# calculus

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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 -

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 -

(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 -

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 -

Remarkable :) LOL

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