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प्रश्न
\[\int\frac{\cos x}{\sqrt{4 - \sin^2 x}} dx\]
बेरीज
उत्तर
` ∫ cos x / \sqrt{4-sin^2 x}`
\[\text{ let }\sin x = t\]
\[ \Rightarrow \text{ cos x dx }= dt\]
Now, ` ∫ cos x / \sqrt{4-sin^2 x}`
\[ = \int\frac{dt}{\sqrt{4 - t^2}}\]
\[ = \int\frac{dt}{\sqrt{2^2 - t^2}}\]
\[ = \sin^{- 1} \left( \frac{t}{2} \right) + C\]
\[ = \sin^{- 1} \left( \frac{\sin x}{2} \right) + C\]
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