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### Course: AP®︎/College Calculus BC>Unit 6

Lesson 13: Using integration by parts

# Integration by parts challenge

## Problem 1

$\int {e}^{x}\mathrm{sin}x\phantom{\rule{0.167em}{0ex}}dx=$

## Problem 2

$\int \left(\mathrm{ln}x{\right)}^{2}\phantom{\rule{0.167em}{0ex}}dx\phantom{\rule{0.167em}{0ex}}=$

## Problem 3

$\int {x}^{2}\mathrm{sin}\left(\pi x\right)\phantom{\rule{0.167em}{0ex}}dx\phantom{\rule{0.167em}{0ex}}=$

## Want to join the conversation?

• what do I do after I've finished all 3 questions? There's no Awesome Show Points button?
• This are "documents" so these ones dont have that, they are completed the moment you click on them even if you dont solve the problems. So you just gotta be happy knowing that you gave the correct answer but thats it
• 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).
• 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.
• 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.
• 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!
• 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
• In Question 1 and 3, why are they fiddling around with the order of v and u in the integration by parts equation?
• 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
• 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)
• You should be able to solve these with just integration by parts. Is there anything in particular you got stuck on?
• For problem 3, why can't you use u-sub for x cos(πx) after doing integration by parts once?
(1 vote)
• Well, you're probably thinking of using u = πx. So, du = π and x = u/π. Substituting these, we get the new function as (u/π)cos(u). From here on, see that you'll still need to integrate by parts.
• 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)
• 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.
• 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)
• So there is nothing wrong with the solution it just a little trick.

You are just the formula for integration by parts which comes from product rule.
(1 vote)
• On problem 2, if u = (lnx)^2, why is du = 2(lnx)/x dx
shouldn't it be only 2(lnx)/x
(1 vote)
• 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)