Question

Please help me solve the following integral (click to see)

Answer 1

\[-\frac{ 1 }{ 3 }\int\limits_{}^{}\sin(3x)\tan(3x)dx\]

Answer 2

say u=3x. du=3dx, dx=du/3
-1/9 (sin(u)tan(u)du) = -1/9 (sin^2(u) / cos(u)) = -1/9 (1-cos^2 u / cosu) = -1/9 (secu -cosu) = -1/9 (ln|secu+tanu| - sinu) = -1/9 (ln|sec(3x)+tan3x| - sin3x) + C

Answer 3

the answer on wolframalpha is much more complicated than that

Answer 4

@P0sitron is correct , wolfram changes the ln(sec +tan| and puts it in a different form using sin and cos