# Math Help - Integrals

1. ## Integrals

Hi,

Evaluate the following integrals using integration by parts:

a) f x sin 5x dx

b) f x^2 e^3x dx

* note: these functions above, are indefinate integrals

thanks,

moon

2. Originally Posted by Moon Hoplite
a) f x sin 5x dx
$\int x \sin 5 x dx = \int x \left( -\frac{1}{5}\cos 5x \right) ' dx = - \frac{x}{5}\cos 5x + \frac{1}{5}\int \cos 5x dx$

b) f x^2 e^3x dx
$\int x^2 \left( \frac{1}{3}e^{3x} \right) ' dx = \frac{x^2}{3}e^{3x} - \frac{2}{3} \int x e^{3x} dx$

Can you continue?

3. Originally Posted by Moon Hoplite
Hi,

Evaluate the following integrals using integration by parts:

a) f x sin 5x dx

b) f x^2 e^3x dx

* note: these functions above, are indefinate integrals

thanks,

moon
Hey!

So, integrating by parts has this really helpful mnemonic device associated with it. ILATE - it's a priority list for what you should make your u. I - inverse trig functions L - log functions A - algebraic functions T - trig functions E - exponential functions. It's a listing of the easiest things to take the derivative of that will also simplify with each successive iteration.

a) $\int x\ sin 5x dx$

Start by making your u = x and then your du = dx
dv = sin(5x) and so your v = -1/5cos(5x)

$-1/5x * cos(5x) - \int -1/5 * cos(5x) dx = -1/5x * cos(5x) + 1/25 sin[5x] + C$

b) $\int x^2\ * e^3x dx$
u = x^2
du = 2x dx

dv = e^3x dx
v = (e^3x)/3

x^2 * (e^3x)/3 - 1/3 * \int e^3x * 2x dx

Use by parts for the integral again.

u = 2x
du = 2 dx

dv = e^3x dx
v = 1/3 * e^3x

x^2 * (e^3x)/3 - 1/3 * ( 2x * 1/3 * e^3x - 2/3 * \int e^3x dx = ((x^2 * e^3x) - 1/3 ( (2x * e^3x)/3 - (2 *e^3x)/9 ) + C