Integration

We want to evaluate the integral:

The function is the greatest integer not exceeding , also known as the floor function. We need to determine the function over different intervals of the integration bounds:

• For , .
• For , .
• For , .

We can therefore split the integral into three parts:

Let's calculate each integral:

1. 
   Calculating:

   Evaluate from to :

3. 
   Calculating:

   Evaluate from to :

Add the results of the three integrals:

Converting to a common denominator:

Thus, the integral evaluates to:

To evaluate the integral , we can first simplify the expression inside the integral.

Recall the trigonometric identity:

Substitute this identity into the integral:

Notice that introduces absolute value because the sine function is negative in certain intervals. Since the integral range is from to , we split this range into intervals where is positive and negative:

• to :
• to :
• to :
• … continue this pattern …
• Making this split, recognize that from to consists of 25 full periods of , each with the integral extending over one full positive and one full negative half-cycle of sine.

Calculate , which is the integral over one period:

Calculate each part separately:

Thus, . Multiply this over 25 periods:

The original integral becomes:

The evaluated result is .