`int e^(3x)(cos2x+sin2x)dx=e^(3x)*g(x)+c` Then `13[g(0)+g((pi)/(4))]=`

int(cos x+3sin x)e^(3x)dx

int e^(3x)cos4xdx=e^(3x)(A sin4x+B cos4x)+c

int(e^(x)+2sin x-3cos x)dx

If int e^(2x)(cos x+7sin x)dx=e^(2x)g(x)+c where c is constant of integration then g(0)+g((pi)/(2))=

int(x(sin x+cos x)+cos x)/(e^(-x)+x sin x+cos x)dx=f(x)+log|g(x)|+c then f(x)+g(x)=

G) int(sin x)/(cos(x-a))dx

int_( Let )^( Let )e^(3x)tan^(-1)e^(x)dx=e^(3x)f(x)+(1)/(6)g(x)+C such that f(0)=(pi)/(12),g(0)=In((2)/(e)) and C is constant of integration,then

(i) int e^(3x) cos 5x dx (ii) int e^(3x) sin 4x dx

int(x sin x)/(cos^(5)x)dx=g(x)+c find g(x)