Question

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

Answer 1

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

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