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प्रश्न

Evaluate:\[\int\frac{\cos \sqrt{x}}{\sqrt{x}} \text{ dx }\]

योग

उत्तर

\[\text{ Let I } = \int\frac{\cos\sqrt{x}}{\sqrt{x}} dx\]
\[\text{ Putting} \sqrt{x} = t\]
\[ \Rightarrow \frac{1}{2\sqrt{x}}dx = dt\]
\[ \Rightarrow \frac{dx}{\sqrt{x}} = 2 \text{ dt}\]
\[ \therefore I = 2\int\text{ cos  t  dt}\]
\[ = 2 \sin t + C ,\text{  where  t }= \sqrt{x}\]
\[ = 2 \sin \sqrt{x} + C\]

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अध्याय 19: Indefinite Integrals - Very Short Answers [पृष्ठ १९८]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Very Short Answers | Q 42 | पृष्ठ १९८

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