Antiderivatives (Indefinite Integrals) solved questions

CHAPTER 19
Antiderivatives (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




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, 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