INTEGRATING TRIGONOMETRIC FUNCTIONS IN THE FORM AX PLUS B

Question 1 :

(i)  ∫cosec (2-x) cot (2-x) dx

(ii)  ∫cosec (4x+2) cot (4x+2) dx

(iii)  ∫cosec (3-2x) cot (3-2x) dx

(iv)  ∫cosec (Lx+m) cot (Lx+m) dx

Solution :

(i) 

∫cosec (2-x) cot (2-x) dx  =  -(1/-1) cosec (2-x) + C

=  cosec (2 - x) + C

(ii)

∫cosec (4x+2) cot (4x+2) dx  =  -(1/4) cosec (4x+2) + C

= -cosec (4x+2)/4 + C

(iii)

∫cosec (3-2x) cot (3-2x) dx  =  -(1/-2) cosec (3-2x) + C

=  cosec (3-2x)/2 + C

(iv)

∫cosec (Lx+m) cot (Lx+m) dx  = -(1/L) cosec (Lx+m) + C

= -cosec (Lx+m)/L + C

Question 2 :

(i)  ∫ sec (3+x) tan (3+x) dx

(ii)  ∫ sec (3x+4) tan (3x+4) dx

(iii)  ∫ sec (4-x) tan (4-x) dx

(iv)  ∫ sec (4-3x) tan (4-3x) dx

(v)  ∫sec (ax+b) tan (ax+b) dx

Solution :

(i)

∫ sec (3+x) tan (3+x) dx  =  (1/1) sec (3-x) + C

=  sec (3-x) + C

(ii)

∫ sec (3x+4) tan (3x+4) dx  =  (1/3) sec (3x+4) + C

=  sec (3x+4)/3 + C

(iii)

∫ sec (4-x) tan (4-x) dx  =  (1/-1) sec (4-x) + C

= - sec (4-x) + C

(iv) 

∫ sec (4-3x) tan (4-3x) dx  =  (1/-3) sec (4-3x) + C

= - sec (4-3x)/3 + C

(v)  

∫sec (a x + b) tan (a x + b) dx = (1/a) sec (a x + b) + C

Question 3 :

(i)  ∫cosec (2-x) cot (2-x) dx

(ii)  ∫cosec (4x+2) cot (4x+2) dx

(iii)  ∫cosec (3-2x) cot (3-2x) dx  

(iv)  ∫cosec (Lx+m) cot (Lx+m) dx

Solution :

(i)

∫cosec (2-x) cot (2-x) dx  = -(1/-1) cosec (2-x) + C

=  cosec (2-x) + C

(ii) 

∫cosec (4x+2) cot (4x+2) dx  =  -(1/4) cosec (4x+2) + C

=  -cosec (4x+2)/4 + C

(iii)

∫cosec (3-2x) cot (3-2x) dx  = -(1/-2) cosec (3-2x) + C

=  cosec (3-2x)/2 + C

(iv) 

∫cosec (Lx+m) cot (Lx+m) dx  =  -(1/L) cosec (Lx+m) + C

= - cosec (Lx+m)/L + C

Question 4 :

(i)  1/cos² (px + a)

(ii)  1/sin² (L - m x)

Solution :

(i)

∫ 1/cos² (px+a) dx  =  ∫sec²(px+a) dx

∫sec²(px+a) dx  =  (1/p) tan (px+a) + C

=  tan (px+a)/p + C

(ii)

∫ 1/sin² (L-mx) dx  =  ∫cosec² (L-mx) dx

∫cosec² (L-m x) dx  =  - (1/-m) cot (L-mx) + C

=  cot (L-mx)/m + C

Question 5 :

∫ (2x - 3)³ dx 

Solution :

∫ (2x - 3)³ dx 

Let t = 2x - 3

Diffrerentiating on both sides, we get

dt = 2(dx)

dx = (1/2) dt

∫ (2x - 3)³ dx = ∫ t³ (1/2)dt 

= (1/2)∫ t³ dt 

= (1/2)[t⁴/4] + C

= (1/8)(2x - 3)⁴ + C

Question 6 :

∫ cos (3x + 5) dx 

Solution :

∫ cos (3x + 5) dx 

Let t = 3x + 5

Differentitating with respect to x, we get

dt = 3(dx)

dx = (1/3) dt

∫ cos (3x + 5) dx = ∫ cos t (1/3) dt

= (1/3)∫ cos t dt

= (1/3) sin t + C

Applying the value of t, we gtet

= (1/3) sin (3x + 5) + C

Question 7 :

∫e^{5x + 2} dx 

Solution :

∫e^{5x + 2} dx

Let t = 5x + 2

Differentiating with respect to x, we get

dt = 5(dx) + 0

dt = 5dx

dx = (1/5) dt

∫e^{5x + 2} dx = ∫e^t (1/5)dt

= (1/5) ∫e^t dt

= (1/5) e^t + C

Applying the value of t, we get

= (1/5) e^{5x + 2} + C

Question 8 :

∫dx/(2x - 1)

Solution :

∫dx/(2x - 1)

Let t = 2x - 1

dt = 2(dx) - 0

dt = 2dx

dx = (1/2) dt

By applying these assumptions, we get

∫dx/(2x - 1) = ∫(1/2)dt/t

= (1/2)∫(1/t) dt

= (1/2) log t + C

Applying the value of t, we get

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

Question 9 :

∫dx/(1 + (5x)²)

Solution :

∫dx/(1 + (5x)²)

Let t = 5x

Differentiating with respect to x, we get

dt = 5(dx)

dx = (1/5) dt

∫dx/(1 + (5x)²) = ∫(1/5) dt/(1 + t²)

= (1/5) ∫1/(1 + t²) dt

∫1/(1 + t²) = tan^-1(t) + C

= (1/5) tan^-1(t) + C

Applying the value of t, we get

= (1/5) tan^-1(5x) + C

Question 10 :

∫sec²(7x + 1) dx

Solution :

∫sec²(7x + 1) dx

Let t = 7x + 1

dt = 7(dx)

dx = (1/7) dt

∫sec²(7x + 1) dx = ∫sec²t (1/7) dt

= (1/7) ∫sec²t dt

= (1/7) tan t + C

= (1/7) tan (7x + 1) + C

Question 11 :

∫cos 5x cos x dx

Solution :

∫cos 5x cos x dx

cos A cos B = (1/2) [cos (A + B) + cos (A - B)]

cos 5x cos x = cos (5x + x) + cos (5x - x)

= cos 6x + cos 4x

∫cos 5x cos x dx = (1/2) ∫(cos 6x + cos 4x] dx

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

Distributing 1/2, we get

= [(1/12) sin 6x + (1/8) sin 4x] + C

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