INTEGRATION PROBLEMS IN THE FORM OF AX PLUS B

∫(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)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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