Question

Evaluate: `int(sec^(3)x-sec x)/(cosec x-cosec^(3)x)dx​

Answer 1

First take $\displaystyle\sf {\sec x}$ and $\displaystyle\sf {\csc x}$ common from numerator and denominator respectively. Then we apply,

• $\displaystyle\sf {\dfrac{\sec x}{\csc x}=\tan x}$
• $\displaystyle\sf {\sec^2x-\tan^2x=1}$
• $\displaystyle\sf {\csc^2x-\cot^2x=1}$

Now the integrand becomes,

$\displaystyle\sf {\longrightarrow\tan x\cdot\dfrac{\tan^2x}{\cot^2x}=\tan^5x.}$

This integrand is written in terms of $\displaystyle\sf {\sec x}$ with its derivative $\displaystyle\sf {\sec x\tan x}$ in the next step. Thus,

$\displaystyle\sf{\longrightarrow \tan^5x=(\tan^2x)^2\cdot\tan x}$

$\displaystyle\sf{\longrightarrow \tan^5x=\dfrac{(\sec^2x-1)^2\sec x\tan x}{\sec x}}$

Now put,

$\displaystyle\sf{\longrightarrow u=\sec x}$

$\displaystyle\sf{\longrightarrow du=\sec x\tan x}$

Now we are integrating $\displaystyle\sf {\dfrac{(u^2-1)^2}{u}}$ wrt $\displaystyle\sf {u.}$

Expanding numerator and simplifying, we get $\displaystyle\sf {u^3-2u+\dfrac{1}{u}}$ and on integrating each term, followed by undoing the substitution $\displaystyle\sf {u=\sec x,}$ the final answer becomes,

$\displaystyle\sf {\longrightarrow\underline {\underline {I=-\left[\dfrac {1}{4}\sec^4x-\sec^2x-\log|\cos x|\right]+C}}}$