`int(1)/(cos(x+a).sin(x+b))dx` का मान ज्ञात कीजिए ।

Options

A. `(1)/(cos(a+b))[log|sin(x+b)|+log|sec(x+a)|]+c`
B. `(1)/(cos(a-b))[log|sin(x+b)|+log|sec(x+a)|]+c`
C. `(1)/(cos(b-a))[log|sin(x+b)|+log|sec(x+a)|]+c`
D. `(1)/(cos(a-b))[log|sin(x-b)|+log|sec(x+a)|]+c`

Answer

B

Explanation

`int(1)/(cos(x+a).sin(x+b))dx`
`" "=(1)/(cos(a-b))int(cos(a-b))/(cos(x+a).sin(x+b))dx`
`" "=(1)/(cos(a-b))int(cos{(x+a)-(x+b)})/(cos(x+a).sin(x+b))dx`
`" "=(1)/(cos(a-b))int(+sin(x+a)sin(x+b))/(cos(x+a)sin(x+b))dx`
`" "=(1)/(cos(a-b))int{(cos(x+a)cos(x+b))/(cos(x+a)sin(x+b))+(sin(x+a)sin(x+b))/(cos(x+a)sin(x+b))}dx`
`" "=(1)/(cos(a-b))int{cot(x+b)+tan(x+a)}dx`
`" "=(1)/(cos(a-b))[log|sin(x+b)|+log|sec(x+a)|]+c`