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

Commented by john_santu last updated on 15/Oct/21

$$B=\int \sqrt{\frac{\sin x+\cos x}{\sin x-\cos x}}dx$$$$=\int \sqrt{\frac{\sin (x+\frac{\pi }{4})}{-\cos (x+\frac{\pi }{4})}}dx$$$$=\int \sqrt{-\tan (x+\frac{\pi }{4})}dx$$$$=\int \sqrt{\cot (x+\frac{3\pi }{4})}d(x+\frac{3\pi }{4})$$$$=\int \frac{dq}{\sqrt{\tan q}}$$$$lety=\sqrt{\tan q}\Rightarrow q={\tan }^{-1}\left(y^2\right)$$$$dq=\frac{2y}{1+y^4}dy$$$$B=\int \frac{1}{y}.\frac{2y}{1+y^4}dy=\int \frac{2dy}{1+y^4}$$$$=\int \frac{(y^2+1)-(y^2-1)}{1+y^4}dy$$$$=\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy-\int \frac{1-y^2}{y^2+\frac{1}{y^2}}dy$$$$=\int \frac{d(y-\frac{1}{y})}{{(y-\frac{1}{y})}^2+2}+\int \frac{d(y+\frac{1}{y})}{2-{(y+\frac{1}{y})}^2}$$$$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{1}{\sqrt{2}}(y-\frac{1}{y})\right)+\frac{1}{\sqrt{2}}\tanh {}^{-1}\left(\frac{1}{\sqrt{2}}(y+\frac{1}{y})\right)+C$$$$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{1}{\sqrt{2}}(\sqrt{\tan q}-\frac{1}{\sqrt{\tan q}})\right)+\frac{1}{\sqrt{2}}\tanh {}^{-1}\left(\frac{1}{\sqrt{2}}(\sqrt{\tan q}+\frac{1}{\sqrt{\tan q}})\right)+C$$$$=\frac{1}{\sqrt{2}}{\tan }^{-1}(\frac{1}{\sqrt{2}}(\sqrt{\tan (x+\frac{3\pi }{4})}-\frac{1}{\sqrt{\tan (x+\frac{3\pi }{4})}})$$$$+\frac{1}{\sqrt{2}}\tanh {}^{-1}(\frac{1}{\sqrt{2}}(\sqrt{\tan (x+\frac{3\pi }{4})}+\frac{1}{\sqrt{\tan (x+\frac{3\pi }{4})}})+C$$

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