The area enclosed between the curve `y=sin^(2)x and y=cos^(2)x` in the interval `0le x le pi` is _____ sq. units.

Options

A. 2 sq unit
B. `(1)/(2)` sq unit
C. 1 sq unit
D. None of these

Answer

C

Explanation

`y=sin^(2)xand y=cso^(2)x`
Solving `sin^(2)x=cos^(2)x`
`therefore" "x=pi//4,3pi//4`
Graph of functions is as shown in the following figure.

From the figure, the required area
`=int_(pi//4)^(3pi//4)(sin^(2)x-cos^(2)x)dx`
`=-int_(pi//4)^(3pi//4)cos2xdx=1`