Integral Problem: Evaluating (4sin(x)+3tan(x))dx

[antiflag403]

Hey everyone,
I can't seem to figure out how to do this integral. If anyone could help that would be great. I would appreciate some hints/advice, rather than just the answer. Thanks!
The question is:
Evaluate the indefinite integral:
(4sin(x)+3tan(x))dx
Ok... So i know you can take the integral of each term seperatly because it is a sum and that is a principle of integrals... so i get the first part to be -4cos(x). However I am not sure how to evalue the antiderivative of 3tan(x). If anyone could help that would be great!
Thanks again.

 

[StatusX]

Noting that tan x = sin x/cos x, you can integrate by subsitution using u=cos x.

 

[antiflag403]

ok... so i tried that.
so with the substitution you end up with du/u whose antiderivative is -ln(absolute value)U. U is cosx so it is -ln(abs)cosx. Moving the negative sign into the logarithm gives ln(abs)secx. So my final answer would be...
4cosx+3ln(abs)secx. I am not to sure this is right though. Could one of you smart math people check and if i made an error somewhere point me in the right direction.
Thanks a lot for the help!

 

[TD]

Watch your signs, the antiderivative of sin(x) is -cos(x).

 

[antiflag403]

Oh yes...
So is the final answer:
-4cosx+3ln(abs)secx?
Thanks for pointing that out. (i think i fixed it)

Another quick question that I can't seem to figure out...
Find the definite integral from 0 to 1 of:
x^2((5x+6)^(1/2))
I thought this may be another substitution question but I can't see what to substitute. Any help would be great!

 

[TD]

I was referring to the term 4cos(x) which should actually be -4cos(x)

 

[Tx]

For the second, let u= 5x + 6 from there it is easy to see that x^2 = (u-6)^2/25, then expand and integrate.

 

[antiflag403]

I don't really understand that Tx. If i let U=5xt6 i end up with x^2(U)^(1/2)(du/5) which doesn't seem to help me in any way. Some more advice would be good! I appreciate the help!