If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Integration by parts challenge

Problem 1

exsinxdx=
Choose 1 answer:

Problem 2

(lnx)2dx=
Choose 1 answer:

Problem 3

x2sin(πx)dx=
Choose 1 answer:

Want to join the conversation?

  • duskpin seedling style avatar for user chang.brian710
    what do I do after I've finished all 3 questions? There's no Awesome Show Points button?
    (34 votes)
    Default Khan Academy avatar avatar for user
  • mr pants teal style avatar for user Grace.Ohwilleke
    In question 2 is it possible to rewrite the equation to lnx * lnx instead of (lnx)^2 and integrate by parts?

    The problem I'm having when I try to use that method is that after I integrate lnx for the first time and substitute it back into the equation I get:

    ∫lnx * lnx dx = x(lnx)^2 - x - ∫(xlnx-x)/x

    I'm not sure how to do the new integral (∫(xlnx-x)/x).
    (10 votes)
    Default Khan Academy avatar avatar for user
    • aqualine ultimate style avatar for user Emmet Costen
      I solved it this way. Simplify that integral by using algebra to:

      (∫(xlnx-x)/x) = ∫(lnx-1)dx

      That should make it easier to work with. I then took that and for work purposes transformed it to:

      ∫(lnx-1)dx = ∫(lnx-1)*1dx

      Once there you should be able to integrate that using another integration by parts.
      (10 votes)
  • blobby green style avatar for user lovingdata
    In problem 2, can't we approach it by taking the integral of (lnx) * (lnx). So u = lnx and dv = lnx? However, when I use this approach I seem to get the wrong answer.
    .............
    In short, for integral( lnx * lnx ), I get lnx * (x lnx +x) - integral( lnx + 1 ), which eventually evaluates to x(lnx)^2 - 2x, but it's the wrong answer
    (5 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Sachin Shukla
    In Question 1 and 3, why are they fiddling around with the order of v and u in the integration by parts equation?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • starky ultimate style avatar for user Aarav Shah
      Sal derived the integration by parts formula as the following: ∫(f(x)g'(x))dx = f(x)g(x) - ∫(f'(x)g(x))dx
      To simplify this formula, we can do a double substitution as such:
      ∫(f(x)g'(x))dx = f(x)g(x) - ∫(f'(x)g(x))dx
      u = f(x) v = g(x)
      du = f'(x)dx dv = g'(x)dx
      Rewriting the formula:
      ∫u dv = uv - ∫v du
      Hope that answered your question!
      (4 votes)
  • leafers seedling style avatar for user colinhill
    Is it normal to think integration is significantly more difficult than differentiation? I swear it got much harder. It's doable, but there's no clear "algorithm" like there usually is for math problems.
    (3 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user Tanner P
      Yes, it is definitely safe to say integration is more difficult. In fact, some functions are easy to differentiate but impossible to integrate. For example, you can find the derivative of e^(-x^2) using the chain rule, but good luck finding the antiderivative!
      (4 votes)
  • leafers seed style avatar for user devore.wes
    For problem 2 and 3, we never talked about compound functions. I am confused. I don't think we've learned how to do these yet.
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user sawyerj
    For problem 3, why can't you use u-sub for x cos(πx) after doing integration by parts once?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • leafers seedling style avatar for user colinhill
    In problem 3, can we use u-substitution after our first integration by parts? I don't need anyone to necessarily look at the work, but I was just wondering if it was a possibility.

    - \int_a^b x^{2}sin( \pi x)dx = (- x^{2} cos( \pi x))/( \pi ) - \int_a^b (-2xcos( \pi x ))/\pi

    \int_a^b (-2xcos( \pi x ))/\pi = -2/ \pi^{2} \int_a^b \pi cos( \pi x)dx

    u = \pi x, du = \pi dx

    -2/ \pi^{2} \int_a^b \pi cos( \pi x)dx = -2/ \pi^{2} \int_a^b cos(u) du = -2/ \pi^{2} (sin(u)) = -2/ \pi^{2} (sin( \pi x))

    So,

    \int_a^b x^{2}sin( \pi x)dx = (- x^{2} cos( \pi x))/( \pi ) + (2sin \pi x) / \ \pi^{2}
    (1 vote)
    Default Khan Academy avatar avatar for user
    • male robot donald style avatar for user Venkata
      You're second step seems a bit off. How did you go from

      $\frac{-2}{\pi}\int\limits_a^b xcos(\pi x) dx$

      to

      $\frac{-2}{\pi^{2}}\int\limits_a^b \pi cos(\pi x) dx$ ?

      Anyway, starting from your previous step of $\frac{-2}{\pi}\int\limits_a^b xcos(\pi x) dx$, you could do a u-sub of u = $\pi x$. This gives you du = $\pi dx$ and $x = \frac{u}{\pi}$. Now, if you substitute these into $\frac{-2}{\pi}\int\limits_a^b xcos(\pi x) dx$, you get $\frac{-2}{\pi}\int\limits_a^b \frac{u}{\pi}cos(u) dx$. See that here, you'll still need to do by parts. No getting around it lol! I also see that you missed a "u" from your calculations. So, small error there.

      Also, just a suggestion. As you are using LaTeX, you can use the \frac{}{} command to enter fractions, and it will render the numerator and denominator one above the other, as opposed to beside each other.

      And a small question for you: how do you render this code without the dollar signs lol!? Couldn't find any software for that.
      (2 votes)
  • blobby green style avatar for user Miray Atar
    I didn't get the solution to the second problem. How can we think of dv as a seperate function and where is the dx part of the antiderivative? If dv is the dx part then doesn't u supposed to be in terms of v? I couldn't get the intuition behind it and it lookks like there would be many algebraic restraints to have such solution. Can you explain it to me please?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Hexuan Sun 9th grade
    On problem 2, if u = (lnx)^2, why is du = 2(lnx)/x dx
    shouldn't it be only 2(lnx)/x
    (1 vote)
    Default Khan Academy avatar avatar for user
    • hopper cool style avatar for user obiwan kenobi
      du/dx is equal to 2(ln x)/x. However, if you want to solve for just du, then you have to multiply both sides by dx:

      du/dx = 2(ln x)/x
      du = 2(ln x)/x dx

      All we did was that we treated du/dx like it was a ratio and then we multiplied the dx to the other side.

      Hope this helps!
      (1 vote)