Find `int(dx)/(sinx cos^(3)x)`.

Explanation

[Here, the power of `sinx` is -1 and that of `cosx` is -3.
Since the sum of powers of `sinx` and `cosx` is -4, which is even and negative, put `z=tanx.`]
Let `z=tanx.` Then `dz=sec^(2)x dx.` Now,
`I=int(dx)/(sinx cos^(3)x)`
`=int(sec^(4)xdx)/(tan x)`
`=int((1+tan^(2)x)sec^(2)x dx)/(tanx)`
Let ` tanx=z." Then, " sec^(2)x dx=dz.` Therefore,
`I=int(1+z^(2))/(z)dz=int((1)/(z)+z)dz`
`= log|z| +(z^(2))/(2)+c`
`=log|tanx|+(tan^(2)x)/(2)+c`