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 `int tan(x-alpha) tan (x+alpha).tan2xdx`
`=[log|sec2x|+qlog|sec(x+a)|+rlog|sec(x-a)|+C`
Here, `tan2x=tan((x-alpha)+(x+alpha))`
`=tan(x-alpha)+tan(x+alpha)`
`1-tan(x-alpha)tan(x+alpha)`
`implies tan2x-tan2xtan(x-alpha)tan(x+alpha)`
`=tan(x-alpha)+tan(x+alpha)`
`impliestan(x-alpha)tan(x+alpha)tan2x`
`impliestan2x-tan(x-alpha)-tan(x+alpha)`
`therefore int tan(x-alpha)tan(x+alpha)tan(2x)dx`
`=int(tan2x-tan(x-alpha)-tan(x+alpha)dx`
`=(log|sec2x|)/(2)-log|sec(x-alpha)|-log|sec(x+alpha)+C`
`=(1)/(2)log|sec2x|+(-1)log|sec(x+alpha)|+(-1)log|sec(x-alpha)+C`
`thereforep=(1)/(2),g=-1 and r=-1`
`thereforep+q+r=(1)/(2)+(-1)+(-1)=(1)/(2)-2=(-3)/(2)`