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\int t-\sin(t)\mathrm{d}t
Evaluate the indefinite integral first.
\int t\mathrm{d}t+\int -\sin(t)\mathrm{d}t
Integrate the sum term by term.
\int t\mathrm{d}t-\int \sin(t)\mathrm{d}t
Factor out the constant in each of the terms.
\frac{t^{2}}{2}-\int \sin(t)\mathrm{d}t
Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t\mathrm{d}t with \frac{t^{2}}{2}.
\frac{t^{2}}{2}+\cos(t)
Use \int \sin(t)\mathrm{d}t=-\cos(t) from the table of common integrals to obtain the result. Multiply -1 times -\cos(t).
\frac{1}{2}\times \left(2\pi \right)^{2}+\cos(2\pi )-\left(\frac{0^{2}}{2}+\cos(0)\right)
The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.
2\pi ^{2}
Simplify.