Question

\[\int\frac{1}{\sin^3 x \cos^5 x} dx\]

Answer 1

\[\int\frac{dx}{\sin^3 x . \cos^5 x}dx\]
`  "Dividing  numerator  and denominator  by" cos^8 x  `
\[ = \int\frac{\frac{1}{\cos^8 x}dx}{\frac{\sin^3 x}{\cos^3 x}}\]
\[ = \int\frac{\sec^8 x}{\tan^3 x}dx\]
\[ = \int\frac{\sec^6 x . \sec^2 x dx}{\tan^3 x}\]
\[ = \int\frac{\left( 1 + \tan^2 x \right)^3 . \sec^2 x dx}{\tan^3 x}\]
\[Let \tan x = t\]
` ⇒  sec^2 x  dx  = dt `
\[Now, \int\frac{\left( 1 + \tan^2 x \right)^3 . \sec^2 x dx}{\tan^3 x}\]
\[ = \int\frac{\left( 1 + t^2 \right)^3}{t^3} . dt\]
\[ = \int\frac{1 + t^6 + 3 t^2 + 3 t^4}{t^3}dt\]
\[ = \int\left( \frac{1}{t^3} + t^3 + \frac{3}{t} + 3t \right)dt\]
\[ = \int t^{- 3} dt + \int t^3 dt + 3\int\frac{dt}{t} + 3\    ∫ t \text{ dt }\]
\[ = \left[ \frac{t^{- 3 + 1}}{- 3 + 1} \right] + \left[ \frac{t^{3 + 1}}{3 + 1} \right] + 3 \log \left| t \right| + \frac{3 t^2}{2} + C\]
\[ = - \frac{1}{2} \left( \tan x \right)^{- 2} + \frac{1}{4} \tan^4 x + 3 \log \left| \tan x \right| + \frac{3}{2} \tan^2 x + C\]