Question

How to integrate cos(t^2) from 0 to 1

Answer 1

First set up the integral with bounds from 0 to 1 with infinitesimal dt. Then just take the derivative in reverse (what gives you cos(t^2) when you take the derivative of it?)

Answer 2

Hey Pawanyadav :)
This does not have a closed form solution.
Are you sure you posted the question correctly?

Or are you using approximation methods maybe?
Or power series or something? :d

Answer 3

Oh ew you're right, this problem has cooties

Answer 4

lol :3

Answer 5

That is
Cosx/√x dx
So by putting. x=t^2
I get this

Answer 6

@zepdrix,@nuttyliaczar

Answer 7

Oh ok that makes more sense :)
I think maybe you missed something in your substitution.

Answer 8

\[\large\rm \int\limits\frac{\cos(x)}{\sqrt x}dx=\int\limits \cos(x)\left(\frac{1}{\sqrt x}dx\right)\]

Answer 9

\[\large\rm x=t^2\qquad\to\qquad \sqrt{x}=t\]Taking derivative,\[\large\rm dx=2t~dt\qquad\to\qquad dx=2\sqrt x~dt\]

Answer 10

So to be clear the function is cos(x/sqrt(x))? With x=t^2

Answer 11

So then you have:\[\large\rm \left(\frac{1}{\sqrt x}dx\right)=2 dt\]Oh no, I guess you did that correctly :)\[\large\rm \int\limits\limits \cos(x)\left(\frac{1}{\sqrt x}dx\right)=\int\limits \cos(t^2)\left(2dt\right)\]Mmm that's no fun >.<

Answer 12

I take 2 outside the integral

Answer 13

Hmm yah I dunno how to deal with this one using normal methods :(
Here is what wolfram provides
https://www.wolframalpha.com/input/?i=integral+of+cos%28x%29%2Fsqrtx
you have to use some other weird trick I think.

Answer 14

Am not getting anything