Question 1 :
Integrate the following with respect to x
∫ [(x + 4)5 + 5/(2 - 5x)4 - cosec2(3x - 1)] dx
Solution :
∫ [(x + 4)5 + 5/(2 - 5x)4 - cosec2(3x - 1)] dx
= ∫ (x + 4)5 dx + ∫(5/(2 - 5x)4) dx - ∫cosec2(3x - 1) dx
= (x + 4)6/6 + 5∫(2 - 5x)-4 dx - ∫cosec2(3x - 1) dx
= (x + 4)6/6 + 5(2-5x)-3/(-3)(-5) dx - cot(3x-1)/(-3) + c
= (1/6)(x + 4)6 - (1/3)(1/(2-5x)3) + (1/3) cot(3x-1) + c
Question 2 :
Integrate the following with respect to x
∫ [4 cos (5-2x) + 9e3x-6 + 24/(6-4x)] dx
Solution :
∫ [4 cos (5-2x) + 9e3x-6 + 24/(6-4x)] dx
= ∫ 4 cos (5-2x) dx + 9 ∫ e3x-6dx + ∫24/(6-4x) dx
= 4 sin (5 - 2x)/(-2) + 9 e3x-6/3 + 24 log (6 - 4x)(-1/4) + c
= -2 sin (5 - 2x) + 3 e3x-6 - 6 log (6 - 4x) + c
Question 3 :
Integrate the following with respect to x
∫ [sec2(x/5) + 18 cos 2x + 10 sec (5x + 3) tan (5x + 3)] dx
Solution :
∫ [sec2(x/5) + 18 cos 2x + 10 sec (5x + 3) tan (5x + 3)] dx
= ∫sec2(x/5)dx+18∫cos 2xdx+10∫sec(5x+3)tan(5x+3)] dx
= tan (x/5)/(1/5)+(18/2)sin 2x+(10/5) sec (5x+3) + c
= 5 tan (x/5) + 9 sin 2x + 2 sec (5x+3) + c
Question 4 :
Integrate the following with respect to x
∫ [8/√(1 - (4x)2) + 27/√(1 - 9x2) - 15/(1+25x2)] dx
Solution :
= ∫8/√(1-(4x)2) dx+∫27/√(1-9x2) dx-∫15/(1+25x2) dx
= 8∫1/√(1-(4x)2)dx+27∫1/√(1-(3x)2) dx-15∫1/(1+(5x)2) dx
= (8/4)sin-1(4x) + (27/3) sin-1(3x) - (15/5) tan-1(5x) + c
= 2 sin-1(4x) + 9 sin-1(3x) - 3 tan-1(5x) + c
Question 5 :
Integrate the following with respect to x
∫ [6/(1 + (3x + 2)2)] - [12/√1 - (3-4x)2] dx
Solution :
= ∫ [6/(1 + (3x + 2)2)] - [12/√1 - (3-4x)2] dx
= 6∫[1/(1 + (3x + 2)2)] - 12 ∫[1 √1 - (3-4x)2] dx
= (6/3) tan-1(3x+2) - (12/(-4)) sin-1(3-4x) + c
= 2 tan-1(3x+2) + 3 sin-1(3-4x) + c
Question 6 :
Integrate the following with respect to x
∫ (1/3) cos ((x/3) - 4)) + 7/(7x+9) + e(x/5) + 3 dx
Solution :
= ∫ (1/3) cos ((x/3) - 4)) dx + ∫7/(7x+9) dx + ∫e(x/5)+3 dx
= (1/3)/(1/3)sin((x/3)-4))+(7/7)log(7x+9) + (1/5)e(x/5) + 3+c
= (1/3)/(1/3)sin((x/3)-4))+log(7x+9) + 5e(x/5) + 3+c
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