Question

Evaluate this equation

Answer 1

Given the integral

`\displaystyle \int \tan(3x) \, dx`

first substitute y = 3x and dy = 3 dx :

`\displaystyle \int \tan(3x) \, dx = \frac13 \int \tan(y) \, dy`

then rewrite tan(y) = sin(y)/cos(y) and substitute z = cos(y) and dz = -sin(y) dy :

`\displaystyle \frac13 \int (\sin(y))/(\cos(y)) \, dy = -\frac13 \int \frac{dz}z`

Recall that 1/z is the derivative of ln|z|. Then back-substitute to get the result in terms of x :

`\displaystyle -\frac13 \int \frac{dz}z = -\frac13 \ln|z| + C`

`\displaystyle \frac13 \int \tan(y) \, dy = -\frac13 \ln|\cos(y)| + C`

`\displaystyle \boxed + C`