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

Last Updated : 25 Jan, 2021

Question 1: âˆ« sin2(2x+5) dx

Solution:

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

â‡’ âˆ«sin2(2x+5)dx= âˆ«(1-cos(4x+10))/2 dx

= 1/2 âˆ«1 dx – 1/2âˆ«cos(4x+10) dx

= x/2 – 1/2 ((sin(4x+10))/4)+C

= x/2 – sin(4x+10)/8 + C

Question 2: âˆ«sin3(2x+1) dx

Solution:

We need to evaluate âˆ«sin3(2x+1)dx

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

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

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

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

Question 3: âˆ«cos42x dx

Solution:

Evaluate the integral as follows

âˆ«cos42xdx= âˆ«(cos22x)2 dx

=âˆ«(1/2(cos4x+1))2dx

=âˆ«(1/4(cos8x+1)/2 + 1/4+ cos4x/2)dx

=âˆ«1/8(cos8x + 3/8 + cos4x/2)dx

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

Question 4: âˆ«sin2bx dx

Solution:

Let I = âˆ«sin2bxdx. Then,

I= âˆ«(1-cos2bx)/2 xdx

=1/2âˆ«dx = 1/2âˆ«cos2bxdx

x/2 – sin(2bx)/4b + c

Therefore, I = x/2 – sin2bx/4b + C

Question 5: âˆ«sin2(x/2) dx

Solution:

Let I= âˆ«sin2(x/2)dx, Then,

I=1/2 âˆ«2 sin2(x/2)dx

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

= 1/2 âˆ«dx – 1/2 âˆ«cosx dx

=x/2-sinx/2 + C

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

Question 6: âˆ«cos2(x/2)dx

Solution:

We have,

âˆ«cos2(x/2)dx = 1/2 âˆ«2cos2(x/2)dx

=1/2 âˆ«1+cosxdx

=1/2 âˆ«dx + 1/2 âˆ«cosx dx

= x/2 + sinx/2 + C

= (x+sin x)/2 + C

Therefore, cos2(x/2) = (x+sinx)/2 + C

Question 7: âˆ«cos2nx dx

Solution:

Let I= âˆ«cos2nx dx. Then,

I= 1/2 âˆ«2cos2nx dx

= 1/2 âˆ«[1+cos2nx]dx

= 1/2 âˆ«[x + sin2nx/2n ] + C

= x/2 + (sin2nx/4n) + C

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

Question 8: âˆ«sinâˆš(1-cos2x) dx

Solution:

Let I = âˆ«sinâˆš(1-cos2x) dx, Then,

I = âˆ«sinx * âˆš(2sin2x * dx

= âˆ«sinx * âˆš2 * sinxdx

= âˆš2 âˆ«2sin2x xdx

= âˆš2 /2âˆ«2sin2xdx

= âˆš2 /2[x – (sin2x)/2]+C

= âˆš2 x/2 – âˆš2 /4 sin2x+C

=x/âˆš2 – sin2x/2âˆš2 +C

Therefore, I= x/âˆš2  – sin2x/2âˆš2  + C

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