# Derivative of an integral $\sqrt{t}\sin t dt$

I need to find the derivative of this function. I know I need to separate the integrals into two and use the chain rule but I am stuck.

$$y=\int_\sqrt{x}^{x^3}\sqrt{t}\sin t~dt~.$$

Let me show you a general method which works in these sorts of situations.

By the Fundamental Theorem of Calculus, we know how to take the derivative of $$F(z):=\int_0^z\sqrt{t}\sin(t)\,dt;$$ in particular, FTC tells us that $$\tag{1} F'(z)=\sqrt{z}\sin(z).$$ Now, note that $$\int_{\sqrt{x}}^{x^3}\sqrt{t}\sin(t)\,dt=\int_0^{x^3}\sqrt{t}\sin(t)\,dt-\int_0^{\sqrt{x}}\sqrt{t}\sin(t)\,dt=F(x^3)-F(\sqrt{x}).$$ So, the derivative you want is $$\frac{d}{dx}\left[F(x^3)-F(\sqrt{x})\right].$$ See if you can use the Chain Rule, and (1), to finish it up from here.

• As written, your first term doesn't have a sine in it. Applying the chain rule gives us $$\frac{d}{dx}\left[F(x^3)-F(\sqrt{x})\right]=3x^2F'(x^3)-\frac{1}{2\sqrt{x}}F'( \sqrt{x}).$$ Now, use (1) to simplify. – Nick Peterson Aug 1 '13 at 0:14
• @Shan Everything is write other than that missing sine in the $\sqrt{x}$ term. – Nick Peterson Aug 1 '13 at 0:15
• I am sorry I typed it in wrong from my paper. I had meant that I had (-4th root(x) sin(sqrt(x)))/(2sqrt(x))+sqrt(x^3)sin(x^3)(3x^2) – Shan Aug 1 '13 at 0:23
• No, not in the denominator. Remember, $F'(z)=\sqrt{z}\sin(z)$; so, $F'(\sqrt{x})=\sqrt[4]{x}\sin(\sqrt{x})$. Plug that (and the similar expression for $F'(x^3)$, in to the statement in my previous comment. – Nick Peterson Aug 1 '13 at 0:25
• Still not quite right. Instead of multiplying your last term by $\sqrt{x}$, you should be multiplying it by $\frac{1}{2\sqrt{x}}$ (i.e. the derivative of $\sqrt{x}$ with respect to $x$), as per the formula I gave in a previous comment. – Nick Peterson Aug 1 '13 at 0:34

Hint

By the chain rule we prove easly:

If $$F(x)=\int_{u(x)}^{v(x)}f(t)dt$$ then $$F'(x)=f(v(x))v'(x)-f(u(x))u'(x)$$