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Q.
If $\int\limits_{\sin x}^{1} t^{2} f(t) d t=1-\sin x, x \in\left(0, \frac{\pi}{2}\right)$ then $f\left(\frac{1}{\sqrt{2}}\right)$ is equal to
Integrals
Solution:
On differentiating both sides with respect to $x$, we get
$0-\sin ^{2} x \cdot f(\sin x) \cos x=-\cos x$
$\Rightarrow f(\sin x)=\frac{1}{\sin ^{2} x} \text {. }$
$\therefore f\left(\frac{1}{\sqrt{2}}\right)=(\sqrt{2})^{2}=2$