derivative of $\int_x^{\sqrt{x}} e^t\sinh(t)dt$ $$\frac{d}{dx}\int_x^{\sqrt{x}} e^t\sinh(t)dt$$
I really don't know how to find this.
All we have learned is FTC1, FTC2, and the Substitution rule,
Please help me. I've got an exam tomorrow and really need to know how to do this 
I tried the substitution rule. I need more assistance. Thanks
 A: Let $F(x) = \int e^x \sinh x  \ dx$. 
Then your integral equals $$\int_{ x}^{\sqrt x} e^t  \sinh t  \ dt  = F(\sqrt x ) - F(x)$$
Differentiating yields 
$$\frac d{dx}  \int_{ x}^{\sqrt x} e^t  \sinh t  \ dt= F'(\sqrt x) \frac 1{2\sqrt x} - F'(x)$$
But $F'(x) = e^x \sinh x$ and you are done!
This is useful when it is too difficult to find an anti derivative (or just impossible). If the integrand would have been $e^{-x^2}$, this would be the only option ;-)
A: $$\int e^t\sinh(t)dt=\int e^t\left(\frac{e^t-e^{-t}}2\right)dt=\int\left(\frac12e^{2t}-\frac12\right)dt=\frac14e^{2t}-\frac12t$$
so
$$\int_x^{\sqrt x}e^t\sinh(t)dt=\left(\frac14e^{2\sqrt x}-\frac12\sqrt x\right)-\left(\frac14e^{2x}-\frac12x\right)=\frac14e^{2\sqrt x}-\frac12\sqrt x-\frac14e^{2x}+\frac12x$$
so
$$\frac d{dx}\int_x^{\sqrt x}e^t\sinh(t)dt=\frac d{dx}\left(\frac14e^{2\sqrt x}-\frac12\sqrt x-\frac14e^{2x}+\frac12x\right)=\cdots$$
Of course Ant's answer is a better way to do problems like this; it's probably the reason this problem was assigned, and you need to learn it. However, you should also know how to find easy integrals like $\int e^t\sinh(t)dt$.
