The value of `intxsinxsec^(3)xdx` is

Options

A. `(1)/(2)[sec^(2)x-tanx]+c`
B. `(1)/(2)[xsec^(2)x-tanx]+c`
C. `(1)/(2)[xsec^(2)x+tanx]+c`
D. `(1)/(2)[sec^(2)x+tanx]+c`

Answer

B

Explanation

`int x sin x sec^(2) x dx = int x sin x(1)/(cos^(3))dx`
`= int x tan x* sec^(2)x dx`
Put `tan x=t rArr sec^(2) x dx=dt`
and `x= tan^(-1)t`
Then, it reduces to
`int tan^(-1) t* t dt =(t^(2))/(2) tan^(-1) t- int (t^(2))/(2(1+t^(2))dt`
`= (x tan^(2) x)/(2)-(1)/(2)t+(1)/(2) tan^(-1)t+c`
`=(x (sec^(2)x-1))/(2)-(1)/(2) tan x+(1)/(2)x+c`
`=(1)/(2)[ xsec^(2)x- tan x] +c`