I would like to know if this is right. Thanks in advance for your time.
Let $f(x)$ be a function such that $\int_{1}^{x^2} (\cos(\pi t)f(t)dt=\sqrt{x^2+x-2}$. Calculate $f(4)$.
$$\int_{1}^{x^2} (\cos(\pi t)f(t)dt=\sqrt{x^2+x-2}$$ $$\frac{d}{dx}\int_{1}^{x^2} (\cos(\pi t)f(t)dt=\frac{d}{dx}\sqrt{x^2+x-2}$$ $$ \cos(\pi x^2)f(x^2)\cdot \frac{d}{dx}(x^2) =\frac{1}{2\sqrt{x^2+x-2}}\cdot \frac{d}{dx}(x^2+x-2)$$ $$ \cos(\pi x^2)f(x^2)\cdot 2x =\frac{1}{2\sqrt{x^2+x-2}}\cdot (2x+1)$$ $$ 2x\cos(\pi x^2)f(x^2)=\frac{2x+1}{2\sqrt{x^2+x-2}}$$ $$\frac{1}{2x\cos(\pi x^2)}\cdot 2x\cos(\pi x^2)f(x^2)=\frac{2x+1}{2\sqrt{x^2+x-2}}\cdot \frac{1}{2x\cos(\pi x^2)}$$ $$f(x^2)=\frac{2x+1}{4x\cos(\pi x^2)\sqrt{x^2+x-2}}$$ $$f(2^2)=\frac{2(2)+1}{4(2)\cos(\pi\cdot 2^2)\sqrt{2^2+2-2}}$$ $$f(4)=\frac{4+1}{8\cos(\pi\cdot 4)\sqrt{4+2-2}}$$ $$f(4)=\frac{5}{8\cos(4\pi)\sqrt{4}}$$ $$f(4)=\frac{5}{8\cdot 1 \cdot 2}$$ $$f(4)=\frac{5}{16}$$