How do you integrate #int e^xsinx# by integration by parts method?

Question

Paisley Johnson · Accepted Answer

#int e^x sin x dx = 1/2 e^x(sin(x) - cos(x)) + C#

Integration by parts can be expressed:

#int u(x)v'(x) dx = u(x)v(x) - int v(x) u'(x) dx#

#color(white)()# Let #u(x) = e^x#, #v(x) = -cos(x)#

Then #u'(x) = e^x#, #v'(x) = sin(x)#

and we find:

#int e^x sin x dx = -e^x cos(x) + int e^x cos(x) dx + C_1#

#color(white)()# Let #u(x) = e^x#, #v(x) = sin(x)#

Then #u'(x) = e^x#, #v'(x) = cos(x)#

and we find:

#int e^x cos(x) dx = e^x sin(x) - int e^x sin(x) dx + C_2#

#color(white)()# Combining these two results, we find:

#int e^x sin x dx = -e^x cos(x) + e^x sin(x) - int e^x sin x dx + (C_1+C_2)#

and hence:

#int e^x sin x dx = 1/2 e^x(sin(x) - cos(x)) + C#

#color(white)()# Footnote

Integration by parts is very useful, but can end up leading you down a rabbit hole if you do not choose the parts appropriately.

In the example above, I would instead tend to find the integral by seeing what happens when you differentiate #e^x sin(x)# and #e^x cos(x)# then combine the results:

#d/(dx) e^x sin(x) = e^x sin(x) + e^x cos(x)#

#d/(dx) e^x cos(x) = e^x cos(x) - e^x sin(x)#

So by subtracting the second from the first of these, we find:

#d/(dx) (e^x (sin(x) - cos(x))) = e^x sin(x) + color(red)(cancel(color(black)(e^x(cos(x))))) - color(red)(cancel(color(black)(e^x(cos(x))))) + e^x sin(x)#

#color(white)(d/(dx) (e^x (sin(x) - cos(x)))) = 2e^x sin(x)#

Hence:

#int e^x sin(x) dx = 1/2 e^x (sin(x) - cos(x)) + C#

Charles Arocha · Answer

undefined To integrate $ \int e^x \sin(x) \, dx $ using integration by parts:

1. Choose $ u $ and $ dv $:
   Let $ u = e^x $ and $ dv = \sin(x) \, dx $.

2. Calculate $ du $ and $ v $:
   $ du = e^x \, dx $ and $ v = -\cos(x) $.

3. Apply the integration by parts formula:
   $$ \int u \, dv = uv - \int v \, du $$

4. Substitute the values:
   $$ \int e^x \sin(x) \, dx = e^x (-\cos(x)) - \int (-\cos(x)) (e^x \, dx) $$

5. Simplify and integrate the remaining integral:
   $$ = -e^x \cos(x) + \int e^x \cos(x) \, dx $$

6. Repeat the integration by parts process for the new integral:
   Let $ u = e^x $ and $ dv = \cos(x) \, dx $.

7. Calculate $ du $ and $ v $:
   $ du = e^x \, dx $ and $ v = \sin(x) $.

8. Apply integration by parts again:
   $$ \int e^x \cos(x) \, dx = e^x \sin(x) - \int \sin(x) e^x \, dx $$

9. Notice that the integral on the right side is the same as the original integral but with the sine and cosine functions swapped. Thus:
   $$ \int e^x \cos(x) \, dx = e^x \sin(x) - \int e^x \sin(x) \, dx $$

10. Substitute this back into the previous equation:
    $$ = -e^x \cos(x) + e^x \sin(x) - \int e^x \sin(x) \, dx $$

11. Add the constant of integration $ C $ to complete the solution.