hellppp calculus!!!!!!!!?

Question

find the indefinite integral of the trig function::
1. integral cos t/ (1+sin t) dt

evaluate each indefinite integral::
1.integral t^3sqrt (t^4 + 2) dt
2.integral cos(6x) dx

Answer 1

(1)
Integral {cos t/ (1+sin t) dt}

It's substitution:
u = 1 + sin t
du = cos t dt ---> dt = du / cos t

Integral {(cos t/ u) (du / cos t) }
=Integral { (cos t/cos t ) (1/u) (du) }
=Integral { (1/u) (du) }
= ln u + C
= ln (1 + sin t) + C


(2)
Integral {(t^3)*sqrt(t^4 + 2) dt}

u = t^4 + 2
du = 4t^3 dt ---> dt =du / 4t^3

Integral {(t^3)*sqrt(u) (du / 4t^3)}
=Integral {(t^3 / 4t^3)*sqrt(u) (du)}
=Integral {(1/4)*sqrt(u) (du)}
= (1/4)*(2/3)*(u)^(3/2) + C
= (1/6)*( t^4 + 2)^(3/2) + C

(3)
Integral {cos(6x) dx}

u = 6x
du = 6 dx ---> dx = du / 6

Integral {cos(u) (du/6)}
=Integral {(1/6) cos(u) (du)}
= (1/6) sin(u) + C
= (1/6) sin(6x + C

Answer 2

The cos(6x) dx is pretty straightforward, Let u = 6x and du = 6 dx. So the function becomes
1/6 cos (u) du which integrates to -1/6 sin(6x)+ c1

The other one is easier than it looks. Again, make a substitution u= t^4+2 an du = 4t^3 dt
Then we have 1/4 sqrt(u) du
when integrated 1/4 (2/3) u^ 3/2 + c1
organizing and substituting
1/6 (t^4+2)^(3/2) + c1

The first will probably work out the same way since d(sin t)= cos t, so you have a du/(1-u) form

Answer 3

1. let u = 1+ sin t , then du = cos t dt

this becomes ⌡du/u = ln |u| + c = ln|1 + sin t| + c

2. Let u = t^4 + 2, then du = 4t^3 dt or 1/4 du = t^3 dt

in terms of u, you now have:

⌡(u)^(1/2) * 1/4 du
1/4 ⌡(u)^(1/2) du
= 1/4 * (2/3) u^(3/2) + C = 1/6 (t^4 + 2)^(3/2) + C

3. Let u = 6x, then du = 6 dx or 1/6 du = dx

in terms of u, you have:

⌡cos u (1/6 du)
= 1/6 int cos u du
= -1/6 sin u + C = -1/6 sin 6x + c