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CalculusHardTechniques of Integration

Techniques of Integration · Hard problem

Two generators feed the same resistor, so the total current is i(t)=3sin⁡(ωt)+4cos⁡(ωt)i(t)=3\sin(\omega t)+4\cos(\omega t) amps. With resistance R=2  ΩR=2\;\Omega, the average power dissipated over one full period is Pavg=1T∫0TR i(t)2 dtP_{\text{avg}}=\dfrac{1}{T}\displaystyle\int_0^T R\,i(t)^2\,dt. Find PavgP_{\text{avg}} (in watts).