$ \int_{\sin x}^{1} t^{2} f(t) d t=1-\sin x \forall x \in(0, \pi / 2) $, then $ f\left(\frac{1}{\sqrt{3}}\right) $ is :

Question

$ \int_{\sin x}^{1} t^{2} f(t) d t=1-\sin x \forall x \in(0, \pi / 2) $, then $ f\left(\frac{1}{\sqrt{3}}\right) $ is :,

MARKS App · Accepted Answer

∫ sin x 1 t 2 f ( t ) d t = 1 − sin x Applying Leibnitz Rule, − sin 2 x f ( sin x ) cos x = − cos x ⇒ f ( sin x ) = 1 sin 2 x ⇒ f ( t ) = 1 t 2 f ( 1 3 ) = 1 ( 1 3 ) 2 = 3

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