Question

Can somebody please demonstrate what the substitution trick is for this?
\[\huge\int\limits \cos(\sqrt{x}) dx\]
(+1 medal for quality :-) )

Answer 1

by parts I think :)

Answer 2

let \(u^2=x\)\[2u\cdot\text du=\text dx\]

\[\Rightarrow\int\limits \cos u \cdot 2\text du\]

Answer 3

***\[\int\limits \cos(u)\cdot 2u \text du\]

Answer 4

\[u=\sqrt x\]\[du=\frac12x^{-1/2}dx=\frac{dx}{2u}\implies dx=2udu\]gives\[\int2u\cos udu\]or am I wrong?

Answer 5

guess not :)

Answer 6

the errors were all mine TT

Answer 7

:P
it happens

Answer 8

\(\int u\text dv=[uv]-\int v\text du\)

Answer 9

^_^ (comes in to look at what's been going on and figure out)

\[2 \int u\cos udu\]
And obviously then u=u so du=du, so dv=cos(u) so v=sin(u)du

I should be able to take it away from here thank you both, not sure whom to give the medal to though :D

Answer 10

du=1

Answer 11

du = 1 du, so sure 1. :-)

Answer 12

Kind of odd how the dx's or du's appear or disappear

Answer 13

yeah, that's the right setup
to be a little picky, I should point out that dv=cos(u)du
du has to go somehwehere when we integrate ;)

@UnkleRhaukus no, I think agentx is right
\[u=u\implies du=du\]

Answer 14

The "with respect to" part is always kind of quirky to me

Answer 15

yes there is an invisible 1 in front, but who cares?

Answer 16

yeah but it's not a full derivative; just a differential

Answer 17

so there is no "with respect to" anything

Answer 18

i got confused , with the u's ., errors

Answer 19

yeah, since we used a u-sub, using the variable u again to mean something else in the by parts formula is confusing
we maybe should have started with a \(\theta\) substitution to avoid confusion :P

Answer 20

...or some other letter

Answer 21

That was a really clever trick back here when you had to find the derivative of \(\sqrt{x}\) and you get 2du = \(\sqrt{x}\) dx and then subsitute \(\sqrt{x}\) back in for u

Answer 22

Thank you both for your time, I'm sure I'll be back. Getting these tips & tricks down is exactly what I was needing. Does no good to just know the answer

Answer 23

a lot of integral that people think are unsolvable, or require a trig sub can be solved with a diligent u-sub as I did above
do not underestimate the power of the u-sub if you follow it all the way through

Answer 24

Got to know HOW or I'll be toast on the exam ^_^ (and not the good tasty toast, the burned smoldering toast)

Answer 25

lol

Answer 26

Turing I may have another along that line here in a moment, in fact I may have had one earlier that was solved via a different method. I'll PM you the link

Answer 27

Err... I can't PM you @TuringTest lol

Answer 28

okay :) I can try
I will fan you first so you can message me

Answer 29

now you can do it I think