# Thread: Help with integration by parts

1. ## Help with integration by parts

So I have two integration by parts problems that I am having trouble with. For both of these problems, I have the answer but I cannot figure out how to get there.

The first one is $\displaystyle \int x^2 cos(x)dx$. First off, I made u = x^2 and dv = cos(x)dx. Next step, I found that du = 2xdx and v = sin(x). So going off of $\displaystyle u*v - \int v*du$, I got $\displaystyle x^2 sin(x) - \int sin(x)*2xdx$. Simplifying that down further leaves me with $\displaystyle x^2 sin(x) + 2cos(x)$. However, the answer is $\displaystyle (x^2 - 2) sin(x) + 2cos(x)$. What did I mess up?

And the second one is $\displaystyle \int x^2 \sqrt{4x-1}$. I made u = x^2 and dv = $\displaystyle \sqrt{4x-1}$. So that meant that du = 2xdx and v = 1/6(4x-1)^3/2. The next step left me with $\displaystyle \frac{1}{6} x^2 (4x-1)^\frac{3}{2} - \int \frac{1}{6}(4x-1)^\frac{3}{2} * 2x$. Simplifying that further left me with $\displaystyle \frac{1}{6} x^2 (4x-1)^\frac{3}{2} - \frac{2}{15}x(4x-1)^\frac{5}{2}$. But, the correct answer is $\displaystyle \frac{x^2(4x-1)^\frac{3}{2}}{14} + \frac{x(4x-1)^\frac{3}{2}}{70} + \frac{(4x-1)^\frac{3}{2}}{420}$. Again, where did I mess up so badly?

2. For each question did you intergrate by parts once or twice?

3. I did once for both of them. If it's twice? What do I separate in the equation? I'm left with a simple square root for one and a trig function for the other, both of which I thought, do not require integration by parts.

4. Originally Posted by TheMathTham
So I have two integration by parts problems that I am having trouble with. For both of these problems, I have the answer but I cannot figure out how to get there.

The first one is $\displaystyle \int x^2 cos(x)dx$. First off, I made u = x^2 and dv = cos(x)dx. Next step, I found that du = 2xdx and v = sin(x). So going off of $\displaystyle u*v - \int v*du$, I got $\displaystyle x^2 sin(x) - \int sin(x)*2xdx$. Simplifying that down further leaves me with $\displaystyle x^2 sin(x) + 2cos(x)$. However, the answer is $\displaystyle (x^2 - 2) sin(x) + 2cos(x)$. What did I mess up?
On the first one, you got this far correctly.

$\displaystyle \displaystyle \int x^2 \cos(x)dx=x^2 \sin(x) - 2\!\! \int x\sin(x)dx$
(I rearranged the integral on the right a little.)

Now use integration by parts (again) to evaluate $\displaystyle \displaystyle \int x\sin(x)dx$ .

Use $\displaystyle u=x$ and $\displaystyle dv=\sin(x)dx$.

5. Originally Posted by TheMathTham
I did once for both of them. If it's twice?
You integrate twice so $\displaystyle x^2$ can be differentiated twice and become constant.