Integral of sin(2x)/(sqrt(1+cos(2x))) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 2 x$.

      Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

      $$\int \frac{\sin{\left(u \right)}}{2 \sqrt{\cos{\left(u \right)} + 1}}\, du$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \frac{\sin{\left(u \right)}}{\sqrt{\cos{\left(u \right)} + 1}}\, du = \frac{\int \frac{\sin{\left(u \right)}}{\sqrt{\cos{\left(u \right)} + 1}}\, du}{2}$$

         1. Let $u = \cos{\left(u \right)} + 1$.

            Then let $du = - \sin{\left(u \right)} du$ and substitute $- du$:

            $$\int \left(- \frac{1}{\sqrt{u}}\right)\, du$$

            1. The integral of a constant times a function is the constant times the integral of the function:

               $$\int \frac{1}{\sqrt{u}}\, du = - \int \frac{1}{\sqrt{u}}\, du$$

               1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

                  $$\int \frac{1}{\sqrt{u}}\, du = 2 \sqrt{u}$$

               So, the result is: $- 2 \sqrt{u}$

            Now substitute $u$ back in:

            $$- 2 \sqrt{\cos{\left(u \right)} + 1}$$

         So, the result is: $- \sqrt{\cos{\left(u \right)} + 1}$

      Now substitute $u$ back in:

      $$- \sqrt{\cos{\left(2 x \right)} + 1}$$

   Method #2

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \frac{2 \sin{\left(x \right)} \cos{\left(x \right)}}{\sqrt{\cos{\left(2 x \right)} + 1}}\, dx = 2 \int \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{\sqrt{\cos{\left(2 x \right)} + 1}}\, dx$$

      1. Rewrite the integrand:

         $$\frac{\sin{\left(x \right)} \cos{\left(x \right)}}{\sqrt{\cos{\left(2 x \right)} + 1}} = \frac{\sqrt{2} \sin{\left(x \right)} \cos{\left(x \right)}}{2 \sqrt{\cos^{2}{\left(x \right)}}}$$

      2. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \frac{\sqrt{2} \sin{\left(x \right)} \cos{\left(x \right)}}{2 \sqrt{\cos^{2}{\left(x \right)}}}\, dx = \frac{\sqrt{2} \int \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{\sqrt{\cos^{2}{\left(x \right)}}}\, dx}{2}$$

         1. Let $u = \cos^{2}{\left(x \right)}$.

            Then let $du = - 2 \sin{\left(x \right)} \cos{\left(x \right)} dx$ and substitute $- \frac{du}{2}$:

            $$\int \left(- \frac{1}{2 \sqrt{u}}\right)\, du$$

            1. The integral of a constant times a function is the constant times the integral of the function:

               $$\int \frac{1}{\sqrt{u}}\, du = - \frac{\int \frac{1}{\sqrt{u}}\, du}{2}$$

               1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

                  $$\int \frac{1}{\sqrt{u}}\, du = 2 \sqrt{u}$$

               So, the result is: $- \sqrt{u}$

            Now substitute $u$ back in:

            $$- \sqrt{\cos^{2}{\left(x \right)}}$$

         So, the result is: $- \frac{\sqrt{2} \sqrt{\cos^{2}{\left(x \right)}}}{2}$

      So, the result is: $- \sqrt{2} \sqrt{\cos^{2}{\left(x \right)}}$

2. Now simplify:

   $$- \sqrt{2} \sqrt{\cos^{2}{\left(x \right)}}$$

3. Add the constant of integration:

   $$- \sqrt{2} \sqrt{\cos^{2}{\left(x \right)}}+ \mathrm{constant}$$

The answer is:

$$- \sqrt{2} \sqrt{\cos^{2}{\left(x \right)}}+ \mathrm{constant}$$