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int tan(x-alpha).tan(x+alpha).tan 2x dx ...

`int tan(x-alpha).tan(x+alpha).tan 2x dx` is equal to

A

`ln|(sqrt(sec 2x).sec (x-alpha))/(sec(x-alpha))|+C`

B

`ln|(sqrt(sec2x))/(sec(x-alpha).sec(x+alpha))|+C`

C

`ln|(sqrt(sec2x).sec (x-alpha))/(sec(x+alpha))|+C`

D

`ln|(sec 2x)/(sec(x-alpha).sec(x+alpha))|+C`

Text Solution

Verified by Experts

The correct Answer is:
B
`tan 2x=tan[(x=alpha)+(x+alpha)]`
`=(tan(x-alpha)+tan(x+alpha))/(1-tan (x-alpha).tan(x+alpha))`
`rArr" "tan(x-alpha).tan(x+alpha). tan 2x = tan 2x - tan (x-alpha)-tan(x+alpha)`
`rArr tan(x-alpha).tan(x-alpha)-tan(x+alpha)]dx`
`=(1)/(2)ln|sec 2x|-ln|sec(x-alpha)|-ln |sec(x+alpha)|+C`
`=ln|(sqrt(sec2x))/(sec(x-alpha).sec(x+alpha))|+C`
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