INTEGRATING TRIGONOMETRIC FUNCTIONS

Here you see formulas used in integrating trigonometric functions.

Question 1 :

∫(20e^x+3x²-2cos x) dx

Solution :

=  ∫(20e^x+3x²-2cos x) dx

=  20∫e^x dx + 3 ∫x²dx - 2∫cos x dx

=  20 e^x + 3 x³/3 - 2 sin x + C

=  20 e^x + x³ - 2 sin x + C

So, the answer is 20 e^x + x³ - 2 sin x + C.

Question 2 :

∫(1+sin x) dx

Solution :

=  ∫(1+sin x) dx

=  ∫1 dx + ∫sin x dx

=  x - cos x + C

So, the answer is x - cos x + C.

Question 3 :

∫ 1/(1 + sin x) dx

Solution :

∫ 1/(1 + sin x) dx

Multiply the given by the conjugate of the denominator.

=  ∫(1-sin x)/[(1-sin x) (1 + sin x)] dx

=  ∫(1-sin x)/[(1²-sin²x)] dx

=  ∫ (1-sin x)/cos² x dx

=  ∫ (1/cos² x - sin x/cos² x) dx

=  ∫sec² x - (sin x/cos x) ⋅ (1/cos x) dx

=  ∫(sec²x - tan x ⋅ sec x) dx

=  ∫sec²x dx - ∫tan x ⋅ sec x dx

=  tan x - sec x + C

So, the answer is tan x - sec x + C.

Question 4 :

∫ 1/(1-cos x) dx

Solution :

=  ∫1/(1-cos x) dx

=  ∫(1+cos x)/[(1-cos x) (1-cos x)] dx

=  ∫(1 - cos x)/(1²- cos²x) dx

=  ∫(1-cos x)/sin²x dx

=  ∫(1/sin² x) dx - ∫(cos x/sin² x) dx

=  ∫cosec² x dx - ∫(cos x/sin x) ⋅ (1/sin x) dx

=  ∫cosec² x dx - ∫cot x ⋅ cosec x dx

=  - cot x - (-cosec x) + C

=  - cot x + cosec x + C

So, the answer is - cot x + cosec x + C.

Question 5 :

(1-cos 2x)/(1+cos 2x)

Solution :

=  ∫(1-cos 2x)/(1+cos 2x) dx

= ∫(2sin²x)/(2 cos²x) dx

=  ∫ (sin²x/cos²x) dx

=  ∫ tan²x dx

=  ∫(sec²x-1) dx

=  tan x - x + C 

So, the answer is tan x - x + C .

Question 6 :

∫(2tan x-3 cot x)² dx

Solution :

=  ∫(2 tan x - 3 cot x)² dx

=  ∫ [ (2 tan x)² - 2 (2 tan x)(3 cot x) + (3 cot x)² ]dx

=  ∫ [ 4 tan² x - 12 tan x cot x + 9 cot² x ]dx

=  ∫ (4 tan² x) dx - ∫ 12 tan x cot x dx + ∫ 9 cot² x dx

=   4 ∫ (tan² x) dx - ∫ 12 dx + 9 ∫ cot² x dx

=  4∫ (sec²x - 1) dx - 12∫dx + 9 ∫ (cosec² x - 1) dx

=  4 [tan x  - x] - 12x + 9 [- cot x - x] + C

=  4 tan x-4x-12x-9 cot x-9 x + C

=  4tan x  - 9cot x - 25 x + C

So, the answer is 4tan x  - 9cot x - 25 x + C.

Question 7 :

∫ (1/cosecx) dx

Solution :

1/cosec x can be written as sin x.

∫(1/cosec x) dx  =  ∫sin dx

= -cos x + C

Question 8 :

∫(tan x/cos x) dx

Solution :

∫(tan x/cos x) dx  =  tan x (1/cos x)

=  tan x sec x

=  ∫tan x sec x dx

=  sec x + C

So, the answer is sec x + C.

Question 9 :

∫(cos x/sin²x) dx

Solution :

cos x/sin²x  =  cos x/(sin x sin x)

=  (cos x/sin x) (1/sin x)

=  cot x cosec x

=  ∫cot x cosec x dx

= -cosec x + C

So, the answer is -cosec x + C.

Question 10 :

∫ (1/cos²x) dx

Solution :

1/cos²x  =  sec²x

=  ∫sec²x dx

=  tan x + C

So, the answer is tan x + C.

Question 11 :

∫ tan x dx

Solution :

∫ tan x dx = ∫ (sin x/cos x) dx

Multiplying negative and negative, we get positive.

= - ∫ (-sin x/cos x) dx

While finding the derivative of denominator, we receive the numerator.

= - log cos x + C

= log (cos x)^-1 + C

= log (1/cos x) + C

= log sec x + C

Question 12 :

∫ sin ²x dx

Solution :

∫ sin ²x dx

sin ²x = 1 - cos 2x

∫ sin ²x dx = ∫ (1 - cos 2x) dx

= x - sin 2x/2 + C

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

Question 13 :

∫ √(1 - cos x) dx

Solution :

cos 2x = 1 - 2sin²x

2sin²x = 1 - cos 2x

Dividing angles by 2 on both sides, we get

2sin²(x/2) = 1 - cos (2x/2)

2sin²(x/2) = 1 - cos x

∫ √(1 - cos x) dx = ∫ √2sin²(x/2) dx

= √2 ∫ sin(x/2) dx

= √2 cos (x/2) (1/2) + C

= (√2/2) cos (x/2) + C

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