Here you see formulas used in integrating trigonometric functions.
∫sin x dx = -cos x + c
∫cos x dx = sin x + c
∫sec x tan x dx = sec x + c
∫sec2 x dx = tan x + c
∫cosec2 x dx = -cot x + c
∫cosec x cot x dx = -cosec x + c
Question 1 :
∫(20ex+3x2-2cos x) dx
Solution :
= ∫(20ex+3x²-2cos x) dx
= 20∫ex dx + 3 ∫x2dx - 2∫cos x dx
= 20 ex + 3 x3/3 - 2 sin x + C
= 20 ex + x3 - 2 sin x + C
So, the answer is 20 ex + 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.
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