ANTIDERIVATIVES (INDEFINITE INTEGRALS)
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Evaluate J (3 sin x + 5 cos AC) dx.
J (3 sin x + 5 cos x) dx = 3(-cos x) + 5 sin x + C = -3 cos x + 5 sin x + C.
Find J (7 sec
2 x — sec * tan x) dx
J (7 sec
2 x — sec x tan x) dx = 1 tan x — sec x + C.
Evaluate / (esc
2 x + 3x
2 ) dx.
J (esc
2 x + 3x
2
) dx = -cot x + x
s + C.
Find
Find
Find / tan
2 x dx.
J tan
2 x dx — J (sec
2 x - 1) dx = tan x - x + C.
Evaluate
Use substitution. Let u = 7.v + 4. Then du=l dx, and
Find
Let M=X — 1. Then du = dx, and
Find J(3x-5)
12 dx.
Let
« = 3x-5,
dw = 3dx.
Then
Evaluate J sin (3x — 1) dx.
Let « = 3x-l, du=3dx. Then
/ (3x - 5)
12 dx = J M
I2 G) du = 1 J u
12 dw = GXi)"'
3 + C =
£(3x-5)
13 + C.
J sin (3x - 1) dx = J sin u (3 du) = 5 / sin u du = 3 (-cos w) + C =
- 3 cos (3x - 1) + C.
Find / sec2 (x/2) dx.
Let
u = x/2,
du = j dx.
Then
/sec
2 (x/2) dx = J" sec
2 « • 2 2 w du = 2 tan w + C =
2 tan (x/2) + C.
Find
Let
Then
Evaluate / (4 -2t
i )
! tdt.
Let u = 4 - 2t\ du = -4t dt. Then
J xVJxdx.