Question

If ${\int }_0^{\sin x}{t}^{2}f\left(t\right)dt=(1-cosx)$, then $f\left(\frac{1}{3}\right)$ is equal to:

• A. $\frac{9}{2\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $2\sqrt{2}$
• D. $3$

Answer 1

The correct option is A $\frac{9}{2\sqrt{2}}$

${\int }_0^{\sin x}{t}^{2}f\left(t\right)dt=(1-\cos x)$
$\Rightarrow \frac{d}{dx}\left({\int }_0^{\sin x}{t}^{2}f\right(t\left)dt\right)=\frac{d}{dx}(1-\cos x)$
Use Leibnitz formula

${\sin }^{2}xf\left(\sin x\right)\cos x=\sin x$
$f\left(\sin x\right)=\frac{\sin x}{{\sin }^{2}x.\cos x}$
$f\left(\sin x\right)=\frac{1}{\sin x.\cos x}$
If we take $\sin x=\frac{1}{3}$
$\Rightarrow \cos x=\sqrt{1-\frac{1}{9}}=\frac{2\sqrt{2}}{3}$
$f\left(\frac{1}{3}\right)=\frac{1}{\frac{1}{3}\times \frac{2\sqrt{2}}{3}}=\frac{9}{2\sqrt{2}}$