Integration (Practice Questions)

Integration is a mathematical process used to find the antiderivative of a function or to calculate the area under a curve. It is essentially the reverse process of differentiation.

Solved Examples

Example 1: Evaluate

• ∫ x⁶ dx  
• ∫1/x⁴ dx  
• ∫³√x dx  
• ∫3^x dx  
• ∫4e^x dx  
• ∫(sin x/cos²x) dx  
• ∫(1/sin²x) dx 
• ∫[1/√(4 - x²)] dx  
• ∫[1/3√(x² - 9)] dx  
• ∫(1 /cos x tan x) dx 

Solution:

 (i)∫x⁶ dx 

= (x⁶⁺¹)/(6 + 1) + C       [∫xⁿ dx = {xⁿ⁺¹/(n+1)} + C    n ≠ -1]

= (x⁷/7) + C 

(ii) ∫1/x⁴ dx 

= ∫x^-4 dx                       [∫xⁿ dx = {xⁿ⁺¹/(n+1)} + C    n ≠ -1]

= (x^-4+1)/(-4 + 1) + C 

= (x^-3/ -3) + C

= -(1/3x³) + C

(iii)  ∫³√x dx 

= ∫x^1/3 dx                     [∫xⁿ dx = {xⁿ⁺¹/(n+1)}+ C    n ≠ -1]

= (x ^(1/3)+1/((1/3)+ 1) + C 

= x^4/3 / (4/3) + C 

=  (3/4)(x^4/3) + C

(iv) ∫3^x dx 

= (3^x / log_e 3) + C        [ ∫a^x dx = (a^x / log_ea) + C]                                            

(v) ∫4e^x dx  

= 4∫e^x dx                     [∫k . f(x) dx = k f(x) dx , where k is constant]

= 4 e^x + C                   [∫e^x dx = e^x + C]

(vi) ∫(sin x/cos²x) dx 

= ∫[(sin x/cos x) .(1/cos x)] dx

= ∫tan x . sec x dx        [ ∫tan x .sec x dx = sec x + C ]       

= sec x + C

(vii) ∫(1/sin²x) dx 

= ∫cosec²x dx               [∫cosec²x dx  = -cot x + C ]

= -cot x  + C

(viii) ∫[1/√(4 - x²)] dx 

= ∫[1/√(2² - x²)] dx      [we know that, ∫[1/√(a² - x²)] dx = sin^-1(x/a) + C]

= sin^-1(x/2) + C

(ix) ∫[1/{3√(x²  - 9)}] dx 

= ∫[1/{3√(x² - 3²)}] dx   [we know that, \int\frac{1}{x\sqrt{x^2-a^2}}dx = (1/a)sec^-1(x/a) + C]

= (1/3)sec^-1(x/3) + C

(x) ∫(1 /cos x tan x) dx 

= ∫[cos x /(cos x sin x)] dx

= ∫(1/ sin x) dx

= ∫cosec x dx                [we know that, ∫cosec x dx =  log |cosec x - cot x| + C]

= log |cosec x - cot x| + C

Example 2: Evaluate ∫{e^9log_e^x + e^8log_e^x}/{e^6log_e^x + e^5log_e^x} dx

Solution:

 Since, e^alog_e^x = x^a

∫{e^9log_e^x + e^8log_e^x}/{e^6log_e^x + e^5log_e^x} dx 
= ∫{x⁹ + x⁸}/{x⁶ + x⁵} dx
= ∫[x⁸(x + 1)]/[x⁵(x + 1)] dx
=∫ x⁸/x⁵ dx
= ∫x³ dx                    [we know that, ∫xⁿ dx = {xⁿ⁺¹/(n+1)} + C    n ≠ -1]
= (x⁴/4) + C 

Example 3: Evaluate ∫ sin x + cos x  dx 

Solution:

  ∫(sin x + cos x) dx 
= ∫sin x dx + ∫cos x dx             [we know that, ∫{f(x) ± g(x)} dx = ∫f(x) dx ± ∫g(x) dx]         
= -cos x + sin x + C                 [we know that, ∫sin x dx = -cos x + C, ∫cos x dx = sin x + C ]

Example 4: Evaluate ∫4^x+2 dx

Solution:

 ∫4^x+2 dx = ∫4^x. 4² dx 
= ∫16. 4^x dx              [ we known that ∫k.f(x) dx = k∫f(x) dx , where k is constant]
= 16∫ 4^x dx               [∫a^x dx = (a^x / log_ea) + C]
= 16 (4^x/log 4) + C

Example 5: Evaluate ∫(x² + 3x + 1) dx

Solution:

 ∫(x² + 3x + 1) dx 
= ∫x² dx+ 3∫x dx + 1∫ x⁰dx         [We know that, ∫xⁿ dx = {xⁿ⁺¹/(n+1)}+ C  n ≠ -1]
= [x²⁺¹/2+1] + 3[[x¹⁺¹/1+1]] + [x⁰⁺¹/0+1] + C
= [x³/3] + 3[x²/2] + x + C

Example 6: Evaluate ∫[4/(1 + cos 2x)] dx 

Solution:

1 + cos 2x = 2cos²x 

∫[4/(1 + cos 2x)] dx 
= ∫[4/(2cos²x)] dx
= ∫(2/cos²x) dx
= ∫2 sec²xdx
= 2∫sec²x dx     [We know that, ∫sec²x dx = tan x + C ]
= 2 tan x + C 

Example 7: Evaluate ∫(3cos x - 4sin x + 5 sec²x) dx

Solution:

∫(3cos x - 4sin x + 5 sec²x) dx 
= ∫3cos x dx - ∫4sin x dx + ∫5sec²x dx   [∫k.f(x) dx = k ∫f(x) dx, where k is constant]
=  3∫cos x dx - 4∫sin x dx + 5∫sec²x dx
= 3sin x - 4(-cos x) + 5 tan x + C
=  3sin x + 4cos x + 5 tan x + C 

Practice Problems

Question 1: ∫ x² dx.

Question 2: ∫ e^x dx.

Question 3: ∫ 1/x dx.

Question 4: ∫ sin (x) dx.

Question 5: ∫ (2x³ + 3x² + x + 1)dx.

Answer Sheet

1. x³/3 + C
2. e^x + C
3. ln |x| + C
4. -cos (x) + C
5. \frac{x(x+2)(x^2+1)}{2}