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

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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,

$I = 1/5e^x(sin2x - 2cos2x) + C$

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