If ∫e2x(cos⁡x+7sin⁡x)dx=e2xg(x)+C where c is a constant of integration, then g(0)+gπ2 is equal to 

If e2x(cosx+7sinx)dx=e2xg(x)+C where c is a constant of integration, then g(0)+gπ2 is equal to 

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    Solution:

    We know that

    eaxsinbxdx=eaxa2+b2(asinbxbcosbx)+Ceaxcosbxdx=eaxa2+b2(acosbx+bsinbx)+Ce2x(cosx+7sinx)dx=e2xcosxdx+7e2xsinxdx=e2x5(2cosx+sinx)+75e2x(2sinxcosx)+C=e2x5(5cosx+15sinx)+C    g(x)=15(5cosx+15sinx)=(cosx+3sinx)    g(0)+gπ2=1+3=2

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