Class 12 RD Sharma Solutions Chapter 19 Indefinite Integrals Exercise 19.6 (original) (raw)

Last Updated : 4 Sep, 2024

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

Summary

Exercise 19.6 in Chapter 19 Indefinite Integrals focuses on evaluating various types of indefinite integrals using different integration techniques. The practice questions cover a range of functions including polynomials, trigonometric functions, exponential functions, and rational functions. Students are required to apply methods such as the power rule, integration by parts, substitution, and integration of trigonometric and rational functions. This exercise aims to enhance students' proficiency in applying various integration techniques and deepen their understanding of indefinite integrals. The problems progress from simpler polynomial and trigonometric integrals to more complex rational and irrational function integrals, providing a comprehensive practice set for students to master the concepts of indefinite integration.