Question

\[\int\frac{1}{\sin^4 x \cos^2 x} dx\]

Answer 1

\[\int\frac{dx}{\sin^4 x . \cos^2 x}\]
`  " Dividing   numerator  and denominator  " by   sin^2 x` 
\[ = \int\frac{\frac{1}{\sin^2 x}}{\sin^4 x . \cot^2 x}dx\]
\[ = \int\frac{{cosec}^6 x}{\cot^2}dx\]

 ` = ∫     { "cosec"^4  x   . "cosec"^2  x  dx}/cot^ 2 x `

` = ∫     { ("1  +cot"^2  x )^2  . "cosec"^2  x  dx}/cot^ 2 x `
\[Let \cot x = t\]
\[ \Rightarrow - {cosec}^2 x = \frac{dt}{dx}\]
\[ \Rightarrow - {cosec}^2 x \text{ dx } = dt\]
Now,` = ∫     { ("1  +cot"^2  x )^2  . "cosec"^2  x  dx}/cot^ 2 x `
\[ = \int \left( \frac{1 + t^2}{t} \right)^2 \left( - dt \right)\]
\[ = - \int\frac{\left( 1 + t^4 + 2 t^2 \right)}{t^2}dt\]
\[ = - \int\left( t^{- 2} + t^2 + 2 \right)dt\]
\[ = - \left[ \frac{t^{- 2 + 1}}{- 2 + 1} + \frac{t^3}{3} + 2t \right] + C\]
\[ = - \left[ - \frac{1}{t} + \frac{t^3}{3} + 2t \right] + C\]
\[ = - \frac{1}{3} t^3 - 2t + \frac{1}{t} + C\]
\[ = - \frac{1}{3} \cot^3 x - 2 \cot x + \frac{1}{\cot x} + C\]
\[ = - \frac{1}{3} \cot^3 x - 2 \cot x + \tan x + C\]