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Question Number 85667 by john santu last updated on 23/Mar/20

$$\int \frac{dx}{\sqrt{1-\sin 2x}}$$

Answered by som(math1967) last updated on 24/Mar/20

$$\int \frac{dx}{(cosx-sinx)}$$$$\int \frac{dx}{\sqrt{2}cos(x+\frac{\pi }{4})}$$$$\frac{1}{\sqrt{2}}\int sec(x+\frac{\pi }{4})dx$$$$\frac{1}{\sqrt{2}}ln\mid tan(\frac{\pi }{4}+\frac{x}{2}+\frac{\pi }{8})\mid +C$$$$\frac{1}{\sqrt{2}}ln\mid tan(\frac{3\pi }{8}+\frac{x}{2})\mid +C$$

Commented by jagoll last updated on 24/Mar/20

$$\mathrm{sir}\sqrt{1-\sin 2\mathrm{x}}=\mid \cos \mathrm{x}-\sin \mathrm{x}\mid$$$$\mathrm{why}\sqrt{1-\sin 2\mathrm{x}}=\cos \mathrm{x}-\sin \mathrm{x}?$$

Commented by jagoll last updated on 24/Mar/20

$$\mathrm{then}\int \sec \mathrm{x}\mathrm{dx}=\ln \mid \sec \mathrm{x}+\tan \mathrm{x}\mid \mathrm{sir}$$

Commented by som(math1967) last updated on 24/Mar/20

$$secx+tanx=\frac{1+sinx}{cosx}=\frac{sin\frac{x}{2}+cos\frac{x}{2}}{cos\frac{x}{2}-sin\frac{x}{2}}$$$$=\frac{tan\frac{x}{2}+1}{1-tan\frac{x}{2}}=tan(\frac{x}{2}+\frac{\pi }{4})$$

Commented by jagoll last updated on 24/Mar/20

$$\mathrm{oo}\mathrm{same}\mathrm{sir}.\mathrm{but}\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{agree}$$$$\mathrm{with}\sqrt{1-\sin 2\mathrm{x}}=\cos \mathrm{x}-\sin \mathrm{x}\mathrm{sir}$$

Commented by jagoll last updated on 24/Mar/20

$$\mathrm{i}\mathrm{got}\mathrm{two}\mathrm{solution}$$

Commented by som(math1967) last updated on 24/Mar/20

$$yestwosolutionpossible$$$$Iuseonly(cosx-sinx)$$

Commented by jagoll last updated on 24/Mar/20

$$\mathrm{yes}\mathrm{sir}.\mathrm{deal}$$

Commented by john santu last updated on 24/Mar/20

$$yeshave2solution$$

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