Intergral problem?

Question

how to do Intergral of cos(sqrare root x)?

I need help..

Answer 1

∫ cos (√x) dx
Let u=√x, du=1/(2√x) dx, dx=2√x du=2u du
∫2u cos u du
Integrate by parts:
2u sin u - ∫2 sin u du
2u sin u + 2 cos u + C
2√x sin (√x) + 2 cos (√x) + C

Edit: fixed constant.

Answer 2

Let sqrt x = y, so x = y^2, so dx = 2y dy
So now you need to integrate cos y. 2y dy.
At this point, you need to use the product rule for differentiation, and rearrange it a bit.
Integral f'g dy = fg - integral fg'dy,
and in this case, f' = cos y, g = 2y, so the problem now becomes
integral 2y.cos y = 2y.sin y - integral 2cos y dy
= 2 y.sin y - 2 sin y + c
Since y = sqrt x, this now gives you the answer
= 2 sqrt x sin sqrt x - 2 sin sqrt x + c
(Check the answer by differentiating this.)