MHB Derivative of ∫ (1+v^3)^10 dv from sinx to cosx | Ashleigh N.

[MarkFL]

Here is the question:

Find the derivative of the integral from sinx to cosx of (1+v^3)^10?

cosx
y = ∫ (1+v^3)^10 dv
sinx

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Ashleigh N.,

We are given to evaluate:

$$\frac{d}{dx}\left(\int_{\sin(x)}^{\cos(x)} \left(1+v^3 \right)^{10}\,dv \right)$$

If we define $F(u)$ to be a function such that:

$$\frac{dF}{du}=f(u)=\left(1+u^3 \right)^{10}$$ then by the FTOC and the chain rule, we may state:

$$\frac{d}{dx}\left(\int_{\sin(x)}^{\cos(x)} \left(1+v^3 \right)^{10}\,dv \right)=\frac{d}{dx}\left(F\left(\cos(x) \right)-F\left(\sin(x) \right) \right)=\frac{d}{dx}F\left(\cos(x) \right)-\frac{d}{dx}F\left(\sin(x) \right)=$$

$$f\left(\cos(x) \right)\frac{d}{dx}\cos(x)-f\left(\sin(x) \right)\frac{d}{dx}\sin(x)=-\sin(x)\left(1+\cos^3(x) \right)^{10}-\cos(x)\left(1+\sin^3(x) \right)^{10}$$