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प्रश्न
\[\int\cos\sqrt{x}\ dx\]
बेरीज
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उत्तर
\[\text{ Let I } = \int\cos \sqrt{x} dx\]
\[ = \int\frac{\sqrt{x} \cdot \cos \sqrt{x}}{\sqrt{x}}dx\]
\[\text{ Let }\sqrt{x} = t\]
\[ \Rightarrow \frac{1}{2\sqrt{x}}dx = dt\]
\[ \Rightarrow \frac{dx}{\sqrt{x}} = 2dt\]
\[ \therefore I = 2\int t_{} \cdot \cos \left( t \right)_{} \cdot dt\]
\[\text{Taking t as the first function and cos t as the second function} . \]
\[ = 2 \left[ t \cdot \sin t - \int1 \cdot \text{ sin t dt }\right]\]
\[ = 2 \left[ t \cdot \sin t + \cos t \right] + C . . . . (1) \]
\[\text{Substituting the value of t in eq} \text{ (1) }\]
\[ = 2 \left[ \sqrt{x} \cdot \sin \sqrt{x} + \cos \sqrt{x} \right] + C\]
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