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प्रश्न
\[\int\frac{5 \cos^3 x + 6 \sin^3 x}{2 \sin^2 x \cos^2 x} dx\]
बेरीज
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उत्तर
\[\int\left( \frac{5 \cos^3 x + 6 \sin^3 x}{2 \sin^2 x \cos^2 x} \right)dx\]
\[ = \int\left( \frac{5 \cos^3 x}{2 \sin^2 x \cos^2 x} + \frac{6 \sin^3 x}{2 \sin^2 x \cos^2 x} \right)dx\]
\[ = \int\left( \frac{5}{2} \frac{\cos x}{\sin^2 x} + 3\frac{\sin x}{\cos^2 x} \right)dx\]
\[ = \frac{5}{2}\int\left( \frac{\cos x}{\sin x} \times \frac{1}{\sin x} \right)dx + 3\int\frac{\sin x}{\cos x} \times \frac{1}{\cos x}dx\]
`= {5}/{2}∫("cosec "x cot x ) dx + 3 ∫ sec x tan x dx`
\[ = \frac{5}{2}\left( - \text{cosec x} \right) + 3 \sec x + C\]
\[ = - \frac{5}{2}\text{cosec x} + 3 \sec x + C\]
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