Thread: Integration by parts and with trig

1. Integration by parts and with trig

I have two problems I don't really know how to work through.
The first is the integral from 0 to 1 of r^3/(sqr(4+r^2))
I started by integrating by parts and eventually ended up with (4+r^2)^(7/2) so I know I'm doing something wrong.

The other problem is: find the integral of cosPIx * cos4PIx
I don't know where to begin here. Any help would be appreciated.

2. For first one you don't need to integrate by parts. Simply sub $\displaystyle u = 4 + r^2.$

For the second one, you can either integrate by parts twice and treat the integral as the unknown or you can use the product to sum identity:

$\displaystyle \cos{a}\cos{b} = \frac{1}{2}(\cos{(a-b)} + \cos{(a+b)})$

The first is the integral from 0 to 1 of r^3/(sqr(4+r^2))
$\displaystyle \int {\frac{{r^3 }} {{\sqrt {4 + r^2 } }}dr}$

define: $\displaystyle 4 + r^2 = x^2 \Rightarrow rdr = xdx$

Now I will do something wrong, but help when I subtitute dx

$\displaystyle \int {\frac{{r^3 }} {x} \cdot \frac{x} {r}} dx = \int {r^2 dx}$

but $\displaystyle r^2 = x^2 - 4$ then

$\displaystyle \int {x^2 - 4dx} = \frac{{x^3 }} {3} - 4x + C$

now restoring we obtain ordered

The other problem is: find the integral of cosPIx * cos4PIx
I don't know where to begin here. Any help would be appreciated.
In this case you can integrating by parts, but whit identiti given for Chop Suey the solution is direct

4. define:

Now I will do something wrong, but help when I subtitute dx

but then

now restoring we obtain ordered
I'm confused by this. Is it double substitution or something? I tried substituting u for 4+r^2 but there's no du present.

I'm confused by this. Is it double substitution or something? I tried substituting u for 4+r^2 but there's no du present.
Link's method is not double substitution and it works.

If you use Chops' method,

$\displaystyle u = 4 + r^2$ yields
$\displaystyle du = 2r \, dr$.

Note that $\displaystyle \int \frac{r^3}{\sqrt{4 + r^2}} dr = \frac{1}{2} \int \frac{2r^3 \, dr}{\sqrt{4 + r^2}} = \frac{1}{2} \int \frac{r^2 \cdot 2r \, dr}{\sqrt{4 + r^2}}$, so you can substitute du for 2r dr and u - 4 for r^2:

$\displaystyle \frac{1}{2} \int \frac{u - 4}{\sqrt{u}}du$

I'm confused by this. Is it double substitution or something? I tried substituting u for 4+r^2 but there's no du present.
How do icemanfan is the correct way, I show my way, because I think that is easier see the change of dr to dx

7. For the second one, I get to the integral of 1/2((cosPIx*cos4PIx + sinPIx*sin4PIx) + (cosPIx*cos4PIx - sinPIx*sin4PIx))dx

I'm pretty sure I use substitution but I don't know what I substitute. Am I allowed to set u=cosPIx du=-PIsinPIx and putting a 4 in front of u for the cos4PIx?

8. Why did you expand the cosines using the addition identity? You're complicating things. Just integrate $\displaystyle \cos{(-3\pi x)} = \cos{(3\pi x)}$ as you would integrate a cos(x).

If the negative sign inside is throwing you off, recall that Cosine is an even function:

$\displaystyle \cos{(-x)} = \cos{(x)}$

9. There's no negative in the cos. I just don't know if I'm "allowed" to take the 4 out of cos(4PIx) and put it in front of the cos. [4cosPIx]

10. No, you can't do that.

$\displaystyle \cos{(\pi x}\cos{(4\pi x)}$

$\displaystyle = \frac{1}{2}(\cos{(\pi x - 4 \pi x)} + \cos{(\pi x + 4 \pi x)})$

$\displaystyle = \frac{1}{2}(\cos{(-3 \pi x)} + \cos{(5 \pi x)})$

$\displaystyle = \frac{1}{2}(\cos{(3 \pi x)}) + \frac{1}{2}(\cos{(5 \pi x)})$

Just integrate it directly!

$\displaystyle \cos \left( {4\pi x} \right) = \cos \left( {2\pi x + 2\pi x} \right) = \cos ^2 \left( {2\pi x} \right) - \sin ^2 \left( {2\pi x} \right)$