Question

How to integrate `int(sqrt(1-cosx)/2) dx` by using u-substitution?

Answer 1

`pm sqrt(2)cos(x/2)+C`

Explanation:

We use that

`cos(x)=cos^2(x/2)-sin^2(x/2)=1-2sin^2(x/2)`
so our integral will be
`pm1/2sqrt(2)intsin(x/2)dx=`

`pmsqrt(2)cos(x/2)+C`

Answer 2

Although Sonnhard has already given a quick identity application that solves the integral, the question asks for a specific method of solution.

However, a "u-substitution" is not a standard reference. There is a common and useful substitution that solves this and many other rational functions of trig functions that is often known as "t-substitution". Is this what was meant? It is useful and instructive and will show the way to solve much more difficult integration problems of this type, so I'll work this example through.

The "t-substitution" is `t=tan(x/2)`. Wikipedia has a derivation of its forms: https://en.wikipedia.org/wiki/Tangent_half-angle_substitution
Note that the line element `dx=2/(1+t^2)dt`

Here we use `cosx=(1-t^2)/(1+t^2)` (see link for the derivation of this). So

`1/2intsqrt(1-cosx)dx=1/2intsqrt(1-(1-t^2)/(1+t^2))*2/(1+t^2)dt=`
`intsqrt(((1+t^2)-(1-t^2))/(1+t^2))*1/(1+t^2)dt=`
`intsqrt((2t^2)/(1+t^2))*1/(1+t^2)dt=`
`sqrt(2)intt/(1+t^2)^(3/2)dt=sqrt(2)intt(1+t^2)^(-3/2)dt`

Notice that we have a function now that has a multiple of the derivative of the function inside the bracket outside the bracket - this is what we'd get from a chain rule differentiation. So we can integrate this directly:

`sqrt(2)intt(1+t^2)^(-3/2)dt=`
`-sqrt(2)(1+t^2)^(-1/2)+C=-sqrt(2/(1+t^2))+C`

Substitute back to our original variable `x`:

`-sqrt(2/(1+t^2))+C=`
`-sqrt(2/(1+tan^2(x/2)))+C`

Recall the Pythagorean identity `1+tan^2x=sec^2x`.

`-sqrt(2/(1+tan^2(x/2)))+C=`
`-sqrt(2)/sec(x/2)+C=`
`-sqrt(2)cos(x/2)+C`, which is the same solution as Sonnhard obtained. It's taken us much longer to get there, but the solution is a much more powerful approach, one worth knowing.

Recall the half-angle identity `cos^2(x/2)=(1+cosx)/2`. We can use this to get back nearer to the original style of the integrand expression.

`-sqrt(2)cos(x/2)+C=`
`-sqrt(2)sqrt((1+cosx)/2)+C=`
`-sqrt(1+cosx)+C`

This is rather unexpectedly beautiful, no? The integral of `sqrt(1-cosx)` is a multiple of `sqrt(1+cosx)`. That doesn't seem at all intuitive to me, but it checks out. Further, the integral of `sqrt(1+cosx)` sends us back to `sqrt(1-cosx)`. Unexpected patterns!