Antiderivatives and definite integrals - 2

Question

Prove that the function $$ f(x)=x^2\int_{0}^{1}t\sin^2(tx)dt $$ is differentiable in $\mathbb{R}$ and determine the formula $f'(x)$ of its derivative.

Motivation: Exactly as in: Antiderivatives and definite integrals

Answer 1

For all $x\in\mathbb{R}$ we have $$ f(x)=x^2\int_{0}^{1}t\sin^2(tx)dt=\int_{0}^{1}x^2t\sin^2(tx)dt=\int_{0}^{1}(xt)\sin^2(tx)xdt $$ Now, let $u=xt$ thus $du=xdt$ ($x$ is considered to be a constant when integrating with respect to $t$). For $t=0,1$, we get $u=0,x$, respectively. Hence $$ f(x)=\int_{0}^{x}u\sin^2 udu, \ \ \ \ \ \ \ \ x\in\mathbb{R} $$ Consequently, $f$ is a differentiable function in $\mathbb{R}$ and its derivative is given by $$ f'(x)=x\sin^2x $$