can someone tell me how to integrate by parts the antideriviative cos(7x)sin(5x) dx?

Question

I'm not that good with this section of calculus

Answer 1

The rule is like this:

Intergral of (u*dv) = u*v - Integral of (v*du) where u*dv
is your old expression you want to be integrated.

you have 2 choises: cos(7x)dx is your new dv, or sin(5x) is your new dv. Try and see what works. You will need to express v from dv and du from u.

Answer 2

Integration by parts is a manipulation of the product rule of differentiation:

d[u*v]/dx = udv + vdu

Integrate both sides to get: uv = S udv + S vdu
Now you have the integration by parts formula: S udv = uv - S vdu

A general rule of thumb is for "u," pick whichever becomes simpler when differentiated. In this case I would select sin(5x) because then we don't get a negative.
u = sin(5x) du = 5cos(5x); dv = cos(7x) v = sin(7x)/7

Now plugging into our formula we have:
sin(5x)*sin(7x)/7 - S sin(7x)/7*5cos(5x) = S sin(5x)cos(7x)

--sorry I made a mistake the first time--
**Unfortunately the middle term isn't straightforward and so we integrate that by parts again, saying u=sin(7x)/7, dv= 5cos(5x), du= cos(7x), v= sin(5x) Then :
S 5/7*sin(7x)*cos(5x) = [ sin(7x)sin(5x)/7 - S sin(5x)cos(7x) ]

Substituting this all back into the first result gives:
sin(5x)sin(7x)/7 - sin(7x)sin(5x)/7 + S sin(5x)cos(7x) = S sin(5x)cos(7x) which appears to be a circular answer. Try switching your u and dv in the second part and see if that comes out better. When I did, I ended up getting a non-circular result with the original integral ending up on both sides of the end eqn. When I then solved for it, my answer became:

S sin(5x)cos(7x)dx = [7sin(5x)sin(7x) + 5cos(5x)cos(7x)]/74 + C

Note that 74 is 49 + 25 or 7^2 + 5^2, there is somewhat of a pattern. Hope this helped.