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प्रश्न
\[\int\frac{\cos x}{\sqrt{\sin^2 x - 2 \sin x - 3}} dx\]
योग
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उत्तर
` ∫ { cos x dx}/{\sqrt{sin^2 x - 2 sin x - 3}}`
` text{ let } \sin x = t`
` ⇒ cos x dx = dt `
`Now,∫ { cos x dx}/{\sqrt{sin^2 x - 2 sin x - 3}}`
\[ = \int\frac{dt}{\sqrt{t^2 - 2t - 3}}\]
\[ = \int\frac{dt}{\sqrt{t^2 - 2t + 1 - 1 - 3}}\]
\[ = \int\frac{dt}{\sqrt{\left( t - 1 \right)^2 - 2^2}}\]
\[ = \text{ log }\left| t - 1 + \sqrt{\left( t - 1 \right)^2 - 2^2} \right| + C\]
\[ = \text{ log }\left| t - 1 + \sqrt{t^2 - 2t - 3} \right| + C\]
\[ = \text{ log } \left| \sin x - 1 + \sqrt{\sin^2 x - 2 \sin x - 3} \right| + C\]
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