Posted by **sarah** on Monday, January 21, 2008 at 7:49pm.

how to take the integral of cos(square root of x) dx.

need to use substitution and integration by parts but I don't know what to substitute

- calculus -
**Damon**, Monday, January 21, 2008 at 8:09pm
let t = x^.5

dt = .5 x^-.5 dx

so dx = 2 x^.5 dt which is dx = 2 t dt

so integral of 2 t cos t dt

that might help

- calculus -
**drwls**, Monday, January 21, 2008 at 8:13pm
(Integral of) cos x^(1/2) dx

Do the substitution part first.

Let x^(1/2) = y

x = y^2

dx = 2y du

So the integral becomes, after substitution

2*(Integral of)y cos y dy

Now use integration by parts. You may have to use it twice.

You should end up with

2 cos y + y sin y

Then substitute back x in terms of y

= 2 cos(sqrt x) + 2sqrt x * sin(sqrt x)

- calculus -
**Damon**, Monday, January 21, 2008 at 8:17pm
integral of 2 t cos t dt

let u = 2 t

so du = 2 dt

let dv = cos t dt

so v = sin t

u v = 2 t sin t

v du = 2 sin t dt

u v - integral v du

= 2 t sin t + 2 cos t

- calculus -
**Damon**, Monday, January 21, 2008 at 8:19pm
Remarkable :) LOL

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