What is the integral of cosine square root of x divided by the square root of x?

Answer 1

The indefinite integral of cos(sqrt(x))/sqrt(x) dx

This is an integral that can be solved by simple u-substitution. This integral in and of itself cannot be solved by any simple antiderivative pattern, but by using u-substitution we can get this integral to resemble something more familiar.

If we set a dummy variable "u" equal to sqrt(x), we could get the integral to look like cos(u). Although this substitution causes the sqrt(x) to disappear inside the cosine function, the variable of integration (dx) is still in terms of x. However, since:

u=sqrt(x)

du=d/dx of sqrt(x) [the derivative of sqrt(x)]

du=(1/2)x(1/2)-1 dx=(1/2)x-1/2 dx=1/(2sqrt(x)) dx

This implies that the 1/(2sqrt(x)) inside of the integral can be replaced by du. All that exists inside of the integral is 1/sqrt(x) though, but by dividing the 1/(sqrt(x)) inside of the integral by the needed 1/(2sqrt(x)), we get that multiplying the entire integral by a constant of 2 would let us solve this integral. So the integral above that was unsolvable in terms of x is now solvable in terms of u:

2 times the integral of cos(u) du

If any limits of integration existed, you could put them through the relationship u=sqrt(x), you can change the limits into terms of u as well. If, for instance, the integral was from x=3 to x=7, the integral in terms of u would be from u=sqrt(3) to u=sqrt(7).

So, the antiderivative of cos(u) would be sin(u). This is a basic antiderivative pattern, one that you should have memorized. So, since the integral in terms of u was multiplied by a constant of 2, the final answer should be as well. So, we get the antiderivative to be:

2sin(u)+C [the "+C" exists since this is an indefinite integral with no limits of integration]

By resubstituting u=sqrt(x) back in, we get the final antiderivative answer in terms of x to be:

2sin(sqrt(x))+C

BUT: realize that if there were limits of integration, you must substitute them into the integral in terms of whatever variable the antiderivative is in. If you were to use the above antiderivative in terms of x, you would have to use limits of integration that were in terms of x. Likewise, if you were to use the antiderivative while it was still in terms of u, you could solve for it at that point by simply using the limits in terms of u as we discussed earlier.