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I tried solving the integral this way $$\int\frac{1}{\sin{\frac{x}{\sqrt{2}}}}dx=\sqrt{2}\int\frac{1}{\sin{\frac{x}{\sqrt{2}}}}d{\frac{x}{\sqrt{2}}}=\sqrt{2}\int\frac{\sin{\frac{x}{\sqrt{2}}}}{\sin^2{\frac{x}{\sqrt{2}}}}d{\frac{x}{\sqrt{2}}}=-\sqrt{2}\int\frac{1}{1-\cos^2{\frac{x}{\sqrt{2}}}}d{\cos\frac{x}{\sqrt{2}}}=-\frac{\sqrt{2}}{2}\ln\left|\frac{\cos{\frac{{x}}{\sqrt{2}}}+1}{\cos{\frac{{x}}{\sqrt{2}}}-1}\right|+C$$ but the answer is different in my corsebook. Could anybody correct me if I am wrong. Thx. P.S. the answer in the coursebook is $$\sqrt{2}\ln{\left| \tan{\frac{x}{2\sqrt{2}}}\right|}+C$$

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To get the coursebook's answer, use the identities

$$2\sin^2 t = 1 - \cos 2t, \quad 2\cos^2 t = 1 + \cos 2t$$

to get

$$-\frac{\sqrt{2}}{2}\ln \left|\frac{\cos \frac{x}{\sqrt{2}} + 1}{\cos \frac{x}{\sqrt{2}} - 1}\right| = -\frac{\sqrt{2}}{2}\ln \left|\frac{2\cos^2 \frac{x}{2\sqrt{2}}}{2\sin^2 \frac{x}{2\sqrt{2}}}\right| = -\sqrt{2}\ln \left|\cot \frac{x}{2\sqrt{2}}\right| = \sqrt{2}\ln\left|\tan \frac{x}{\sqrt{2}}\right|.$$

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  • $\begingroup$ thx, opened my eyes ;P $\endgroup$ – dimaastronom Feb 24 '15 at 19:46

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