If $\phi(x)=\sin x\int_{^0}^{x}\cos t dt+2\int_{^0}^{x}t dt+\cos^2x-x^2$.If $x^2 – 2x=3\ge\phi(x) \forall x\in R$ , then prove that greatest area bounded by $x\phi(x)$ and ordinate $x = 0$ and $x = 5$ is $25$
I simplified $\phi(x)=\sin^2x+x^2+\cos^2x-x^2=1$ but i cannot understand the question further.