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Question Number 211796 by Spillover last updated on 21/Sep/24

Answered by Ghisom last updated on 21/Sep/24

$$\int \sqrt{\frac{\cos (x-a)}{\sin (x+a)}}dx=$$$$=\int \sqrt{\frac{\cos a\cos x+\sin a\sin x}{\sin a\cos x+\cos a\sin x}}dx=$$$$[t=\tan x]$$$$=\int \sqrt{\frac{t\sin a+\cos a}{t\cos a+\sin a}}\times \frac{dt}{t^2+1}=$$$$[u=\sqrt{\frac{t\sin a+\cos a}{t\cos a+\sin a}}]$$$$=2\int \frac{(1-{2\cos }^{2}a)u^2}{u^4-4u^2\cos a\sin a+1}du=$$$$=I+J$$$$I=-\frac{\sqrt{2}(\cos a-\sin a)}{2}\int \frac{u}{u^2-\sqrt{2}(\cos a+\sin a)u+1}du$$$$J=\frac{\sqrt{2}(\cos a-\sin a)}{2}\int \frac{u}{u^2+\sqrt{2}(\cos a+\sin a)u+1}du$$$$\mathrm{let}a=b-\frac{\pi }{4}$$$$I=-\cos b\int \frac{u}{u^2-2u\sin b+1}du$$$$J=\cos b\int \frac{u}{u^2+2u\sin b+1}du$$$$I=-\sin b\arctan \frac{u-\sin b}{\cos b}-\frac{\cos b}{2}\ln (u^2-2u\sin b+1)$$$$J=-\sin b\arctan \frac{u+\sin b}{\cos b}+\frac{\cos b}{2}\ln (u^2+2u\sin b+1)$$$$I+J=$$$$=\sin b\arctan \frac{2u\cos b}{u^2-1}+\frac{\cos b}{2}\ln \frac{u^2+2u\sin b+1}{u^2-2u\sin b+1}$$$$\mathrm{now}\mathrm{insert}b=a+\frac{\pi }{4}\mathrm{and}u=\sqrt{\frac{\cos (x-a)}{\sin (x+a)}}$$

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