If `int(1)/((sinx+4)(sinx-1))dx`
`=A(1)/("tan"(x)/(2)-1)+B"tan"^(-1){f(x)}+C`. Then,

Options

A. `A=(1)/(5),B=(-2)/(5sqrt(15)),f(x)=(4tanx+3)/(sqrt(15))`
B. `A=-(1)/(5),B=(1)/(sqrt(15)),f(x)=("4tan"((x)/(2))+1)/(sqrt(15))`
C. `A=(2)/(5),B=(-2)/(5),f(x)=(4tanx+1)/(5)`
D. `A=(2)/(5),B=(-2)/(5sqrt(15)),f(x)=("4 tan"(x)/(2)+1)/(sqrt(15))`

Answer

D

Explanation

let `l=int(1)/((six+4)(six-1))dx`
`=(1)/(5)int((sinx+4)-(sinx-1))/(((sinx+4)(sinx-1))dx`
`=(1)/(5)int(1)/(sinx-1)dx-(1)/(5)int(1)/(sinx+4)dx`
`=(1)/(5)int(secx^(2)x//2)/(2tanx//2-1-tan^(2)x//2)dx`
`-(1)/(5)int(sec^(2)x//2dx)/(2tanx//2+4+4tan^(2)x//2)`
Put `"tan"(x)/(2)=t`
`implies sec^(2)(x)/(2)dx=2t dt`
`int(2dt)/([2t+4(1+t^(2))])`
`thereforel=-(2)/(5)int(dt)/(2t-1-t^(2))-(1)/(10)`
`int(dt)/(t^(2)+(1)/(2)t+1)`
`=(-2)/(5)int(1)/((t-1)^(2))dt-(1)/(10)`
`int(dt)/((t+(1)/(4))^(2)+(sqrt(15))/(4))^(2))`
`=(2)/(5)(t)/((t-1))-(2)/(5sqrt(15))"tan"^(-1)((4t+1)/(sqrt(15)))+c`
`=(2)/(5)(1)/("tan"(x)/(2)-1)-(2)/(5sqrt(15))`
`tan^(-1)(4"tan"(x)/(2)+1)/(sqrt(15)))+c`