`IfI=int(dx)/(secx+cos e cx),t h e nIe q u a l s` `1/2(cosx+sinx-1/(sqrt(2))log(cos e cx-cosx))+C` `1/2(sinx-cosx-1/(sqrt(2))log|cos e cx+cotx|)+C` `1/(sqrt(2))(sinx+cosx+1/2"log"|cos e cx-cosx|)` `1/2[sinx-cosx]-1/(sqrt(2))"log"|cos e c(x+pi/4)`

Options

A. `(1)/(2)(cosx+sinx-(1)/(sqrt(2))log("cosec"x-cosx))+C`
B. `(1)/(2)(sinx-cosx-(1)/(sqrt(2))log|"cosec"x-cotx|)+C`
C. `(1)/(sqrt(2))(sinx+cosx+(1)/(2)log|"cosec"x-cosx|)+C`
D. `(1)/(2)[sinx-cosx]-(1)/(sqrt(2))log|"cosec"(x+pi//4)-cot(x+pi//4)|+C`

Answer

D

Explanation

`I=int(sinx cosx)/(sinx+cosx)dx`
`=(1)/(2)int((sinx+cosx)^(2)-1)/(sinx+cosx)dx`
`=(1)/(2)int[sinx+cosx-(1)/(sqrt(2)sin(x+pi//4))]dx`
`=(1)/(2)[sinx-cosx]-(1)/(2sqrt(2))log|"cosec"(x+pi//4)-cot(x+pi//4)|+C`