The value of `int_(pi//4)^(pi//2)e^(x)(logsinx+cotx)dx` is

Options

A. `e^(pi//4)log2`
B. `-e^(pi//4)log2`
C. `(1)/(2)e^(pi//4)log2`
D. `-(1)/(2)e^(pi//4)log2`

Answer

C

Explanation

Let `I = int _(pi//4)^(pi/2)e^(x) ( log sin x+ cot x)dx`
`rArr I= int_(pi//4)^(pi//2)e^(x) log sin x dx +int_(pi//4)^(pi//2) e^(x) cot x dx`
`int_(pi//4)^(pi//2) e^(x) log sin x dx +[ e^(x) log sin x]_(pi//4)^(pi//2)`
`int_(pi//4)^(pi//2) e^(x) log sin x dx`
` =e^(pi//2) log "sin" (pi)/(2) - e^(pi//4) log "sin" (pi)/(4)`
`=- e^(pi//4) log ((1)/(sqrt(2)))`
`=(1)/(2)e^(pi//4) log 2`