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int(cos 4x-1)/(cot x-tanx)dx is equal to...

`int(cos 4x-1)/(cot x-tanx)dx` is equal to

A

`-1/2 "cos" 4x + c`

B

`-1/4 "cos" 4x + c`

C

`-1/2 sin 2x + c`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
D
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Knowledge Check

  • int(1+cos 4x)/(cotx-tanx)dx=

    A
    `-(1)/(8)cos 4x`
    B
    `(1)/(8)cos 4x`
    C
    `(1)/(8)sin 4x`
    D
    `-(1)/(8)sin 4x`

  • int (cos 4x + 1)/(cot x - tan x) dx is given by

    A
    `-1/2 "cos" 4x + c`
    B
    `-1/4 "cos" 4x + c`
    C
    `-1/2 sin 2x + c`
    D
    `-1/8 "cos 4x + c`

  • If int(cos 4x+1)/(cotx-tanx)dx=a cos 4x+c, then a=

    A
    `-(1)/(2)`
    B
    `-(1)/(8)`
    C
    `-(1)/(4)`
    D
    `(1)/(2)`

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