Calculus Integration by Parts -- two problems. Please help?

Question

Evaluate the following integrals:

e^(6x)sin(5x)dx

(sqrt(t))*ln t*dt

For the first one... what do I do after i make u = e^6?

Answer 1

For the first one you need to generate a reduction formula.

Let u = sin5x , hence u' = 5cos5x

and v' = e^6x hence v= 1/6e^6x

Int[uv'] = uv - Int[vu']

Therefore Integral = 1/6e^6xsin5x - 5/6Int [ cos5xe^6x] dx (i took the 5/6 out)

Do integration by parts again with the [ cos5xe^6x] dx

u = cos5x, u' = -5sin5x

v' = e^6x, v = 1/6e^6x

Integral = 1/6e^6xsin5x - 5/6{1/6e^6xcos5x + 5/6Int [ e^6xsin5x ] dx }

Now see the recurrence of e^6xsin5x , replace it by I and rearrange for I.

Therefore

I = 1/6e^6xsin5x - 5/36e^6xcos5x - 25/36 I

Therefore

I + 25/36 I = 1/6e^6xsin5x - 5/36e^6xcos5x

and hence

I (61/63) = 1/6e^6xsin5x - 5/36e^6xcos5x

Finally

I (integral) = [1/6e^6xsin5x - 5/36e^6xcos5x] / [61/63]

Then just take some common factors out to make it look nicer.

I = 63/366 e^6x ( sin5x - 5/6cos5x ) + c

Answer 2

you don't know the formula ?? int of { u dv } = u v - int { v du}...thus compute du and v...do it twice like I previously indicated....on the right you likely get the original integral * 36 / 25

for the second u = ln t....not as difficult as the 1st

Answer 3

a) ∫ e^(6x) sin(5x) dx =

integrate it by parts, assuming:

sin(5x) = u → (differentiating both sides) → 5cos(5x) dx = du

e^(6x) dx = dv → (integrating both sides) → (1/6) e^(6x) = v

thus, integrating by parts, you get:

∫ u dv = u v - ∫ v du →

∫ e^(6x) sin(5x) dx = (1/6) e^(6x) sin(5x) - ∫ (1/6) e^(6x) [5cos(5x)] dx →

∫ e^(6x) sin(5x) dx = (1/6) e^(6x) sin(5x) - (5/6) ∫ e^(6x) cos(5x) dx →

a further integration by parts being needed, let again e^(6x) dx = dv:

e^(6x) dx = dv → (1/6) e^(6x) = v

cos(5x) = u → 5 [-sin(5x)] dx = du → - 5 sin(5x) dx = du

thus, integrating the remaining integral, you get:

∫ u dv = u v - ∫ v du →

∫ e^(6x) sin(5x) dx = (1/6) e^(6x) sin(5x) - (5/6) ∫ e^(6x) cos(5x) dx →

∫ e^(6x) sin(5x) dx = (1/6) e^(6x) sin(5x) - (5/6) {(1/6) e^(6x) cos(5x) -

∫ (1/6)e^(6x) [- 5 sin(5x)] dx} →

∫ e^(6x) sin(5x) dx = (1/6) e^(6x) sin(5x) - (5/6) {(1/6) e^(6x) cos(5x) +

(5/6) ∫ e^(6x) sin(5x) dx } →

∫ e^(6x) sin(5x) dx = (1/6) e^(6x) sin(5x) - (5/36) e^(6x) cos(5x) - (25/36) ∫ e^(6x) sin(5x) dx →

having got the same unknown integral at both sides, shift

- (25/36) ∫ e^(6x) sin(5x) dx to left:

∫ e^(6x) sin(5x) dx + (25/36) ∫ e^(6x) sin(5x) dx = (1/6) e^(6x) sin(5x) -

(5/36) e^(6x) cos(5x) →

[1 + (25/36)] ∫ e^(6x) sin(5x) dx = (1/6) e^(6x) sin(5x) - (5/36) e^(6x) cos(5x) →

[(36 + 25)/36] ∫ e^(6x) sin(5x) dx = (1/6) e^(6x) sin(5x) - (5/36) e^(6x) cos(5x) →

(61/36) ∫ e^(6x) sin(5x) dx = (1/6) e^(6x) sin(5x) - (5/36) e^(6x) cos(5x) →

thus, finding out the unknown integral,

∫ e^(6x) sin(5x) dx = (36/61)[(1/6) e^(6x) sin(5x) - (5/36) e^(6x) cos(5x)] + C →

∫ e^(6x) sin(5x) dx = (36/61)(1/6) e^(6x) sin(5x) - (36/61)(5/36) e^(6x) cos(5x) + C →

in conclusion,

∫ e^(6x) sin(5x) dx = (6/61) e^(6x) sin(5x) - (5/61) e^(6x) cos(5x) + C

b) ∫ √t ln t dt =

let √t dt = dv (integrate both sides) → {t^[(1/2)+1]} /[(1/2)+1] = v →

[t^(3/2)] /(3/2) = v → (2/3)√t³ = v → (2/3) t√t = v

ln t = u (differentiate both sides) → (1/ t) dt = du

then, integrating by parts, you get:

∫ u dv = u v - ∫ v du →

∫ √t ln t dt = (2/3) t√t ln t - ∫ (2/3) t√t (1/ t) dt =

t canceling out,

(2/3) t√t ln t - ∫ (2/3) √t dt =

(2/3) t√t ln t - (2/3) ∫ √t dt =

(2/3) t√t ln t - (2/3) ∫ t^(1/2) dt =

(2/3) t√t ln t - (2/3) {t^[(1/2)+1]} /[(1/2)+1] + C =

(2/3) t√t ln t - (2/3) [t^(3/2)] /(3/2) + C =

(2/3) t√t ln t - (2/3)(2/3) t^(3/2) + C =

(2/3) t√t ln t - (2/3)(2/3)√t³ + C =

(2/3) t√t ln t - (2/3)(2/3) t√t + C

thus, in conclusion:

∫ √t ln t dt = (2/3) t√t [ln t - (2/3)] + C

I hope it helps...

Bye!