# The integral of x... Simple right?

Today, my friend asked me when to use trig sub in integration. Sarcastically, I said, "always" to which he replied, "Really? What about when you integrate x." So I thought for a second and this is what resulted.

$$\displaystyle\int x~dx$$

$$x=\sin(u)$$

$$dx=\cos(u)~du$$

$$\displaystyle\int \cos(u)\sin(u)~du=\frac{1}{2}\displaystyle\int \sin(2u)~du$$

$$w=2u$$

$$\frac{1}{4}\displaystyle\int \sin(w) ~dw=-\frac{1}{4} \cos(2u)=-\frac{1}{4}(1-2\sin^2(u))$$

$$-\frac{1}{4}(1-2x^2)=-\frac{1}{4}+\frac{x^2}{2}+c$$

Which, because of the constant of integration, is equivalent to $$\frac{x^2}{2}+c$$

Moral of the story: always use trig sub.

Note by Trevor Arashiro
3 years, 1 month ago

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But if you put sin u =x then the x can only take values from -1 to 1

- 3 years, 1 month ago

Integration of 1 will have more profound effect.

- 3 years, 1 month ago

Would it change how you integrated it at all?

- 3 years, 1 month ago

$$\int \sin^2 x + \cos^2x \, dx$$.

Staff - 3 years, 1 month ago

Hey Trevor! How's it goin'? Are you swamped with homework already? And yeah, trig sub is my fave too :)

- 3 years, 1 month ago

Hey Brian :) Glad to "see" you again. Yes, I'm overflowed with work from school. The main reason I haven't been posting much recently is because I haven't been able to think of many good problems. I'm still solving problems occasionally but there just haven't been many inspirations.

I have around 10 problems just sitting on my phone that I don't wanna post simply because they're not good enough. I guess it's because my problem posting standards have gotten much higher as last year I would have posted them without hesitation. I'll be posting a somewhat useful integration identity soon once I finish all my work that I have to make up from being sick.

- 3 years, 1 month ago

Sorry to hear that you've been sick. :( Catching up can be tough; it always seems to take twice as long to cover the same amount of material. As for posting problems, I basically took the summer off, (although I kept posting solutions now and then), and only posted my first question in two months a few days ago. I'm having issues with inspiration and standards too, but I'm hoping that just getting a few more posted will "prime the pump" and get me back into the flow of things. For you, school takes precedence, as it should, and I assume that any spare time you have for Brilliant gets devoted to your responsibilities as a moderator and collaborator. Anyway, to cheer you up I thought I'd create a problem in your honor. Enjoy. :)

- 3 years, 1 month ago

Thanks for the problem. Was fun to solve! Btw, good news and bad news.

Bad news: the note on the integration thingy I was gonna post has been delayed.

Good news: while I was posting it, I found a more general form of the integral and another case that works so I'll have to delay posting the note until I fully figure out this new case.

- 3 years, 1 month ago

I'm glad you enjoyed "Trevorsaurus Rex". It took me quite a while to LaTeX that one. :P

I'm continuing to look forward to your next posting; I'm sure that it will be worth the wait. Hope that you've fully recovered from your illness and have caught up with all your coursework. :)

P.S.. How about that Jason Day?!?! He's in another league right now. Fun to watch, and he seems to be such a great guy and role model. I think we may be entering another 'golden age' of men's professional golf with all these young, exceptional players.

- 3 years, 1 month ago

Haha, I just looked at the latex. I can't imagine doing that.

The integral note I'm going to post has to do with integrals of the form $$\dfrac{1}{x^a \prod (x^b+k)^c}$$ and a trick that you can use to get rid of 1, or sometimes (but rarely) more terms in the denominator. You can always get rid of the $$x^a$$ assuming the the powers in the denominator fit a certain case. If the powers don't fit a the case, then the integral will have logs in its final form. However, I had to delay it because I found another case and I'm working with when a or c are negative (the negative case of a and c will probably be posted in a separate note cuz it changes A LOT).

The 59 was amazing! I unfortunately wasn't watching it but I saw the highlights video (it was practically all 59 shots). Golf is heading towards another surge in popularity, maybe not quite a golden age but with all the young phenoms coming in who knows?

- 3 years, 1 month ago

This works for integrals like $$\displaystyle\int \sin(x) \, dx$$ as well.

- 3 years, 1 month ago

You wrote -1/4(cos2u)=1/4(1-2sin^2u) It should be -1/4(cos2u)=-1/4(1-2sin^2u)

- 3 years, 1 month ago

Can help on this question ? : https://brilliant.org/discussions/thread/extremely-weird-integration/?ref_id=946409

- 3 years, 1 month ago

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