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∫ Cos X √ Sin 2 X − 2 Sin X − 3 D X - Mathematics

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प्रश्न

\[\int\frac{\cos x}{\sqrt{\sin^2 x - 2 \sin x - 3}} dx\]
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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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पाठ 19: Indefinite Integrals - Exercise 19.18 [पृष्ठ ९९]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.18 | Q 15 | पृष्ठ ९९

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