Integral of lntg(2x)+cosln(x) dx

1. Integrate term-by-term:

   1. Use integration by parts:

      $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

      Let $u{\left(x \right)} = \log{\left(\tan{\left(2 x \right)} \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$.

      Then $\operatorname{du}{\left(x \right)} = \frac{2 \tan^{2}{\left(2 x \right)} + 2}{\tan{\left(2 x \right)}}$.

      To find $v{\left(x \right)}$:

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

         $$\int 1\, dx = x$$

      Now evaluate the sub-integral.

   2. There are multiple ways to do this integral.

      Method #1

      1. Rewrite the integrand:

         $$\frac{x \left(2 \tan^{2}{\left(2 x \right)} + 2\right)}{\tan{\left(2 x \right)}} = \frac{2 x \tan^{2}{\left(2 x \right)} + 2 x}{\tan{\left(2 x \right)}}$$

      2. Rewrite the integrand:

         $$\frac{2 x \tan^{2}{\left(2 x \right)} + 2 x}{\tan{\left(2 x \right)}} = 2 x \tan{\left(2 x \right)} + \frac{2 x}{\tan{\left(2 x \right)}}$$

      3. Integrate term-by-term:

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

            $$\int 2 x \tan{\left(2 x \right)}\, dx = 2 \int x \tan{\left(2 x \right)}\, dx$$

            1. Don't know the steps in finding this integral.

               But the integral is

               $$\int x \tan{\left(2 x \right)}\, dx$$

            So, the result is: $2 \int x \tan{\left(2 x \right)}\, dx$

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

            $$\int \frac{2 x}{\tan{\left(2 x \right)}}\, dx = 2 \int \frac{x}{\tan{\left(2 x \right)}}\, dx$$

            1. Don't know the steps in finding this integral.

               But the integral is

               $$\int \frac{x}{\tan{\left(2 x \right)}}\, dx$$

            So, the result is: $2 \int \frac{x}{\tan{\left(2 x \right)}}\, dx$

         The result is: $2 \int \frac{x}{\tan{\left(2 x \right)}}\, dx + 2 \int x \tan{\left(2 x \right)}\, dx$

      Method #2

      1. Rewrite the integrand:

         $$\frac{x \left(2 \tan^{2}{\left(2 x \right)} + 2\right)}{\tan{\left(2 x \right)}} = 2 x \tan{\left(2 x \right)} + \frac{2 x}{\tan{\left(2 x \right)}}$$

      2. Integrate term-by-term:

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

            $$\int 2 x \tan{\left(2 x \right)}\, dx = 2 \int x \tan{\left(2 x \right)}\, dx$$

            1. Don't know the steps in finding this integral.

               But the integral is

               $$\int x \tan{\left(2 x \right)}\, dx$$

            So, the result is: $2 \int x \tan{\left(2 x \right)}\, dx$

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

            $$\int \frac{2 x}{\tan{\left(2 x \right)}}\, dx = 2 \int \frac{x}{\tan{\left(2 x \right)}}\, dx$$

            1. Don't know the steps in finding this integral.

               But the integral is

               $$\int \frac{x}{\tan{\left(2 x \right)}}\, dx$$

            So, the result is: $2 \int \frac{x}{\tan{\left(2 x \right)}}\, dx$

         The result is: $2 \int \frac{x}{\tan{\left(2 x \right)}}\, dx + 2 \int x \tan{\left(2 x \right)}\, dx$

   1. Let $u = \log{\left(x \right)}$.

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

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

      1. Use integration by parts, noting that the integrand eventually repeats itself.

         1. For the integrand $e^{u} \cos{\left(u \right)}$:

            Let $u{\left(u \right)} = \cos{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

            Then $\int e^{u} \cos{\left(u \right)}\, du = e^{u} \cos{\left(u \right)} - \int \left(- e^{u} \sin{\left(u \right)}\right)\, du$.

         2. For the integrand $- e^{u} \sin{\left(u \right)}$:

            Let $u{\left(u \right)} = - \sin{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

            Then $\int e^{u} \cos{\left(u \right)}\, du = e^{u} \sin{\left(u \right)} + e^{u} \cos{\left(u \right)} + \int \left(- e^{u} \cos{\left(u \right)}\right)\, du$.

         3. Notice that the integrand has repeated itself, so move it to one side:

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

            Therefore,

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

      Now substitute $u$ back in:

      $$\frac{x \sin{\left(\log{\left(x \right)} \right)}}{2} + \frac{x \cos{\left(\log{\left(x \right)} \right)}}{2}$$

   The result is: $x \log{\left(\tan{\left(2 x \right)} \right)} + \frac{x \sin{\left(\log{\left(x \right)} \right)}}{2} + \frac{x \cos{\left(\log{\left(x \right)} \right)}}{2} - 2 \int \frac{x}{\tan{\left(2 x \right)}}\, dx - 2 \int x \tan{\left(2 x \right)}\, dx$

2. Now simplify:

   $$x \log{\left(\tan{\left(2 x \right)} \right)} + \frac{\sqrt{2} x \sin{\left(\log{\left(x \right)} + \frac{\pi}{4} \right)}}{2} - 2 \int \frac{x}{\tan{\left(2 x \right)}}\, dx - 2 \int x \tan{\left(2 x \right)}\, dx$$

3. Add the constant of integration:

   $$x \log{\left(\tan{\left(2 x \right)} \right)} + \frac{\sqrt{2} x \sin{\left(\log{\left(x \right)} + \frac{\pi}{4} \right)}}{2} - 2 \int \frac{x}{\tan{\left(2 x \right)}}\, dx - 2 \int x \tan{\left(2 x \right)}\, dx+ \mathrm{constant}$$

The answer is:

$$x \log{\left(\tan{\left(2 x \right)} \right)} + \frac{\sqrt{2} x \sin{\left(\log{\left(x \right)} + \frac{\pi}{4} \right)}}{2} - 2 \int \frac{x}{\tan{\left(2 x \right)}}\, dx - 2 \int x \tan{\left(2 x \right)}\, dx+ \mathrm{constant}$$