Question

Please help me with trig substitution in integration?
The question is attached below :)

Answer 1

use the given hint :
\(\large u = \sin^2 x\)

Answer 2

I know we have to use the hint and i've attempted to try and solve the problem but i'm stuck .
Using the substitution method is new to me so yeah.. Not very strong at it.

Answer 3

Here, I'll show you how to work these kinds of things out with a similar example, \[\Large \int\limits xe^{x^2} dx\] So first thing you do when you see this is what? Freak out uncontrollably and run away. No. Just try stuff, play around with it, it's just harmless lines on paper, really.

I'll give you a hint, x^2=u just like they did in your other problem. So now the task in a substitution is that you can take the derivative of this, and use it to solve for "dx", here like this: \[\Large u = x^2 \\ \Large \frac{du}{dx}=2x \\ \Large \frac{1}{2x} du = dx\] Treating these like normal algebra is kind of a trick, and won't work for higher dimensional integrals, but don't worry about that right now, I just want to let you understand it's a trick.

Now we plug in everywhere to the integral what we can "dx" and "x" to get them all in terms of u: \[\Large \int\limits x e^{x^2}dx = \int\limits x e^{(u)} (\frac{1}{2x} du)\] I put the stuff I substituted in in parentheses. Notice however we still have x's in there! Luckily they will divide each other and go away this time, but if for some reason they didn't just take the original substitution and rearrange it: \[\Large u=x^2 \\ \Large \sqrt{u}=x\] Now you can plug it in wherever. So let's clean it up a little more: \[\Large \frac{1}{2} \int\limits e^u du = \frac{1}{2}e^u +C\] The problem was originally in x's and we made up u's so we gotta get rid of them, so just plug back in, and you're done! \[\Large \int\limits xe^{x^2}dx = \frac{1}{2}e^{x^2}+C\] Get in the habit of always checking yourself by doing a quick and simple chain rule to make sure that the derivative of our answer is the original integral!

Answer 4

Wow... @Kainui
Thank you SOOOOOOO incredibly much for taking the time to type all of that out for me and explaining it thoroughly!
I'll read it and apply the concept to my question and see if i can work it out.
Thanks a bunch :)

Answer 5

It looks like you might have to do some trickier stuff than what I said, like possibly use some trig identities to get your integral in terms of only u this time, hopefully that helps!

Answer 6

Start with\[u =1+\sin ^{2}x\]

Answer 7

The question only said u=sin^2 x :)

Answer 8

yeah you may use textbook hint for now
you will get clever ideas over time as you do more integrals :)

Answer 9

Yes you could do that but it's simpler if you start with u = 1 + sin^2 x.

Answer 10

True, but the problem specifically states to use that worse substitution. @calculusfunctions

Answer 11

I mean it's really only marginally worse, it's no really a big deal. You either deal with the 1 up front or later, it's not a big deal or anything haha.

Answer 12

Yes it does but as a teacher I know first hand that textbooks aren't perfect! Hence it could be a typo.

Answer 13

Anyway, either way it's really no big deal. But substitutions work.

Answer 14

hmm.. Okey dokey.
Thank you both for your input xD
Will take it all on board! @calculusfunctions @Kainui

Answer 15

Yeah, totally. There's a lot of good "integrators" on OS actually, they're kinda like fun puzzles after a while.

Answer 16

Oops! I meant both substitutions work.

Answer 17

alright so i attempted to work it out.
Feeling like i did something wrong xD
Not sure where to go from here.

Answer 18

@Kainui

Answer 19

It looks perfect. =D

Answer 20

but how do we integrate du/1+u ? :O

Answer 21

Well one way to think of it is what has the form: \[\Large \frac{f'(x)}{f(x)}\] You could also try doing another substitution like 1+u=v. I just don't want to give you the answer when you got so close on your own! =D

Answer 22

i completely forgot how to integrate logarithms so it took me a while xD
But i ended up with the answer being ln (7/6).
@Kainui . Thanks a million :) !