x + \frac{π}{4})}} d x

= \int \sqrt{- \tan (x + \frac{π}{4})} d x

= \int \sqrt{\cot (x + \frac{3 π}{4})} d (x + \frac{3 π}{4})

= \int \frac{d q}{\sqrt{\tan q}}

l e t y = \sqrt{\tan q} \Rightarrow q = \tan^{- 1} (y^{2})

d q = \frac{2 y}{1 + y^{4}} d y

B = \int \frac{1}{y} . \frac{2 y}{1 + y^{4}} d y = \int \frac{2 d y}{1 + y^{4}}

= \int \frac{(y^{2} + 1) - (y^{2} - 1)}{1 + y^{4}} d y

= \int \frac{1 + \frac{1}{y^{2}}}{y^{2} + \frac{1}{y^{2}}} d y - \int \frac{1 - y^{2}}{y^{2} + \frac{1}{y^{2}}} d y

= \int \frac{d (y - \frac{1}{y})}{{(y - \frac{1}{y})}^{2} + 2} + \int \frac{d (y + \frac{1}{y})}{2 - {(y + \frac{1}{y})}^{2}}

= \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{1}{\sqrt{2}} (y - \frac{1}{y})) + \frac{1}{\sqrt{2}} \tanh^{- 1} (\frac{1}{\sqrt{2}} (y + \frac{1}{y})) + C

= \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{1}{\sqrt{2}} (\sqrt{\tan q} - \frac{1}{\sqrt{\tan q}})) + \frac{1}{\sqrt{2}} \tanh^{- 1} (\frac{1}{\sqrt{2}} (\sqrt{\tan q} + \frac{1}{\sqrt{\tan q}})) + C

= \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{1}{\sqrt{2}} (\sqrt{\tan (x + \frac{3 π}{4})} - \frac{1}{\sqrt{\tan (x + \frac{3 π}{4})}})

+ \frac{1}{\sqrt{2}} \tanh^{- 1} (\frac{1}{\sqrt{2}} (\sqrt{\tan (x + \frac{3 π}{4})} + \frac{1}{\sqrt{\tan (x + \frac{3 π}{4})}}) + C