1. ## integration by parts

Evaluate the integral.

i dont understand how this problem will end up finishing itself using integration by parts i guess i just dont understand it!

2. Originally Posted by ahawk1
Evaluate the integral.

i dont understand how this problem will end up finishing itself using integration by parts i guess i just dont understand it!
u = sin 2x v = (1/5) e^5x
du = 2 cos 2x dx dv = e^5x dx

{sin 2x e^5x dx = (1/5) e^5x (sin 2x) - 2/5 {e^5x cos 2x dx

Important note: whenever you're stuck with another "integration by parts" expression in the end, always try to use similar u, du, v, and dv identities. In this case, the u/du side appears to be trigonometric and the v/dv sides appears to be e-related. So...

u = cos 2x v = (1/5)e^5x
du = -2 sin 2x dv = e^5x

{cos 2x e^5x dx = (1/5)e^5x cos 2x + 2/5{e^5x sin 2x dx

So, returning to the original integration by parts expression...

{sin 2x e^5x dx = (1/5) e^5x (sin 2x) - 2/5 {e^5x cos 2x dx

Substitution gives us...

{sin 2x e^5x dx = (1/5) e^5x (sin 2x) - 2/5 [(1/5)e^5x cos 2x + 2/5 {e^5x sin 2x dx]

{sin 2x e^5x dx = (1/5) e^5x (sin 2x) - (2/25)e^5x cos 2x - 4/25{e^5x sin 2x dx

Adding the same integral to both sides....

(29/25){sin 2x e^5x dx = (1/5) e^5x (sin 2x) - (2/25)e^5x cos 2x

{sin 2x e^5x dx = (25/145)e^5x (sin2x) - (50/725)e^5x cos 2x

e^5x [(5/29)sin 2x - (2/29)cos 2x] + C

3. Originally Posted by ahawk1
Evaluate the integral.

i dont understand how this problem will end up finishing itself using integration by parts i guess i just dont understand it!
Strong hint:

One way is to apply integration by parts twice. The first time you end up with an integral involving $\displaystyle e^{5 \theta} \cos(2 \theta)$.

Apply integration by parts to this second integral and you now have an integral involving $\displaystyle e^{5 \theta} \sin(2 \theta)$.

It looks like you are in an infinite loop. But no! You have an equation involving $\displaystyle \int e^{5 \theta} \sin(2 \theta)$ which you can solve for the value of the integral.