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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