INTEGRATION OF TRIGONOMETRIC FUNCTIONS WITH INDEFINITE INTEGRALS

Integrate the following :

(1)  cos²2 x- sin 6 x

(2)  1/(1+sin x)

(3)  1/(1-cos x)

(4)  √(1 - sin2x)

(5)  cos x/cos (x - a)

(6)  √tan x/sin x cos x

(7)  sin √x/√x

(8)  sin 2x/(a cos²x + b sin²x)

(9)  sin²3 x + 4 cos 4 x

Problem 1 :

cos²2 x- sin 6 x

Solution :

cos²x = (1 + cos 2x)/2

=  ∫ (cos²x- sin 6 x ] dx

=  ∫cos²x dx  - ∫[ sin 6 x ] dx

=  ∫ [(1 + cos 2x)/2] dx  - ∫[sin 6x] dx

=  (1/2) [x + (sin 2x/2)] - [-cos 6x/6] + C

=  (1/2) [x + (sin 2x/2)] + (cos 6 x/6) + C

Problem 2 :

1/(1+sin x)

Solution :

Multiplying both numerator and denominator by the conjugate of denominator.

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

=  ∫[(1-sin x)/(1+sin x)(1-sin x)] dx

=  ∫[(1-sin x)/(1²-sin²x)] dx

=  ∫[(1-sin x)/cos²x] dx

=  ∫ [(1/cos²x) - (sin x/cos²x)] dx

=  ∫[sec²x - (sin x/cosx) (1/cos x)] dx

=  ∫[sec²x - tan x sec x] dx

=  ∫sec²x dx - ∫ tan x sec x dx

=  tan x - sec x + C

Problem 3 :

 1/(1-cos x)

Solution :

=  ∫[1/(1-cos x)] ⋅ (1+cos x)/(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/sin²x) + (cos x/sin²x)] dx

=  ∫[cosec²x + (cos x/sin x) (1/sin x)] dx

=  ∫[cosec²x + cot x cosec x] dx

=  ∫cosec²x dx + ∫ cot x cosec x dx

=  -cot x -cosec x + C

Problem 4 :

√(1 - sin2x)

Solution :

sin 2x  =  2 sin x cos x

=  ∫√(1 - sin2x) dx

=  ∫√(cos²x + sin²x - 2sin x cos x) dx

=  ∫√[(cos x - sin x)²] dx

=  ∫ (cos x - sin x) dx

=  ∫cos x dx - ∫ sin x dx

=  sin x - (-cos x) + C

=  sin x + cos x + C

Problem 5 :

cos x/cos (x - a)

Solution :

=  ∫cos x/cos (x - a) dx

=  ∫cos (x+a-a)/cos (x-a) dx

By comparing the numerator with the formula

cos (A- B) = cos A cos B - sin A sin B

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

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

=  ∫cos a dx - ∫tan (x + a) sin a dx

=  x cos a - sin a log cos (x-a) + C

Problem 6 :

√tan x/sin x cos x

Solution:

=  ∫√tan x/sin x cos x dx

=  ∫[√tan x/sin x cos x] [√tan x/√tan x] dx

=  ∫[tan x/√tan x sin x cos x] dx

=  ∫[(sin x/cos x)/√tan x sin x cos x] dx

=  ∫[(sec²x)/√tan x]  dx

let t  =  tan x

differentiating with respect to x.

dt  =  sec²x dx

=  ∫[(sec² x)/√tan x]  dx

=  ∫[1/√t]  dt

=  ∫ t^(-1/2)  dt

=  2 √t + C

=  2√tan x + C

Problem 7 :

sin √x/√x

Solution :

=  ∫sin√x/√x dx

let t  =  √x

differentiating with respect to x

dt  =  (1/2√x) dx

2 dt  =  (1/√x) dx

=  ∫sin√x/√x dx

=  ∫sin t (2 dt)

=  2∫sin t dt

=  2 (-cos t) + C

=  -2cos t + C

now we are going to apply the value of t

=  -2cos √x + C

Problem 8 :

sin 2x/(a cos²x + b sin²x)

Solution :

∫sin 2x/(a cos²x + b sin²x) dx

t  =  (a cos²x + b sin²x)

dt  =  2 a cos x (-sin x) + 2 b sin x cos x

dt  =  -2 a sin x cos x + 2 b sin x cos x

dt  =  - a sin 2x + b sin 2x

dt  =  sin 2x(b - a) dx

dt/(b-a)  =  sin 2x dx

∫sin 2x/(a cos²x + b sin²x) dx  =  1/(b-a) ∫1/t dt

=  1/(b-a) ln t + C

=  1/(b-a) ln (a cos²x + b sin²x) + C

Problem 9 :

sin²3 x + 4 cos 4 x

Solution :

∫ (sin²3 x + 4 cos 4x) dx

sin²x  =  (1-cos 2x)/2

sin² 3x  =  [1 - cos 2(3x)]/2

=  (1 - cos 6x)/2

=  ∫[sin²3x + 4 cos 4x] dx

=  ∫[(1- cos 6x)/2 + 4 cos 4 x] dx

=  ∫[(1- cos 6x)/2] dx + ∫ 4 cos 4 x dx

=  (1/2)∫[(1- cos 6x)] dx + 4 ∫ cos 4 x dx

=  (1/2)[x - (sin 6x/6)] + 4 (sin 4 x/4) + C

=  (1/2)[x - (sin 6x/6)] + sin 4 x + C

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