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I-0-pi-2-2304cosx-cos4x-8cos2x-15-2-dx-




Question Number 144069 by SOMEDAVONG last updated on 21/Jun/21
I=∫_0 ^(π/2) ((2304cosx)/((cos4x−8cos2x+15)^2 ))dx=?
$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2304cosx}}{\left(\mathrm{cos4x}−\mathrm{8cos2x}+\mathrm{15}\right)^{\mathrm{2}} }\mathrm{dx}=? \\ $$

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