INTEGRATION BY PARTS WITH TRIGONOMETRIC FUNCTIONS

Problem 1 :

x tan2 x

Solution :

∫ x tan2 x dx

sec2x - 1  =  tan2 x

∫ x tan 2x dx  =  ∫ x(sec2x - 1) dx

=  ∫ (x sec2x - x) dx

=  ∫ x sec2x dx - ∫ x dx

u  =  x          dv  =  sec² x

du  =  dx        v  =  tan x

=  x (tan x) - ∫ tan x dx - ∫ x dx

=  x (tan x) - ∫ (sin x/cos x) dx - ∫ x dx

=  x (tan x) - log (cos x) - (x2/2) + C

Problem 2 :

x cos2 x

Solution :

∫ x cos2 x dx

∫ x cos2 x dx  =  ∫ x (1+cos 2x)/2 dx

=  (1/2) ∫ x (1+cos 2x) dx 

=  (1/2) ∫ (x + x cos 2x) dx 

=  (1/2) [∫ x dx + ∫ x cos 2x dx]

=  (1/2) [(x2/2)+ ∫ x cos 2x dx]

Let I  =  ∫x cos 2x dx

u  =  x             dv  =  cos 2x

du  =  dx           v  =  sin 2x/2

=  x (sin 2x)/2 - ∫(sin 2x/2) dx

=  x (sin 2x)/2 - (1/2) ∫ sin 2x dx

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

=  (x/2) (sin 2x) + (1/4) (cos 2x) + C

=  (1/2) [(x²/2)+ (x/2) (sin 2 x) + (1/4) (cos 2x) + C

Problem 3 :

x cos 5 x cos 2 x

Solution :

∫ x cos 5x cos 2x dx

now we are going to apply the trigonometric formula 2 cos A cos B

∫ x cos 5 x cos 2 x dx  =  (2/2)∫ x cos 5 x cos 2 x dx

=  (1/2)∫ x 2 cos 5 x cos 2 x dx  

=  (1/2)∫ x [cos (5x+2x) + cos (5x-2x)] dx  

=  (1/2)∫x [cos 7x + cos 3x] dx  

=  (1/2){∫ [x cos 7x] dx + ∫ [x cos 3x] dx}

∫ x cos 7x dx

u  =  x          dv  =  cos 7 x

du  =  dx        v  =  sin 7x/7

=  x (sin 7x/7) - ∫(sin 7x/7) dx

=  (x/7) (sin 7x) - (1/7)∫sin 7x dx

=  (x/7) (sin 7x) + (1/7) (cos 7x/7) + C

=  (x/7)(sin 7x) + (1/49) (cos 7x) + C

∫ [x cos 3x] dx

u  =  x          dv  =  cos 3 x

du  =  dx        v  =  sin 3x/3

=  x (sin 3x/3) - ∫(sin 3x/3) dx

=  (x/3) (sin 3x) - (1/3)∫sin 3x dx

=  (x/3) (sin 3x) + (1/3) (cos 3x/3) + C

=  (x/3)(sin 3x) + (1/9) (cos 3x) + C

=  (1/2){(x/7)(sin 7x) + (1/49) (cos 7x)+(x/3)(sin 3x)+(1/9)(cos 3x)}+ C

Problem 5 :

x2 cos 3x

Solution :

∫ x2 cos 3x dx

u  =  x2                 dv = cos 3 x

du  =  2x dx           v = sin 3 x/3

=  (x2 sin 3x/3) - ∫ [sin 3x/3] 2 x dx

=  (x2/3)sin 3x - (2/3)∫ x [sin 3 x] dx

u  =  x       dv  =  sin 3x

du  =  dx      v  =  -cos 3x/3

=  (x2/3)sin 3x - (2/3){[x (-cos 3 x/3)] - ∫ [-cos 3x/3] dx}

=  (x2/3)sin 3x - (2/3)[x (-cos 3 x/3)] - (2/3)∫ [cos 3x/3] dx

=  (x2/3)sin 3x + (2/9)[x cos 3 x] - (2/9)∫ [cos 3x] dx

=  (x2/3)sin 3x + (2/9)[x cos 3 x] - (2/27)[sin 3x] + C

Problem 6 :

cosec3x

Solution :

∫ cosec3x dx = ∫ cosec x (cosec2x) dx

now we are going to apply partial differentiation

u  =  cosec x                     dv  =  cosec2x

du  =  - cosec x cot x           v  =  - cot x

∫ u dv  =  u v - ∫v du

=  (cosec x)(- cot x) - ∫ - cot x (- cosec x cot x) dx

=  -cosec x cot x - ∫ cosec x cot2x dx

=  -cosec x cot x - ∫ cosec x (cosec2x - 1) dx

=  -cosec x cot x - ∫ cosec3x dx + ∫ cosec x dx

∫cosec3x dx = -cosec x cot x - ∫ cosec³x dx + ∫ cosec x dx

∫cosec3xdx + ∫ cosec3x dx = -cosec x cot x + log tan (x/2) + C

2∫cosec3x dx = -cosec x cot x + log tan (x/2) + C

∫ cosec3x dx = (1/2)[-cosec x cot x + log tan (x/2)] + C

∫ cosec3x dx = (1/2)[-cosec x cot x] + (1/2)[log tan (x/2)] + C

Problem 7 :

eax cos bx

Solution :

∫ eax cos bx dx

u  =  cos b x           dv  =  eax

du = - bsin bx        v = eax/a

∫ u dv = u v - ∫ v du

=  (cos b x)(eax/a) - ∫(eax/a) (- b sin bx) dx

=  (cos b x)(eax/a) + (b/a) ∫ eax (sin bx) dx--------(1)

∫ eax (sin bx) dx

 u  =  sin bx                 dv  =  eax

 du  =  b cos bx             v  =  eax/a

=  (sin bx)(eax/a) - ∫(eax/a)(b cos bx) dx

=  (sin bx)(eax/a) - (b/a) ∫eax cos bx dx

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

∫eax cos bx dx+ (b²/a²) ∫eax cos bx dx

= (cos b x)(eax/a)+(b/a)(sin bx)(eax/a)

∫(1 + (b²/a²)) eax cos bx dx = (cos b x)(eax/a)+(b/a)(sin bx)(eax/a)

∫((a²+b²)/a²) eax cos bx dx = (cos b x)(eax/a)+(b/a)(sin bx)(eax/a)

∫eaxcos bx dx  =  [a²/(a²+b²)](cos b x)(eax/a)+(b/a)(sin bx(eax /a)

=  [eax/(a²+b²)][a cos b x + b sin bx]  

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