Question

How do you evaluate the integral `int sinx/(root5(cosx))dx` from 0 to `pi`?

Answer 1

`~~ 2.261 - 0.735i`

Explanation:

Let `I=int_0^pi sinx/(root(5)cosx)dx.`

We take, `cosx=t`, so that, `-sinxdx=dt`, or, `sinxdx=-dt.`

Also, `x=0 rArr t=cos0=1, x=pi rArr t=-1.`

`:. I =-int_1^-1 1/(root(5)t)dt=int_-1^1t^(-1/5)dt`,

`=[t^(-1/5+1)/(-1/5+1)]_-1^1`,

`=5/4[t^(4/5)]_-1^1`

`=5/4[cancel(1^(4/5))^(1) - (-1)^(4/5)]`,

`= 5/4(1 - ((-1)^(1/2))^(8/5))`

`= 5/4 - 5/4i^(8/5)`

`~~ color(blue)(2.261 - 0.735i)`

Answer 2

`0`

Explanation:

The integral

`int_(x=0)^(x=pi) sin x/(cosx)^(1/5) dx = int_(y = -pi/2)^(y = pi/2) sin(y+pi/2)/(cos(y+pi/2))^(1/5) dy = -int_(y = -pi/2)^(y = pi/2) (cos(y)/sin(y)^(1/5)) dy = 0` because

`cos(y)/sin(y)^(1/5)` is an odd function.

Note: We consider `root(5)sin y equiv "sign"(y)root(5)abs sin y` for `-pi/2 le y le pi/2`