The integral of the function `tan^(4)x` is

Options

A. `(tan^(3)x)/(3)-tanx-x+C`
B. `(tan^(3)x)/(3)+tanx-x+C`
C. `(tan^(3)x)/(3)-tanx+x+C`
D. `(tan^(3)x)/(3)+tanx+x+C`

Answer

C

Explanation

Let `l=int tan^(4)x dx = int (tan^(2)x)^(2)dx`
`rArr" "l=int(tan^(2)x)(tan^(2)x)dx`
`=int(sec^(2)x-1)(tan^(2)x)dx`
`=intsec^(2)x tan^(2)x dx- int tan^(2) xdx`
`=int sec^(2) x tan^(2) x dx - int [sec^(2)x-1]dx`
`=int sec^(2)x tan^(2)xdx - [int sec^(2)x dx-int 1dx]`
Now, let `l_(1)=int sec^(2)x tan^(2)x dx and l_(2)=int sec^(2) x dx - int 1 dx`
Then, `" "l=l_(1)-l_(2)" ...(i)"`
Put`" "tan x= t rArr sec^(2)x=(dt)/(dx)`
`rArr" "dx=(dt)/(sec^(2)x)" "therefore" "l_(1)=int sec^(2)x t^(2).(dt)/(sec^(2)x)`
`rArr" "l_(1)=int t^(2)dt=(t^(3))/(3)+C_(1)=(Tan^(3)x)/(3)+C_(1)`
`" "l_(2)=int sec^(2)x dx-int 1dx=tanx-x+C_(2)`
`therefore" Putting the values of "l_(1) and l_(2) " in Eq. (i), we get "`
`l=(tan^(3)x)/(3)+C_(1)-(tanx-x+C_(2))`
`rArr" "l=(tan^(3)x)/(3)-tanx+x+C" "[because C_(1)-C_(2)=C]`