substitution method?

[SilentSymphony]

i have two problems to do on substitution to find integrals. the only problem is, i dont have a clue as to how to go about them

(a) (integral) sin (6x-5)dx (b):: (integral) dx/(4+3x)

heres my work:

u = 6x-5 u= 4+3x
du = sin 6 dx du= dx
(1/u)du
sin (u) du ln |u| du ... ???

= ???

i have no clue, so again, any guidance would be great. i had to miss class this day, and im paying for it bigtime![/img]

 

[galactus]

\(\displaystyle \L\\\int{sin(6x-5)}dx\)

Let \(\displaystyle \L\\u=6x-5, \;\ du=6dx, \;\ \frac{1}{6}du=dx\)

Now, make the subs:

\(\displaystyle \L\\\frac{1}{6}\int{sin(u)}du\)

Now, integrate.

 

[SilentSymphony]

okay ill try that. thanks for all your help : )