Integral of cos(x)*cos(x)*ln(x) dx

1. Use integration by parts:

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

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

   Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$.

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

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

         1. Let $u = 2 x$.

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

            $$\int \frac{\cos{\left(u \right)}}{2}\, du$$

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

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

               1. The integral of cosine is sine:

                  $$\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$$

               So, the result is: $\frac{\sin{\left(u \right)}}{2}$

            Now substitute $u$ back in:

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

         So, the result is: $\frac{\sin{\left(2 x \right)}}{4}$

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

         $$\int \frac{1}{2}\, dx = \frac{x}{2}$$

      The result is: $\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}$

   Now evaluate the sub-integral.

2. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \frac{1}{x}$.

      Then let $du = - \frac{dx}{x^{2}}$ and substitute $- \frac{du}{4}$:

      $$\int \left(- \frac{u \sin{\left(\frac{2}{u} \right)} + 2}{4 u^{2}}\right)\, du$$

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

         $$\int \frac{u \sin{\left(\frac{2}{u} \right)} + 2}{u^{2}}\, du = - \frac{\int \frac{u \sin{\left(\frac{2}{u} \right)} + 2}{u^{2}}\, du}{4}$$

         1. Rewrite the integrand:

            $$\frac{u \sin{\left(\frac{2}{u} \right)} + 2}{u^{2}} = \frac{\sin{\left(\frac{2}{u} \right)}}{u} + \frac{2}{u^{2}}$$

         2. Integrate term-by-term:

            1. Let $u = \frac{1}{u}$.

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

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

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

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

                  SiRule(a=2, b=0, context=sin(2*_u)/_u, symbol=_u)

                  So, the result is: $- \operatorname{Si}{\left(2 u \right)}$

               Now substitute $u$ back in:

               $$- \operatorname{Si}{\left(\frac{2}{u} \right)}$$

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

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

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

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

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

            The result is: $- \operatorname{Si}{\left(\frac{2}{u} \right)} - \frac{2}{u}$

         So, the result is: $\frac{\operatorname{Si}{\left(\frac{2}{u} \right)}}{4} + \frac{1}{2 u}$

      Now substitute $u$ back in:

      $$\frac{x}{2} + \frac{\operatorname{Si}{\left(2 x \right)}}{4}$$

   Method #2

   1. Rewrite the integrand:

      $$\frac{\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}}{x} = \frac{2 x + \sin{\left(2 x \right)}}{4 x}$$

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

      $$\int \frac{2 x + \sin{\left(2 x \right)}}{4 x}\, dx = \frac{\int \frac{2 x + \sin{\left(2 x \right)}}{x}\, dx}{4}$$

      1. Let $u = \frac{1}{x}$.

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

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

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

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

            1. Rewrite the integrand:

               $$\frac{u \sin{\left(\frac{2}{u} \right)} + 2}{u^{2}} = \frac{\sin{\left(\frac{2}{u} \right)}}{u} + \frac{2}{u^{2}}$$

            2. Integrate term-by-term:

               1. Let $u = \frac{1}{u}$.

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

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

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

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

                     SiRule(a=2, b=0, context=sin(2*_u)/_u, symbol=_u)

                     So, the result is: $- \operatorname{Si}{\left(2 u \right)}$

                  Now substitute $u$ back in:

                  $$- \operatorname{Si}{\left(\frac{2}{u} \right)}$$

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

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

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

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

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

               The result is: $- \operatorname{Si}{\left(\frac{2}{u} \right)} - \frac{2}{u}$

            So, the result is: $\operatorname{Si}{\left(\frac{2}{u} \right)} + \frac{2}{u}$

         Now substitute $u$ back in:

         $$2 x + \operatorname{Si}{\left(2 x \right)}$$

      So, the result is: $\frac{x}{2} + \frac{\operatorname{Si}{\left(2 x \right)}}{4}$

   Method #3

   1. Rewrite the integrand:

      $$\frac{\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}}{x} = \frac{1}{2} + \frac{\sin{\left(2 x \right)}}{4 x}$$

   2. Integrate term-by-term:

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

         $$\int \frac{1}{2}\, dx = \frac{x}{2}$$

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

         $$\int \frac{\sin{\left(2 x \right)}}{4 x}\, dx = \frac{\int \frac{\sin{\left(2 x \right)}}{x}\, dx}{4}$$

         SiRule(a=2, b=0, context=sin(2*x)/x, symbol=x)

         So, the result is: $\frac{\operatorname{Si}{\left(2 x \right)}}{4}$

      The result is: $\frac{x}{2} + \frac{\operatorname{Si}{\left(2 x \right)}}{4}$

3. Now simplify:

   $$- \frac{x}{2} + \frac{\left(2 x + \sin{\left(2 x \right)}\right) \log{\left(x \right)}}{4} - \frac{\operatorname{Si}{\left(2 x \right)}}{4}$$

4. Add the constant of integration:

   $$- \frac{x}{2} + \frac{\left(2 x + \sin{\left(2 x \right)}\right) \log{\left(x \right)}}{4} - \frac{\operatorname{Si}{\left(2 x \right)}}{4}+ \mathrm{constant}$$

The answer is:

$$- \frac{x}{2} + \frac{\left(2 x + \sin{\left(2 x \right)}\right) \log{\left(x \right)}}{4} - \frac{\operatorname{Si}{\left(2 x \right)}}{4}+ \mathrm{constant}$$