I have a couple of problems that I'm trying to work through. I'm a tad stuck on 2. Here is what I have?

  1. $\int t \cdot e^{-3t} dt$

so let's say:

$$u = t \quad \text{and} \quad du = dt$$

$$dv = e^{-3t} \quad \text{and} \quad v = \frac{e^{-3t}}{-3}$$

so according to integration by parts:

$$\begin{align*} \int t \cdot e^{-3t} dt &= t \cdot \frac{e^{-3t}}{-3} - \int \frac{e^{-3t}}{-3} dt \newline &= \frac{t}{-3} \cdot e^{-3t} - \frac{-1}{3} \frac{e^{-3t}}{-3} \newline &= \left(\frac{t}{3} - \frac{1}{9} \right) e^{-3t} \end{align*}$$

Is this right?

  1. $\int t^2 \sin (\beta t) dt$

Is $\beta $ a constant? What is this notation?

  1. $\int \ln \sqrt{x} dx$

$$\int \ln \sqrt{x} dx = \int \ln x^{\frac{1}{2}} dx$$

so let's try:

$$u = \ln{x^{\frac{1}{2}}} \quad \text{and} \quad du = \frac{1}{x^{\frac{1}{2}}} \cdot \frac{1}{2} \frac{1}{x^{\frac{1}{2}}} = \frac{1}{2x}$$


$$dv = dx \quad \text{and} \quad v = x$$

so $$\begin{align*} \int \ln \sqrt{x} dx = \ln x^{\frac{1}{2}} \cdot x - \int x dx = x \ln{x^{\frac{1}{2}}} - \frac{x^2}{2} \end{align*}$$

How does that look?

  • 1
    $\begingroup$ You pick $u$ and $dv$ in order to make $\int v\,du$ easier to calculate than $\int u\,dv$. So you shouldn’t pick $dv$ to be something you can’t integrate, since you need $v$. And you should generally pick $u$ to be something whose derivative $du$ is “simpler.” Use these ideas to think about (2) and (3). $\endgroup$ Jan 30 '19 at 5:14
  • 1
    $\begingroup$ Note $\log(x^{1/2})=\frac12\log(x)$. Then let $u=\log(x)$ and $v=x$. $\endgroup$
    – Mark Viola
    Jan 30 '19 at 5:14
  • $\begingroup$ math.stackexchange.com/questions/768332/liate-how-does-it-work $\endgroup$ Jan 30 '19 at 5:21
  1. You seemed to have missed out a minus sign on the integral on line 2.

$$- \int \frac{e^{-3t}}{-3} dt = - \frac{1}{-3}\frac{e^{-3t}}{-3} + c$$

  1. If it doesn't state it in the question, then yes, $\beta$ is a constant. As long as it doesn't say that $\beta$ is some function of $t$, since the integral is with respect to $t$. For this question you need to integrate by parts twice, choosing to differentiate the $t$ term every time.

  2. You seemed to have substituted the variables wrong.

$$= x\ln x^{\frac{1}{2}} - \int \frac{x}{2x} dx$$

Remember that $\int k\cdot f(x) dx$, where $k$ is a constant is equal to $k\int f(x) dx$. In this case realise that $\log x^n = n\log x$. This should make it a bit less messy.

  • $\begingroup$ How did you see this for 2? Can you show me? $\endgroup$
    – Jwan622
    Jan 30 '19 at 13:43
  • $\begingroup$ Something in the form of $\int x^n \sin(mx)dx$ or $\int x^n \cos(mx)dx$ is a fairly common question to test how well you know integration by parts. $\endgroup$
    – Infiaria
    Jan 30 '19 at 13:52
  1. I'm going to check your work by integrating it too, but I'm' going to do it my way. Here's my solution:

$$ \begin{align} \int xe^{-3x} dx &=-\frac{1}{3}\int x\left(e^{-3x}\right)' dx\\ &=-\frac{1}{3}\left(xe^{-3x}-\int e^{-3x} x'dx\right)\\ &=-\frac{1}{3}\left(xe^{-3x}-\int e^{-3x} dx\right)\\ &=-\frac{1}{3}\left(xe^{-3x}-\left(-\frac{1}{3}\right)\cdot\int\left(e^{-3x}\right)'dx\right)\\ &=-\frac{1}{3}\left(xe^{-3x}+\frac{1}{3}e^{-3x}\right)\\ &=-\frac{1}{3e^{3x}}\left(x+\frac{1}{3}\right)\\ &=-\frac{1}{3e^{3x}}\cdot\frac{1}{3}\cdot\left(3x+1\right)\\ &=-\frac{3x+1}{9e^{3x}}+C. \end{align} $$

WolframAlpha's solution is identical to mine: link. Your answer is not correct. And please remember that there always has to be a constant of integration, $C$, at the end of your answer.

  1. Yes, it's a constant, but I don't know what you mean by "What is this notation?".

3. $$ \begin{align} \int \ln \left(\sqrt{x}\right) dx &=\int \ln\left(\sqrt{x}\right) x'dx\\ &=x\ln\left(\sqrt{x}\right)-\int x \left[\ln\left(\sqrt{x}\right)\right]' dx\\ &=x\ln\left(\sqrt{x}\right)-\int x \frac{1}{\sqrt{x}}\frac{1}{2\sqrt{x}} dx\\ &=x\ln\left(\sqrt{x}\right)-\frac{1}{2}\int dx\\ &=x\ln\left(\sqrt{x}\right)-\frac{x}{2}\\ &=x\left(\ln\left(\sqrt{x}\right)-\frac{1}{2}\right)+C. \end{align} $$

To do integration by parts, I just directly used this formula which is equivalent to the other formula you're using (the one with $v$'s and $u$'s):

$$\int f(x) g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx$$

  • $\begingroup$ Regarding number 1... in wolfram's example doesn't the -1/9 distribute to both parts inside the parenthesis? Isn't it identical to mine then? $\endgroup$
    – Jwan622
    Jan 30 '19 at 13:55
  • $\begingroup$ In that respect, yes, it is identical. But you have a problem with the minus sign. $\endgroup$ Jan 30 '19 at 13:58
  • $\begingroup$ Isn't it supposed to be (-1/9) in the answer? $\endgroup$
    – Jwan622
    Jan 30 '19 at 14:04
  • $\begingroup$ Yes, see my answer. I edited it a little bit. There is a $-1/9$ in front of the rest of the answer. Can you see it? $\endgroup$ Jan 30 '19 at 14:06
  • $\begingroup$ Nope? What do you mean? $\endgroup$
    – Jwan622
    Jan 30 '19 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.