Giải thích các bước giải:

Ta có:

$f'(x)=\cos^2x$

$\to f(x)=\int\cos^2xdx$ 

$\to f(x)=\int\dfrac{2+\cos2x}2dx$ 

$\to f(x)=\int\dfrac{2+\cos2x}4d(2x)$ 

$\to f(x)=\dfrac12x+\dfrac14\sin2x+C$

Mà $f(0)=\dfrac12$

 $\to \dfrac12\cdot 0+\dfrac14\cdot\sin(2\cdot 0)+C=\dfrac12$

$\to C=\dfrac12$

$\to f(x)=\dfrac12x+\dfrac14\sin2x+\dfrac12$

$\to \int^{\pi}_0f(x)dx=\int^{\pi}_0\dfrac12x+\dfrac14\sin2x+\dfrac12dx$

$\to \int^{\pi}_0f(x)dx=\dfrac{x^2}4-\dfrac18\cos2x+\dfrac12x\bigg|^{\pi}_0$

$\to \int^{\pi}_0f(x)dx=\dfrac{2\pi+\pi^2}{4}$

$\to C$