Integral of (ch3x-4tg(5x-1)) dx

1. Integrate term-by-term:

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

      $$\int \left(- 4 \tan{\left(5 x - 1 \right)}\right)\, dx = - \int 4 \tan{\left(5 x - 1 \right)}\, dx$$

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

         $$\int 4 \tan{\left(5 x - 1 \right)}\, dx = 4 \int \tan{\left(5 x - 1 \right)}\, dx$$

         1. Rewrite the integrand:

            $$\tan{\left(5 x - 1 \right)} = \frac{\sin{\left(5 x - 1 \right)}}{\cos{\left(5 x - 1 \right)}}$$

         2. There are multiple ways to do this integral.

            Method #1

            1. Let $u = \cos{\left(5 x - 1 \right)}$.

               Then let $du = - 5 \sin{\left(5 x - 1 \right)} dx$ and substitute $- \frac{du}{5}$:

               $$\int \frac{1}{25 u}\, du$$

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

                  $$\int \left(- \frac{1}{5 u}\right)\, du = - \frac{\int \frac{1}{u}\, du}{5}$$

                  1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

                  So, the result is: $- \frac{\log{\left(u \right)}}{5}$

               Now substitute $u$ back in:

               $$- \frac{\log{\left(\cos{\left(5 x - 1 \right)} \right)}}{5}$$

            Method #2

            1. Let $u = 5 x - 1$.

               Then let $du = 5 dx$ and substitute $\frac{du}{5}$:

               $$\int \frac{\sin{\left(u \right)}}{25 \cos{\left(u \right)}}\, du$$

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

                  $$\int \frac{\sin{\left(u \right)}}{5 \cos{\left(u \right)}}\, du = \frac{\int \frac{\sin{\left(u \right)}}{\cos{\left(u \right)}}\, du}{5}$$

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

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

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

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

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

                        1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

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

                     Now substitute $u$ back in:

                     $$- \log{\left(\cos{\left(u \right)} \right)}$$

                  So, the result is: $- \frac{\log{\left(\cos{\left(u \right)} \right)}}{5}$

               Now substitute $u$ back in:

               $$- \frac{\log{\left(\cos{\left(5 x - 1 \right)} \right)}}{5}$$

         So, the result is: $- \frac{4 \log{\left(\cos{\left(5 x - 1 \right)} \right)}}{5}$

      So, the result is: $\frac{4 \log{\left(\cos{\left(5 x - 1 \right)} \right)}}{5}$

   1. Let $u = 3 x$.

      Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

      $$\int \frac{\cosh{\left(u \right)}}{9}\, du$$

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

         $$\int \frac{\cosh{\left(u \right)}}{3}\, du = \frac{\int \cosh{\left(u \right)}\, du}{3}$$

         HyperbolicRule(func='cosh', arg=_u, context=cosh(_u), symbol=_u)

         So, the result is: $\frac{\sinh{\left(u \right)}}{3}$

      Now substitute $u$ back in:

      $$\frac{\sinh{\left(3 x \right)}}{3}$$

   The result is: $\frac{4 \log{\left(\cos{\left(5 x - 1 \right)} \right)}}{5} + \frac{\sinh{\left(3 x \right)}}{3}$

2. Now simplify:

   $$\frac{4 \log{\left(\cos{\left(5 x - 1 \right)} \right)}}{5} + \frac{\sinh{\left(3 x \right)}}{3}$$

3. Add the constant of integration:

   $$\frac{4 \log{\left(\cos{\left(5 x - 1 \right)} \right)}}{5} + \frac{\sinh{\left(3 x \right)}}{3}+ \mathrm{constant}$$

The answer is:

$$\frac{4 \log{\left(\cos{\left(5 x - 1 \right)} \right)}}{5} + \frac{\sinh{\left(3 x \right)}}{3}+ \mathrm{constant}$$