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

Question

26392 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-14T14:15:39

# 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 #