So I was doing a integral question and I stumbled upon this question.
$\displaystyle{\int\sqrt{1 + \sin\left(\frac x2\right)}\,dx}$
In order to solve it I did the following:
I took $u = \frac12x$
Then $\frac {du}{dx}$
Which gave me $2 du = dx$.
After that I substituted u in the equation to get $2\int \sqrt{1 + \sin(u)} du$.
After this I was stuck as I am new to integration of trigonometry so I checked my textbook which did the same just the same but the step after this was this one
$2\int{\sqrt{\sin^2 \frac12u + \cos^2\frac12u + 2\sin \frac 12u\cos \frac12u}\text{ du}}$
I am in a complete awe how the textbook got $2\int{\sqrt{\sin^2 \frac12u + \cos^2\frac12u + 2\sin \frac 12u\cos \frac12u}\text{ du}}$ from $2\int \sqrt{1 + \sin(u)} \text{ du} $
Can someone please explain me how the this is achieved? I am totally stuck
Best Answer
First, they used the double angle formula:
$$\sin u = \sin\left(2\frac{u}{2}\right) = 2\sin\frac{u}{2}\cos\frac{u}{2}.$$
Then they replaced the $1$ with $\sin^2\frac{u}{2}+\cos^2\frac{u}{2}.$