Let $$f$$ be a differentiable function in $$\left(0, \frac{\pi}{2}\right)$$. If $$\int_{\cos x}^{1} t^2 f(t) dt = \sin^3 x + \cos x$$, then $$\frac{1}{\sqrt{3}} f'\left(\frac{1}{\sqrt{3}}\right)$$ is equal to

Given

$$\int_{\cos x}^{1} t^2f(t)\,dt=\sin^3x+\cos x.$$

Differentiating both sides with respect to

$$x$$

using Leibniz rule,

$$\sin x\cdot (\cos^2x)\,f(\cos x)=3\sin^2x\cos x-\sin x.$$

Hence

$$\cos^2x\,f(\cos x)=3\sin x\cos x-1.$$

Let

$$t=\cos x.$$

Since

$$\sin x=\sqrt{1-t^2},$$

we get

$$t^2f(t)=3t\sqrt{1-t^2}-1.$$

Therefore,

$$f(t)=\frac{3t\sqrt{1-t^2}-1}{t^2}.$$

Differentiating,

$$f'(t)=\frac{\left(3t\sqrt{1-t^2}-1\right)'t^2-2t\left(3t\sqrt{1-t^2}-1\right)}{t^4}.$$

Now

$$\left(3t\sqrt{1-t^2}-1\right)'=\frac{3(1-2t^2)}{\sqrt{1-t^2}}.$$

At

$$t=\frac1{\sqrt3},$$

we have

$$\sqrt{1-t^2}=\sqrt{\frac23}=\frac{\sqrt2}{\sqrt3}.$$

Thus

$$3t\sqrt{1-t^2}-1=\sqrt2-1,$$

and

$$\left(3t\sqrt{1-t^2}-1\right)'=-\frac{\sqrt3}{\sqrt2}.$$

Substituting in the derivative formula,

$$f'\left(\frac1{\sqrt3}\right) =\frac{-\frac{\sqrt3}{\sqrt2}\cdot\frac13-\frac{2}{\sqrt3}(\sqrt2-1)}{\left(\frac13\right)^2}.$$

Simplifying,

$$f'\left(\frac1{\sqrt3}\right) =-\frac{3\sqrt3}{\sqrt2}-6\sqrt3(\sqrt2-1).$$

Hence

$$\frac1{\sqrt3}f'\left(\frac1{\sqrt3}\right) =-\frac3{\sqrt2}-6(\sqrt2-1).$$

Therefore,

$$\frac1{\sqrt3}f'\left(\frac1{\sqrt3}\right)=6-\frac{15}{\sqrt2}=6-\frac{15\sqrt2}{2}.$$