## 12th Harvard MIT Calculus Problem 2

The differentiable function $F$:$\mathbb{R}\to \mathbb{R}$ satisfied $F(0)=-1$ and $\frac{d}{dx}F(x)=\sin (\sin (\sin (\sin (x))))\cos (\sin (\sin (x)))\cos (\sin (x))\cos (x)$

Find $F(x)$ as a function of $x$.

Solution

Substitution $u=\sin (\sin (\sin (x)))$

then $du={{(\sin (\sin (x)))}^{'}}\cos (\sin (\sin (x)))dx=\cos (\sin (\sin (x)))\cos (\sin x)\cos xdx$

So we get $F(x)=\int{\sin udu=-\cos u+c}$

since we know that $F(0)=-1$ then we get $c=0$

Therefore $F(x)=\cos (\sin (\sin (\sin (x))))$

😀 Thank you Pablo for comment about $F(x)=\cos x$ 😀