# What is the value of F'(x) if F(x) = int_0^sinxsqrt(t)dt ?

Jan 21, 2017

$\therefore F ' \left(x\right) = \left(\sqrt{\sin} x\right) \left(\cos x\right) .$

#### Explanation:

$F \left(x\right) = {\int}_{0}^{\sin} x \sqrt{t} \mathrm{dt}$

$\because , \int \sqrt{t} \mathrm{dt} = \int {t}^{\frac{1}{2}} \mathrm{dt} = {t}^{\frac{1}{2} + 1} / \left(\frac{1}{2} + 1\right) = \frac{2}{3} {t}^{\frac{3}{2}} + c ,$

$\therefore F \left(x\right) = {\left[\frac{2}{3} {t}^{\frac{3}{2}}\right]}_{0}^{\sin} x$

$\therefore F \left(x\right) = \frac{2}{3} {\sin}^{\frac{3}{2}} x$

$\therefore F ' \left(x\right) = \frac{2}{3} \left[{\left\{\left(\sin x\right)\right\}}^{\frac{3}{2}}\right] '$

Using the Chain Rule, $F ' \left(x\right) = \frac{2}{3} \left[\frac{3}{2} {\left(\sin x\right)}^{\frac{3}{2} - 1}\right] \frac{d}{\mathrm{dx}} \left(\sin x\right)$

$= {\left(\sin x\right)}^{\frac{1}{2}} \left(\cos x\right)$

$\therefore F ' \left(x\right) = \left(\sqrt{\sin} x\right) \left(\cos x\right) .$

Enjoy Maths.!