Question

`int "cosec"^4x  dx` = ______.

Options

• `cotx + 1/3 cot^3x + c`

• `cotx - 1/3 cot^3x + c`

• `-(cotx + 1/3 cot^3x) + c`

• `-(cotx - 1/3 cot^3x) + c`

Answer 1

`int "cosec"^4x  dx` = `underlinebb(-(cotx + 1/3 cot^3x) + c)`.

Explanation:

I = `int "cosec"^4x  dx`

= `int "cosec"^2x*"cosec"^2x  dx`

= `int (1 + cot^2x) "cosec"^2x  dx`

Put cot x = t

`\implies` cosec²x dx = –dt

∴ I = `-int (1 + t^2)dt`

= `-[t + t^3/3] + c`

= `-[cot x + 1/3 cot^3x] + c`