∫(ax+b)ⁿ dx = (1/a) [(ax+b)(n+1)/(n + 1)] + c
∫1/(ax+b) dx = (1/a) log (ax+b) + c
∫sin(ax+b) dx = -(1/a) cos (ax+b) + C
∫cos(ax+b) dx = (1/a) sin (ax+b) + C
∫sec2 (ax+b) dx = (1/a) tan (ax+b) + C
∫sec (ax+b) tan (ax+b) dx = (1/a) sec (ax+b) + C
∫cosec2 (ax+b) dx = -(1/a) cot (ax+b) + c
∫cosec (ax+b) cot (ax+b) dx = -(1/a) cosec (ax+b) + c
Question 1 :
(i) ∫(x+3)5 dx
(ii) ∫(3x+4)6 dx
(iii) ∫(4-3x)7 dx
(iv) ∫(Lx+m)8 dx
Solution :
(i)
∫(x+3)5 dx = (x+3)(5+1)/(5+1)
= (x+3)6/6 + C
= (1/6)(x+3)6 + C
(ii)
∫(3x+4)6 dx = (3x+4)(6+1)/(6+1)
= (1/3)[(3x+4)7/7] + C
= (1/21)(3x+4)7 + C
(iii)
∫(4-3x)7 dx = (4-3x)(7+1)/(7+1)
= (-1/3)[(4-3x)8/8] + C
= (-1/24)(4-3x)8 + C
Question 2 :
(i) ∫ 1/(x+5)4 dx
(ii) ∫1/(2x+3)5 dx
(iii) ∫ 1/(4-5x)7dx
(iv) ∫ 1/(ax+b)8 dx
Solution :
(i) ∫ 1/(x+5)4 dx
t = x+5 ==> dt = dx
∫1/(x+5)4 dx = ∫ (1/t4) dt
= ∫ t-4 dt
= t(-4+1)/(-4+1) + C
= t-3/(-3) + C
= -1/3t3 + C
(ii)
∫1/(2x+3)5 dx
t = 2x+3 ==> dt = 2dx
dx = dt/2
∫ 1/ (2x+3)5 dx = ∫ (1/t5) (dt/2)
= ∫ t-5 (dt/2)
= (1/2)t(-5+1)/(-5+1) + C
= (1/2)t-4/(-4) + C
= -1/8t4 + C
= -1/8 (2x+3)4 + C
(iii)
∫ 1/(4-5x)7dx
t = 4-5x ==> dt = - 5 dx ==> dx = -dt/5
∫ 1/ (4-5x)7 dx = ∫ (1/t7) (-dt/5)
= ∫ t-7 (-dt/5)
= (1/5)t(-7+1)/(-7+1) + C
= (1/5)t-6/(-6) + C
= -1/30 t6 + C
= -1/30 (4-5x)6 + C
(iv)
∫ 1/(ax+b)8 dx
t = ax+b
dt = a dx
dx = dt/a
∫ 1/ (ax+b)8 dx = ∫ (1/t8 ) (dt/a)
= ∫ t-8 (dt/a)
= (1/a)t(-8+1)/(-8 + 1) + C
= (1/a)t-7/(-7) + C
= 1/7a t7 + C
= 1/7a (ax+b)7 + C
Question 3 :
(i) ∫ sin (x+3) dx
(ii) ∫ sin (2x+4) dx
(iii) ∫ sin (3-4x) dx
(iv) ∫ cos (4x+5) dx
(v) ∫ cos (5-2x) dx
Solution :
(i) ∫ sin (x+3) dx
∫ sin (x + 3) dx = -cos (x+3) + C
(ii) ∫ sin (2x+4) dx
∫ sin (2x+4) dx = -(1/2) cos (2x+4) + C
= - cos (2x+4)/2 + C
(iii) ∫ sin (3-4x) dx
∫ sin (3-4x) dx = -(1/4) cos (3-4x) + C
= -cos (3-4x)/4 + C
(iv) ∫cos (4x+5) dx
∫ cos (4x+5) dx = (1/4) sin (4 x + 5) + C
= sin (4 x + 5)/4 + C
(v) ∫ cos (5-2x) dx
∫cos (5-2x) dx = (1/2) sin (5-2x) + C
= sin (5-2x)/2 + C
Question 4 :
(i) ∫ sec2 (2-x) dx
(ii) ∫ cosec2 (5+2x) dx
(iii) ∫ sec2 (3+4x) dx
(iv) ∫cosec2 (7-11x) dx
Solution :
(i) ∫ sec2 (2-x) dx
∫ sec2 (2-x) dx = (1/-1) tan (2-x) + C
= - tan (2-x) + C
(ii) ∫ cosec2 (5+2x) dx
∫ cosec2 (5+2x) dx = -(1/2) cot (5+2x) + C
= -cot (5+2x)/2 + C
(iii) ∫ sec2 (3+4x) dx
∫ sec2 (3+4x) dx = (1/4) tan (3+4x) + C
= tan (3+4x)/4 + C
(iv) ∫cosec2 (7-11x) dx
∫cosec2 (7-11x) dx = -(1/-11) cot (7-11x) + C
= cot (7-11x)/11 + C
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