RD Sharma Class 12 Ex 19.6 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.6 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.6 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

RD Sharma Class 12 Ex 19.6 Solutions Chapter 19 Indefinite Integrals

Question 1: ∫ sin²(2x+5) dx

Solution:

sin²(2x+5)= (1-cos2(2x+5)/)2 = (1-cos(4x+10))/2

⇒ ∫sin²(2x+5)dx= ∫(1-cos(4x+10))/2 dx

                          = 1/2 ∫1 dx – 1/2∫cos(4x+10) dx

                          = x/2 – 1/2 ((sin(4x+10))/4)+C

                          = x/2 – sin(4x+10)/8 + C

Question 2: ∫sin³(2x+1) dx

Solution:

We need to evaluate ∫sin³(2x+1)dx

By using the formula :  sin3A = -4sin³A + 3sinA

Therefore, sin3(2x+1)= (3sin(2x+1) – sin3(2x+1))/4

∫sin³(2x+1)dx = ∫(3sin(2x+1) – sin3(2x+1))/4 dx

                        = -3cos(2x+1)/8+ cos3(2x+1)/24+C

Question 3: ∫cos⁴2x dx

Solution:

Evaluate the integral as follows 

∫cos⁴2xdx= ∫(cos²2x)² dx

=∫(1/2(cos4x+1))²dx

=∫(1/4(cos8x+1)/2 + 1/4+ cos4x/2)dx

=∫1/8(cos8x + 3/8 + cos4x/2)dx

= sin8x/64 + 3x/8 + sin4x/8 + C

Question 4: ∫sin²bx dx

Solution:

Let I = ∫sin²bxdx. Then,

I= ∫(1-cos2bx)/2 xdx

=1/2∫dx = 1/2∫cos2bxdx

x/2 – sin(2bx)/4b + c

Therefore, I = x/2 – sin2bx/4b + C

Question 5: ∫sin²(x/2) dx

Solution:

Let I= ∫sin²(x/2)dx, Then,

     I=1/2 ∫2 sin²(x/2)dx

      = 1/2  ∫(1-cosx)dx                                                        [ cos2x = 1-2sin²x ]

      = 1/2 ∫dx – 1/2 ∫cosx dx

       =x/2-sinx/2 + C

Therefore, I= (x-sinx)/2 + C

Question 6: ∫cos²(x/2)dx

Solution:

We have,

∫cos²(x/2)dx = 1/2 ∫2cos²(x/2)dx

                      =1/2 ∫1+cosxdx

                      =1/2 ∫dx + 1/2 ∫cosx dx

                      = x/2 + sinx/2 + C

                      = (x+sin x)/2 + C

Therefore, cos²(x/2) = (x+sinx)/2 + C

Question 7: ∫cos²nx dx

Solution:

Let I= ∫cos²nx dx. Then,

     I= 1/2 ∫2cos²nx dx

      = 1/2 ∫[1+cos2nx]dx

      = 1/2 ∫[x + sin2nx/2n ] + C

      = x/2 + (sin2nx/4n) + C

Therefore, I= x/2 + (sin2nx/4n) + C

Question 8: ∫sin√(1-cos2x) dx

Solution:

Let I = ∫sin√(1-cos2x) dx, Then,

     I = ∫sinx * √(2sin²x * dx

       = ∫sinx * √2 * sinxdx

       = √2 ∫2sin²x xdx

       = √2 /2∫2sin²xdx

       = √2 /2[x – (sin2x)/2]+C

       = √2 x/2 – √2 /4 sin2x+C

       =x/√2 – sin2x/2√2 +C

Therefore, I= x/√2  – sin2x/2√2  + C

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