# 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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