# Class 12 RD Sharma Solutions- Chapter 19 Indefinite Integrals – Exercise 19.6

• Last Updated : 25 Jan, 2021

### 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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