Integral of $$$t \sin^{2}{\left(\omega \right)}$$$ with respect to $$$t$$$
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Your Input
Find $$$\int t \sin^{2}{\left(\omega \right)}\, dt$$$.
Solution
Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=\sin^{2}{\left(\omega \right)}$$$ and $$$f{\left(t \right)} = t$$$:
$${\color{red}{\int{t \sin^{2}{\left(\omega \right)} d t}}} = {\color{red}{\sin^{2}{\left(\omega \right)} \int{t d t}}}$$
Apply the power rule $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:
$$\sin^{2}{\left(\omega \right)} {\color{red}{\int{t d t}}}=\sin^{2}{\left(\omega \right)} {\color{red}{\frac{t^{1 + 1}}{1 + 1}}}=\sin^{2}{\left(\omega \right)} {\color{red}{\left(\frac{t^{2}}{2}\right)}}$$
Therefore,
$$\int{t \sin^{2}{\left(\omega \right)} d t} = \frac{t^{2} \sin^{2}{\left(\omega \right)}}{2}$$
Add the constant of integration:
$$\int{t \sin^{2}{\left(\omega \right)} d t} = \frac{t^{2} \sin^{2}{\left(\omega \right)}}{2}+C$$
Answer
$$$\int t \sin^{2}{\left(\omega \right)}\, dt = \frac{t^{2} \sin^{2}{\left(\omega \right)}}{2} + C$$$A