Question

\[\int\frac{5 \cos^3 x + 6 \sin^3 x}{2 \sin^2 x \cos^2 x} dx\]

Answer 1

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