If `int tan(x-alpha)tan(x+alpha)*tan2xdx=plog|sec2x|+qlog|sec(x+alpha)|+rlog|sec(x-alpha)|+c` then p+q+r= . . ..

Options

A. `(-3)/(2)`
B. `(-5)/(2)`
C. `(5)/(2)`
D. `(3)/(2)`

Answer

A

Explanation

We have,
`int tan (x - alpha) tan (x + alpha). Tan 2x dx`
= plog `|sec 2x| + q log| sec (x + a)| + r log | sec (x - a)| + C`
Here, `tan 2x = tan ((x - alpha) + (x + alpha))`
`= tan (x - alpha) + tan (x + alpha)`
`1- tan (x - alpha) tan (x + alpha)`
`implies tan 2x - tan 2x than (X - alpha) tan (x + alpha)`
`= tan (x - alpha) + tan (x + alpha)`
`implies tan(x - alpha) tan (x + alpha) tan 2x`
`implies 2x - tan (x - alpha) - tan (x + alpha)`
`:. int tan (x - alpha) tan (x + alpha) - tan (x + alpha) dx`
`= (log|sec 2x|)/(2) - log| sec (x - alpha)|-log|sec(x + alpha) + c`
`(1)/(2) log |sec 2x| + (-) log| sec (x + alpha)| + (-1) log |sec (x - alpha + C`
`:. P = (1)/(2), g = -1` and `r = - 1`
`:. P + q + r = (1)/(2) + (-1) + (-1) =(1)/(2) - 2 = (-3)/(2)`