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