Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.3 | Set 2

Question 11. Integrate

Solution:

Let I =

On simplifying the above equation, we get

I =

=

= ∫ tan² x/2 dx

=

On integrating the equation(1), we get

=

Hence, I = 2 tan x/2 – x + c

Question 12. Integrate

Solution:

Let I =

Now multiply with the conjugate,

=

On integrating the equation, we get

=

= 2 tan x/2 + 2 sec x/2 +c

Hence, I = 2 (tan x/2 + sec x/2) + c

Question 13. Integrate

Solution:

Let I =

Now multiply with the conjugate,

= ∫ 1/(1 + cos 3x) × (1 – cos 3x)/(1 – cos 3x) dx

= ∫ (1 – cos 3x)/ (1 – cos² 3x) dx

= ∫ (1 – cos 3x)/ (sin² 3x) dx

= ∫ (1/ sin² 3x) – (cos 3x/ sin² 3x) dx

= ∫ (cosec² 3x – cosec3x cot3x) dx -(1)

On integrating the equation(1), we get

= – cot 3x/3 + cosec 3x/3 + c

= (-1/3) × (cos 3x/ sin 3x) + (1/3) × (1/sin 3x) + c

= (1 – cos 3x) / 3 sin 3x + c

Therefore, I = (1 – cos 3x) / 3 sin 3x + c

Question 14. Integrate ∫(e^x+ 1)² e^x dx

Solution:

Let I = ∫ (e^x + 1)² e^x dx -(1)

(e^x + 1) = t -(2)

On differentiating the above equation, we get

e^x dx = dt -(3)

Now, put the eq(2) and (3) in eq(1)

= ∫ (t²) dt -(4)

On integrating the equation(4), we get

= (t³ /3) + c

Therefore, I = (e^x + 1)³/3 + c

Question 15. Integrate ∫ (e^x + (1 + e^x))² dx

Solution:

Let I = ∫ (e^x + (1/e^x))² dx

= ∫ (e^2x + (1/e^2x) + 2)dx -(1)

On integrating the equation(1), we get

= (e^2x/2) – (1/2 e^-2x) + 2x + c

Therefore, I = (e^2x/2) – (1/2 e^-2x) + 2x + c

Question 16. Integrate

Solution:

Let I =

=

On simplifying the above equation,

= ∫ cos²2x. (sin2x / cos2x) dx

= ∫ cos 2x. sin2x dx

= 1/2∫ sin (2x + 2x) + sin (2x – 2x) dx

= 1/2∫(sin 4x + sin 0) dx

= 1/2∫(sin 4x + 0) dx

= 1/2 ∫sin 4x dx -(1)

On integrating the equation(1), we get

= (-1/2) ((cos 4x)/4) + c

Therefore, I = (-1/8) (cos 4x) + c

Question 17. Integrate

Solution:

Let I =

Now multiply with the conjugate,

=

= ∫(x +3)^1/2 + (x + 2)^1/2 dx -(1)

On integrating the equation(1), we get

=

= (2/3)(x + 3)^3/2 +(2/3) (x + 2)^3/2 + c

Hence, I = (2/3){(x + 3)^3/2 + (x + 2)^3/2} +c

Question 18. Integrate ∫ tan²(2x – 3) dx

Solution:

Let I = ∫ tan²(2x – 3) dx

= ∫ sec² (2x – 3) – 1 dx -(1)

Now put, 2x – 3 = t -(2)

2dx = dt -(3)

Put eq(3) and (2) in eq(1)

= 1/2∫sec² t dt – ∫1dx -(4)

On integrating the equation(4), we get

= 1/2 tan t – x + c

= 1/2 tan(2x – 3) – x + c

Therefore, I = 1/2 tan(2x – 3) – x + c