Question

the integral of tan(3x) dx with upper limit pie/3 and lower limit pie/4

Answer 1

Put, u = 3x and du = 3dx .The integrand simplifies to
\[1/3\int\limits \tan(u) du\]
= -1/3 log(cos(u))+constant
= -1/3 log(cos(3 x))+constant(Substituting back)

Answer 2

Now you can just plug in the limits.

Answer 3

With limits it will become \[-\ln (2)/6 ~~0.115525\]

Answer 4

−ln(2)/6= -0.115525