Integration Round Two?

Question

Evaluate the integral of (cos square root of x) / (square root of x) dx

Answer 1

Let u = sqrt(x). Then du = 0.5 * dx/(sqrtx). We have dx/sqrt(x), which is 2 * (0.5 dx/sqrtx).

The integral above is equal to the integral of:
2 * cosu du

Integrate that, and we have:
2sinu + C.

Don't forget to replace (sqrt x) back in!
2 sin(sqrt x) + C.

Answer 2

S(cos square root of x) / (square root of x) dx

let u = sqrt(x)
du = 2 sqrt(x) dx giving you

S(2 sqrt(x) cos(u) du)/u

remember that u = sqrt(x) and replace bottom u with sqrt(x)
S(2 sqrt(x) cos(u) du)/sqrt(x)

S2 cos(u) du

integrate
2 sin(u) + C
2 sin(sqrt(x)) + C

Answer 3

2*sin(sqrt(x))