Evaluate: `int_0^(100pi)sqrt((1-cos2x))dxdot`

Answer

`200sqrt(2)`

Explanation

We have `f(x)=sqrt((1-cos2x))=sqrt(2sin^(2)x)=sqrt(2)|sinx|`
Now, `f(x+pi)=sqrt(2)|sin(x+pi)|=sqrt(2)|sinx|=f(x)`
i.e. `f(x)` is periodic function with period `pi`.
`int_(0)^(100pi) sqrt((1-cos2x))dx=int_(0)^(100pi) sqrt(2)|sinx|dx`
`=100 sqrt(2)int_(0)^(pi)|sinx|dx`
`100 sqrt(2)int_(0)^(pi) sin x dx`
`=100 sqrt(2) [-cos x]_(0)^(pi) =200sqrt(2)`