Science:Math Exam Resources/Courses/MATH105/April 2015/Question 01 (g)/Solution 1

Recall the fundamental theorem of calculus: if $F$ is an anti-derivative of $f$, then $F(y)-F(a)=\int _{a}^{y}f(t)\,dt.$

Let $f(t)=t^{6}+8,$ and let $F$ be an anti-derivative of $f$ (i.e., $F'=f$).

Then, applying the fundamental theorem of calculus with $y=\sin x$ and $a=0,$ we have

$\int _{0}^{\sin x}f(t)\,dt=F(\sin x)-F(0).$

Now, taking the derivative of both sides of the equation (and remembering to use the chain rule), we have

${\frac {d}{dx}}\left(\int _{0}^{\sin x}f(t)\,dt\right)={\frac {d}{dx}}(F(\sin(x))-F(0))=F'(\sin x)\cdot (\sin x)'-0=f(\sin x)\cos x=\color {blue}(\sin ^{6}x+8)\cos x$.