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Evaluate: ∫ Sec 2 √ X √ X D X - Mathematics

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प्रश्न

Evaluate:\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} \text{ dx }\]

 

योग
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उत्तर

\[\text{ Let I }= \int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx\]
\[\text{ Let }\sqrt{x} = t\]
\[ \Rightarrow \frac{dx}{2\sqrt{x}} = dt\]
\[ \Rightarrow \frac{dx}{\sqrt{x}} = 2\text{  dt}\]
\[\text{ Putting}\ \sqrt{x} = t  \text{ and} \frac{dx}{\sqrt{x}} = \text{ 2 dt }\]
\[ \therefore I = 2\int \sec^2 + dt\]
\[ = 2 \tan t + C\]
\[ = 2 \tan \left( \sqrt{x} \right) + C \left( \because t = \sqrt{x} \right)\]

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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 40 | पृष्ठ १९८

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