Question

If n is a positive integer, then `int_0^(2pi) (sin^(2n) x)/(sin^(2n) x + cos^(2n) x)  dx` is equal to

विकल्प

• `pi/2`

• `pi`

• 4pi`

• `2/pi`

Answer 1

`pi`

Explanation:

I = `int_0^(2pi) (sin^(2n) x)/(sin^(2n) x + cos^(2n) x) dx`  .....(i)

= `int_0^(2pi) (sin^(2n) (2pi -x))/(sin^(2n) (2pi - x) + cos^(2n) (2pi - x)) dx`

= `int_0^(2pi) (sin^(2n) x)/(sin^(2n) x + cos^(2n) x) dx`

= `4int_0^(pi/2) (sin^(2n) x)/(sin^(2n) x + cos^(2n) x) dx`

= `4int_0^(pi/2) (sin^(2n) (pi/2 - x))/(sin^(2n) (pi/x - x) + cos^(2n) (pi/2 - x)) dx`

I = `4int_0^(pi/2) (cos^(2n) x)/(cos^(2n) x + sin^(2n) x) dx`

Adding equations (i) and (ii), we get

`2I = 4int_0^(pi/2) 1. dx = 4[x]_0^(pi/2)`

`2I = 4 . pi/2` ⇒ `I = pi`