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If int(k)/(sin(x-a)sin(x-b))dx=log[(sin(...

If `int(k)/(sin(x-a)sin(x-b))dx=log[(sin(x-a))/(sin(x-b))]+c,` then `k=`

A

`sin(a-b)`

B

`cos(a-b)`

C

`sin(a+b)`

D

`cos(a-b)`

Text Solution

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The correct Answer is:
A
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Knowledge Check

  • If int(k)/(sin(x-a)cos(x-b))dx=log[(sin(x-a))/(cos(x-b))]+c , then k=

    A
    `sin(a-b)`
    B
    `cos(a-b)`
    C
    `tan(a-b)`
    D
    `sec(a-b)`
  • int(dx)/(sin(x+a)sin(x+b))=

    A
    `(1)/(cos(b-a))log|sin(x+a)/sin(x+b)|+c`
    B
    `(1)/(cos(b-a))log|sin(x+b)/sin(x+a)|+c`
    C
    `(1)/(sin(b-a))log|sin(x+b)/sin(x+a)|+c`
    D
    `(1)/(sin(b-a))log|sin(x+a)/sin(x+b)|+c`
  • int(dx)/(cos(x-a)sin(x-b))=

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