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प्रश्न
\[\int\frac{\sin^3 x - \cos^3 x}{\sin^2 x \cos^2 x} dx\]
योग
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उत्तर
\[\int\left( \frac{\sin^3 x - \cos^3 x}{\sin^2 x \cdot \cos^2 x} \right)dx\]
\[ = \int\frac{\sin^3 x}{\sin^2 x \cdot \cos^2 x}dx - \int\frac{\cos^3 x}{\sin^2 x \cdot \cos^2 x}dx\]
\[ = \int\frac{\sin x}{\cos^2 x}dx - \int\frac{\cos x}{\sin^2 x}dx\]
\[ = \int\frac{\sin x}{\cos x} \times \frac{1}{\cos x}dx - \int\frac{\cos x}{\sin x} \times \frac{1}{\sin x}dx\]
`=∫ sec x tan x dx - ∫ "cosec" x cot x dx`
\[ = \sec x - \left( - \text{cosec x} \right) + C\]
\[ = \sec x + \text{cosec x }+ C\]
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