How do you use substitution to integrate #e^x * sin2xdx #?

Question

Emily Ammerman · Accepted Answer

# int \ e^(x)sin2x \ dx = 1/5e^x(sin2x - 2cos2x) + C #

We seek the integral: # I =int \ e^(x)sin2x \ dx # There is no suitable substitution, however, We can apply Integration By Parts: Let # { (u,=sin2x, => (du)/dx,=2cos2x), ((dv)/dx,=e^x, => v,=e^x ) :}# Then plugging into the IBP formula: # int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx # We have: # int \ (sin2x)(e^x) \ dx = (sin 2x)(e^x) - int \ (e^x)(2cos 2x) \ dx # # :. I = e^(2x)sin2x - 2 \ int \ e^xcos 2x \ dx # Now consider the integral given by: # I_2 = int \ e^xcos 2x \ dx # We will now need to apply IBP again: Let # { (u,=cos2x, => (du)/dx,=-2sin2x), ((dv)/dx,=e^x, => v,=e^x ) :}# Then plugging into the IBP formula we have:: # int \ (cos2x)(e^x) \ dx = (cos2x)(e^x) - int \ (e^x)(-2sin2x) \ dt # # I_2 = e^xcos2x + 2 \ int \ e^xsin2x \ dt # # \ \ \ = e^xcos2x + 2I # And so combining the results we find that: # I = e^xsin2x - 2{e^xcos2x + 2I} # # \ \ = e^xsin2x - 2e^xcos2x - 4I # # :. 5I = e^x(sin2x - 2cos2x) # # :. I = 1/5e^x(sin2x - 2cos2x) # And not forgetting the constant of integration, # I = 1/5e^x(sin2x - 2cos2x) + C #

Christopher Agresta · Answer

undefined To integrate e^x * sin(2x) dx using substitution, let u = sin(2x), then du = 2cos(2x) dx. Rearranging, dx = du / (2cos(2x)). Now, substitute u and dx in terms of u into the integral:

∫ e^x * sin(2x) dx = ∫ e^x * u * (1/2cos(2x)) du
                  = (1/2) ∫ e^x * u / cos(2x) du
                  = (1/2) ∫ e^x * u / (1 - 2sin^2(x)) du
                  = (1/2) ∫ e^x * u / (1 - 2u^2) du

This integral can be further simplified using partial fractions or other methods, but the initial substitution is the key step in solving this integral.