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## Integral Calculus

### Course: Integral Calculus>Unit 1

Lesson 17: 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). • 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? • 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. • 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) • For problem 3, why can't you use u-sub for x cos(πx) after doing integration by parts once?
(1 vote) • 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.  