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.

## Comments

Sort by:

TopNewestBut if you put sin u =x then the x can only take values from -1 to 1 – Usama Khidir · 1 year, 1 month ago

Log in to reply

Integration of 1 will have more profound effect. – Pi Han Goh · 1 year, 1 month ago

Log in to reply

– Trevor Arashiro · 1 year, 1 month ago

Would it change how you integrated it at all?Log in to reply

– Calvin Lin Staff · 1 year, 1 month ago

\( \int \sin^2 x + \cos^2x \, dx \).Log in to reply

Hey Trevor! How's it goin'? Are you swamped with homework already? And yeah, trig sub is my fave too :) – Brian Charlesworth · 1 year, 1 month ago

Log in to reply

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. – Trevor Arashiro · 1 year, 1 month ago

Log in to reply

problem in your honor. Enjoy. :) – Brian Charlesworth · 1 year, 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 aLog in to reply

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. – Trevor Arashiro · 1 year, 1 month ago

Log in to reply

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. – Brian Charlesworth · 1 year, 1 month ago

Log in to reply

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? – Trevor Arashiro · 1 year, 1 month ago

Log in to reply

This works for integrals like \(\displaystyle\int \sin(x) \, dx\) as well. – Pi Han Goh · 1 year, 1 month ago

Log in to reply

You wrote -1/4(cos2u)=1/4(1-2sin^2u) It should be -1/4(cos2u)=-1/4(1-2sin^2u) – Sanchit Nayyar · 1 year, 1 month ago

Log in to reply

Can help on this question ? : https://brilliant.org/discussions/thread/extremely-weird-integration/?ref_id=946409 – 柯 南 · 1 year, 1 month ago

Log in to reply