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प्रश्न

\[\int\frac{\sin\sqrt{x}}{\sqrt{x}} dx\]
योग
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उत्तर

\[\int\frac{\sin \sqrt{x}}{\sqrt{x}}dx\]
\[\text{Let} \sqrt{x} = t\]
\[ \Rightarrow \frac{1}{2\sqrt{x}} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{dx}{\sqrt{x}} = 2dt\]
\[Now, \int\frac{\sin \sqrt{x}}{\sqrt{x}}dx\]
\[ = 2\int\text{sin t dt}\]
\[ = 2 \left[ - \cos t \right] + C\]
\[ = - 2 \cos \sqrt{x} + C\]

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अध्याय 19: Indefinite Integrals - Exercise 19.09 [पृष्ठ ५९]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.09 | Q 47 | पृष्ठ ५९

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