Integral of sin2xe^cos2x dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 2 x$.

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

      $$\int \frac{e^{\cos{\left(u \right)}} \sin{\left(u \right)}}{4}\, du$$

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

         $$\int \frac{e^{\cos{\left(u \right)}} \sin{\left(u \right)}}{2}\, du = \frac{\int e^{\cos{\left(u \right)}} \sin{\left(u \right)}\, du}{2}$$

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

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

            $$\int 1\, du$$

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

               $$\int \left(-1\right)\, du = - \int 1\, du$$

               1. The integral of a constant is the constant times the variable of integration:

                  $$\int 1\, du = u$$

               So, the result is: $- u$

            Now substitute $u$ back in:

            $$- e^{\cos{\left(u \right)}}$$

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

      Now substitute $u$ back in:

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

   Method #2

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

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

      1. Rewrite the integrand:

         $$e^{\cos{\left(2 x \right)}} \sin{\left(x \right)} \cos{\left(x \right)} = e^{2 \cos^{2}{\left(x \right)} - 1} \sin{\left(x \right)} \cos{\left(x \right)}$$

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

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

         $$\int \frac{u e^{2 u^{2}}}{e^{2}}\, du$$

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

            $$\int \left(- \frac{u e^{2 u^{2}}}{e}\right)\, du = - \frac{\int u e^{2 u^{2}}\, du}{e}$$

            1. Let $u = e^{2 u^{2}}$.

               Then let $du = 4 u e^{2 u^{2}} du$ and substitute $\frac{du}{4}$:

               $$\int \frac{1}{16}\, du$$

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

                  $$\int \frac{1}{4}\, du = \frac{\int 1\, du}{4}$$

                  1. The integral of a constant is the constant times the variable of integration:

                     $$\int 1\, du = u$$

                  So, the result is: $\frac{u}{4}$

               Now substitute $u$ back in:

               $$\frac{e^{2 u^{2}}}{4}$$

            So, the result is: $- \frac{e^{2 u^{2}}}{4 e}$

         Now substitute $u$ back in:

         $$- \frac{e^{2 \cos^{2}{\left(x \right)}}}{4 e}$$

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

2. Add the constant of integration:

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

The answer is:

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