integrate t^(1/2)sintdt. let lower bound=(x^1/2) upper bound=(x^3). use the chain rule and fundamental theorem calculas part1.
It seems to me that $\displaystyle x$ should be at least $\displaystyle x>0$
It seems you wish to integrate the function,
$\displaystyle f(x)=\int_{x^{1/2}}^{x^3}t^{1/2}\sin tdt$
Let use find whether $\displaystyle x^{1/2}=x^3$
The only possible solution is $\displaystyle x=0,1$ but $\displaystyle x>0$ thus, $\displaystyle x=1$.
In that case,
$\displaystyle f(1)=0$
If, $\displaystyle 0<x<1$ then, $\displaystyle x^{1/2}>x^3$
Thus,
$\displaystyle \int_{x^{1/2}}^{x^3}t^{1/2}\sin t dt=-\int_{x^3}^{x^{1/2}}t^{1/2}\sin tdt$
Therefore, there exists a point in the interval $\displaystyle [x^3,x^{1/2}]$
Using the integral subdivision rule,
$\displaystyle -\int_{x^3}^c t^{1/2}\sin tdt-\int_c^{x^{1/2}}t^{1/2}\sin tdt$
Thus,
$\displaystyle \int_c^{x^3} t^{1/2}\sin tdt-\int_c^{x^{1/2}}t^{1/2}\sin tdt$
Let,
$\displaystyle \mu (x)=\int_c^x t^{1/2}\sin tdt$
Therefore, those two functions can be expressed as composition functions,
$\displaystyle \mu (x^3)-\mu (x^{1/2})$
But by the Fundamental theorem,
$\displaystyle \mu ' (x)=x^{1/2}\sin x$
And since $\displaystyle x^3,x^{1/2}$ are both differenciable on $\displaystyle (0,1)$ we can use the chain rule.
$\displaystyle 3x^2 \mu' (x^3)+\frac{1}{2}x^{-1/2}\mu'(x^{1/2})=3 x^{7/2}\sin(x^3)+\frac{1}{2}x^{-1/2}\sin (x^{-1/2})$
Something interesting to note.
$\displaystyle \lim_{x\to 1}$ of this expression if not equal to $\displaystyle f(1)$ which means the function you mentioned is not everywhere differenciable.
The other part of the problem is to consider the case $\displaystyle x>1$. Since it is late now I cannot complete the solution but I think it will be the same as for $\displaystyle (0,1)$