integration?

Question

evaluate

int{(3sinx+4cosx)
/(3+2cosx)}dx

hint:-

int{(3sinx+4cosx)
/(3+2cosx)}dx
=int((3sinx)/
(3+2cosx))dx+
int((4cosx)/
(3+2cosx))dx

please show working!

there has been a correct
answer submitted however,
the answerer never
included the constant of
integration!

Answer 1

becomes 3sinx/(3 +2cosx) + 4cosx/(3 + 2cosx)
= Int1 + Int 2

Int1 = -3/2 log(3 +2 cosx)

To obtain Int 2 have to divide 4cosx by 3 + 2 cosx
This comes to 2 - 6/(3 + 2cosx)

To find int of - 6/(3 + 2cosx) have to make the substitution tan(x/2) = t
Which leads to
Int 2/(5 + t²) =
2/root5(invtan(tan(x/2)/root5))

Almost impossible to type out all steps but can safely say that final answer is as given below:-
-3/2(log(3 + 2cosx) + 2x -
(12/root5)invtan(tan(1/2)x) /root5)

Hope you can follow this but there is rather a lot to show on this site.

Answer 2

Int (3+2cos x) dx = 3x+2sin x

Int( 3sinx +4cosx)dx= -3cosx +4sinx

Answer 3

/[3+2cosx]dx=/3dx+/2(cos x)dx
=3x+2(sin x)+c

/[3sinx+4cosx]dx=/3(sin x)dx+/4(cos x)dx
=3/(sin x)dx+4/(cos x) dx
=3(-cos x)+4(sin x)+c
here ''c'' is an arbitrary constant & ''/'' mark is used instead integration mark.

Answer 4

Integration by using areas. The function is a manufactured from 2 purposes: (x) * (cosx) call one in each and every of them "u" and the different "dv" decide for the "u" so as that its by-product simplifies u = x du = a million*dx dv = cosx v = sinx this is for the formulation setup, it relatively is u*v - indispensable (vdu) =xsinx - indispensable(sinx * a million * dx) =xsinx - (-cosx) + C = xsinx + cosx + C

Answer 5

I hope you were not expected to do this by hand.

2x - [(12 ArcTan[Tan[x/2]/SqareRoot(5)]
/SqareRoot(5)]-(3/2)Log[3 + 2 Cos(x)
+ Constant of integration

Answer 6

Yeah! It did my head in as well!