Use the chain rule to evaluate $\frac{\mathrm{d}}{\mathrm{d}x}\displaystyle\int_{x^2-1}^{\sin(x)} \cos(t) \, \mathrm{d}t $

Question

Doesn't the derivative of that integral just equal $\cos(x)$? What does it mean to use the Chain Rule? I know for sure it has nothing to do with $u$-substitution. Any help would be appreciated, thanks.

Answer 1

For some real $c$, one obtains $$\frac{d}{dx}\int_{x^2-1}^{\sin x}\cos t \: dt = \frac{d}{dx}\int_c^{\sin x}\cos t \: dt-\frac{d}{dx}\int_c^{x^2-1} \cos t \: dt.$$ If $\phi(s) = \int_c^s \cos \: dt$, then $\int_{x^2-1}^{\sin x}\cos t \: dt = \phi(\sin x) - \phi(x^2 -1)$, thus $$\frac{d}{dx}\int_{x^2-1}^{\sin x}\cos t \: dt = \frac{d}{dx}\phi(\sin x) - \frac{d}{dx}\phi(x^2-1).$$ Now you see where the chain rule comes in?

Answer 2

The fundamental theorem of calculus write $$\frac d{dx}\int_{a(x)}^{b(x)} f(t)\,dt=f(b(x))\, b'(x)-f(a(x))\, a'(x)$$ Now, for your case $$a(x)=x^2-1\implies a'(x)=?$$ $$b(x)=\sin(x)\implies b'(x)=?$$

Answer 3

Note that $\int \cos t \, \mathrm{d}t = \sin t + \mathcal{C}$. You want the integral between $x^2 - 1$ and $\sin x$. That is, you want:

$$\frac{\mathrm{d}}{\mathrm{d}x}(\sin (\sin x) - \sin (x^2-1))$$

where you need to use the chain rule to differentiate this expression.

Answer 4

Notice:

$$\frac{\partial}{\partial x}\left[\int_{y(x)}^{z(x)}f(t)\space\text{d}t\right]=f(z(x))\cdot\frac{\partial z(x)}{\partial x}-f(y(x))\cdot\frac{\partial y(x)}{\partial x}=f(z(x))z'(x)-f(y(x))y'(x)$$

So, we get:

$$\frac{\partial}{\partial x}\left[\int_{x^2-1}^{\sin(x)}\cos(t)\space\text{d}t\right]=$$ $$\cos(\sin(x))\cdot\frac{\text{d}}{\text{d}x}\left(\sin(x)\right)-\cos(x^2-1)\cdot\frac{\text{d}}{\text{d}x}\left(x^2-1\right)=$$ $$\cos(x)\cos(\sin(x))-2x\cos(x^2-1)$$