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Antiderivatives (Indefinite Integrals) solved questions
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CHAPTER 19Antiderivatives (Indefinite Integrals)
19.1 Evaluate $ (g(x)Yg'(x) dx.
By the chain rule, ,x((g(X)Y+1) = (r + l)(g(X)Y-g'(X)Hence
where C is an arbitrary constant.
19.2 Evaluate J xr dx for r ^ — I.
sinceDz(x'+l) = (r+!)*'.
19.3 Find $(2x3-5x2 + 3x + l)dx.
19.4 Find
19.5 Find
19.( Evaluate
19.7 Find
19.8 Find
19.9 Find
19.10 Find
142
, ANTIDERIVATIVES (INDEFINITE INTEGRALS) 143
19.11 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.
19.12 Find J (7 sec2 x — sec * tan x) dx
J (7 sec2 x — sec x tan x) dx = 1 tan x — sec x + C.
19.13 Evaluate / (esc2 x + 3x2) dx.
J (esc2 x + 3x2) dx = -cot x + xs + C.
19.14 Find J xVJxdx.
19.15 Find
19.16 Find / tan 2 x dx.
J tan2 x dx — J (sec2 x - 1) dx = tan x - x + C.
19.17 Evaluate
Use substitution. Let u = 7.v + 4. Then du=l dx, and
19.18 Find
Let M = X — 1. Then du = dx, and
19.19 Find J(3x-5) 1 2 dx.
Let « = 3x-5, dw = 3dx. Then / (3x - 5)12 dx = J M I 2 G) du = 1 J u 12 dw = G X i ) " ' 3 + C =
13
£(3x-5) + C.
19.20 Evaluate J sin (3x — 1) dx.
Let « = 3x-l, du=3dx. Then J sin (3x - 1) dx = J sin u ( 3 du) = 5 / sin u du = 3 (-cos w) + C =
- 3 cos (3x - 1) + C.
19.21 Find / sec2 (x/2) dx.
Let u = x/2, du = j dx. Then /sec2 (x/2) dx = J" sec2 « • 2 </« = 2 J sec2 w du = 2 tan w + C =
2 tan (x/2) + C.
19.22 Find
Let Then
19.23 Evaluate / (4 -2ti)!tdt.
Let u = 4 - 2t\ du = -4t dt. Then
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