find the integral of y=x^2 +sin^2(x) with respect to x between the limits 0 and pi

intergal₀^pi(x^2 + sin^2(x))dx = integral₀^pi(x^2 + 1/2 - cos(2x)/2)dx since sin^2(x) + cos^2(x) = 1cos^2(x) - sin^2(x) = cos(2x)2sin^2(x) = 1 - cos(2x)thereforeintegral₀^pi(x^2 +sin^2(x))dx = [(x^3)/3 + x/2 - sin(2x)/4]₀^pi= ((pi^3)/3 + pi/2 - 0/4) - (0)=(pi^3)/3 + pi/2