Charles Kingsley
2021-12-18
Answered

Evaluate the integral.

$\int t{\mathrm{sin}}^{2}tdt$

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Determine this real integral with the Residue-theorem.

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}\frac{\mathrm{sin}x}{{x}^{4}-6{x}^{2}+10}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}\frac{\mathrm{sin}x}{{x}^{4}-6{x}^{2}+10}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$