Verified answer

Let's say that the integral of e^(7x) * cos(4x) * dx = t

u = e^(7x)

du = 7 * e^(7x) * dx

dv = cos(4x) * dx

v = (1/4) * sin(4x)

t = uv - int(v * du)

t = (1/4) * e^(7x) * sin(4x) - (7/4) * int(e^(7x) * sin(4x) * dx)

Integrate by parts again:

u = e^(7x)

du = 7 * e^(7x) * dx

dv = sin(4x) * dx

v = (-1/4) * cos(4x)

t = (1/4) * e^(7x) * sin(4x) - (7/4) * (uv - int(v * du))

t = (1/4) * e^(7x) * sin(4x) - (7/4) * ((-1/4) * e^(7x) - (-7/4) * int(e^(7x) * cos(4x) * dx))

t = (1/4) * e^(7x) * sin(4x) - (7/4) * (-1/4) * e^(7x) * cos(4x) - (7/4) * (7/4) * int(e^(7x) * cos(4x) * dx)

remember, we set the integral of (e^(7x) * cos(4x) * dx) = t, so:

t = (1/4) * e^(7x) * sin(4x) + (7/16) * e^(7x) * cos(4x) - (49/16) * t

16t = 4 * e^(7x) * sin(4x) + 7 * e^(7x) * cos(4x) - 49t

65t = e^(7x) * (4 * sin(4x) + 7 * cos(4x))

t = (1/65) * e^(7x) * (4 * sin(4x) + 7 * cos(4x))

EDIT:

A note to remember is that you should be very careful of your signs. I almost made a few mistakes and luckily I caught myself quickly enough. It can get a little tricky, especially after you integrate by parts the second time around.

because of fact the spinoff of tan(x) is sec²(x), ?tan(x) sec²(x) dx = (a million/2)tan²(x) + C ---- replace: the reason which you're turning out to be maximum of distinctive finding solutions decrease than is by the fact a million+tan²(x) = sec²(x). in specific, think that C = a million/2 + D Then the respond could be (a million/2)tan²(x) + a million/2 + D = (a million/2)(tan²(x) + a million) + D = (a million/2)sec²(x) + D = (a million/2)(a million/cos²(x)) + D those solutions are all equivalent. it truly is between the few circumstances the place that consistent you customarily tack on after integration impacts what the top result sounds like. So many times, to verify if 2 integrations of the comparable function are equivalent, you should subtract the two consequences and if the type is a relentless, you already know that the two solutions are purely as valid (or purely as incorrect, for that count).

The above answers are wrong because exponents come first by default. Integration by parts is much easier to do in terms of functions and derivatives rather than differentials, yet the above answers have chosen the more complicated way.

Integrate the original integrand by parts:

∫ ℮⁷xcos(4x) dx = ℮⁷ ∫ xcos(4x) dx

Let f'(x) = cos(4x)

f(x) = sin(4x) / 4

Let g(x) = x

g'(x) = 1

∫ f'(x)g(x) dx = f(x)g(x) - ∫ f(x)g'(x) dx

∫ xcos(4x) dx = xsin(4x) / 4 - ∫ sin(4x) / 4 dx

∫ xcos(4x) dx = xsin(4x) / 4 + cos(4x) / 16 + C

∫ xcos(4x) dx = [4xsin(4x) + cos(4x)] / 16 + C

∫ ℮⁷xcos(4x) dx = ℮⁷[4xsin(4x) + cos(4x)] / 16 + C

Use the table of integrals

You are correct that:

∫ e^(7x)cos(4x) = (1/7)e^(7x)cos(4x) + 4/7 ∫ e^(7x)sin(4x) dx.

At this point, integrate ∫ e^(7x)sin(4x) dx by parts again with:

u = sin(4x) ==> du = 4cos(4x) dx

dv = e^(7x) dx ==> v = (1/7)e^(7x).

This yields:

∫ e^(7x)cos(4x) = (1/7)e^(7x)cos(4x) + 4/7 ∫ e^(7x)sin(4x) dx

= (1/7)e^(7x)cos(4x) + (4/7)(uv - ∫ v du)

= (1/7)e^(7x)cos(4x) + (4/7)[(1/7)e^(7x)sin(4x) - 4/7 ∫ e^(7x)cos(4x) dx]

= (1/7)e^(7x)cos(4x) + (4/49)e^(7x)sin(4x) - 16/49 ∫ e^(7x)cos(4x) dx.

Note that we now have ∫ e^(7x)cos(4x) dx on the right side. By adding 16/49 ∫ e^(7x)cos(4x) dx to both sides, we get:

65/49 ∫ e^(7x)cos(4x) dx = (1/7)e^(7x)cos(4x) + (4/49)e^(7x)sin(4x)

==> 65/49 ∫ e^(7x)cos(4x) dx = e^(4x)[7cos(4x) + 4sin(4x)]/49, by factoring.

Multiplying both sides by 49/65 and tacking on the integration constant yields:

∫ e^(7x)cos(4x) dx = e^(4x)[7cos(4x) + 4sin(4x)]/65 + C.

I hope this helps!