Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.1 | Set 1

Chapter 20 of RD Sharma's Class 12 textbook focuses on Definite Integrals a fundamental concept in calculus. This chapter introduces the concept of definite integrals including their properties, evaluation, and applications. Understanding definite integrals is crucial for solving problems involving the areas under curves and physical quantities in mathematics.

Definite Integrals

The Definite integrals are used to compute the exact area under a curve between the two specific points on the x-axis. They are defined as the limit of a Riemann sum as the number of the subdivisions approaches infinity. The value of a definite integral is obtained using the Fundamental Theorem of the Calculus which connects differentiation and integration. This provides a powerful tool for evaluating and interpreting the accumulated quantities.

Evaluate the following definite integrals:

Question 1. \int_{4}^{9} \frac{1}{\sqrt{x}}dx

Solution:

We have,

I = \int_{4}^{9} \frac{1}{\sqrt{x}}dx

I = \left[\frac{x^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\right]^9_4

I = \left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]^9_4

I = \left[2\sqrt{x}\right]^9_4

I = 2[√9 - √4 ] 

I = 2 (3 − 2)

I = 2 (1)

I = 2

Therefore, the value of \int_{4}^{9} \frac{1}{\sqrt{x}}dx       is 2.

Question 2. \int_{-2}^{3} \frac{1}{x+7}dx

Solution:

We have,

I = \int_{-2}^{3} \frac{1}{x+7}dx

I = \left[log(x+7)\right]^3_{-2}

I = log (3 + 7) − log (−2 + 7)

I = log 10 − log 5

I = log\frac{10}{5}

I = log 2

Therefore, the value of \int_{-2}^{3} \frac{1}{x+7}dx       is log 2.

Question 3. \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1-x^2}}dx

Solution:

We have,

I = \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1-x^2}}dx

Let x = sin t, so we have, 

=> dx = cos t dt

Now, the lower limit is,

=> x = 0 

=> sin t = 0

=> t = 0

Also, the upper limit is,

=> x = 1/2

=> sin t = 1/2

=> t = π/6

So, the equation becomes,

I = \int_{0}^{\frac{\pi}{6}} \frac{1}{\sqrt{1-sin^2t}}costdt

I = \int_{0}^{\frac{\pi}{6}} \frac{1}{\sqrt{cos^2t}}costdt

I = \int_{0}^{\frac{\pi}{6}} (\frac{1}{cost})costdt

I = \int_{0}^{\frac{\pi}{6}} 1dt

I = \left[t\right]_0^{\frac{\pi}{6}}

I =  π/6 - 0

I = π/6​

Therefore, the value of \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1-x^2}}dx       is π/6.

Question 4. \int_{0}^{1} \frac{1}{1+x^2}dx

Solution:

We have,

I = \int_{0}^{1} \frac{1}{1+x^2}dx

I = \left[tan^{-1}x\right]_0^1

I = tan^{-1}1-tan^{-1}0

I = \frac{\pi}{4}-0

I = π/4

Therefore, the value of \int_{0}^{1} \frac{1}{1+x^2}dx       is π/4.

Question 5. \int_{2}^{3} \frac{x}{x^2+1}dx

Solution:

We have,

I = \int_{2}^{3} \frac{x}{x^2+1}dx

Let x² + 1 = t, so we have,

=> 2x dx = dt

=> x dx = dt/2

Now, the lower limit is, x = 2

=> t = x² + 1

=> t = (2)² + 1

=> t = 4 + 1

=> t = 5

Also, the upper limit is, x = 3

=> t = x² + 1

=> t = (3)² + 1

=> t = 9 + 1

=> t = 10

So, the equation becomes,

I = \int_{5}^{10} \frac{1}{2t}dt

I = \frac{1}{2}\int_{5}^{10} \frac{1}{t}dt

I = \frac{1}{2}\left[logt\right]^{10}_5

I = 1/2[log10 - log5] 

I = 1/2[log10/5]

I = 1/2[log2]

I = log√2

Therefore, the value of \int_{2}^{3} \frac{x}{x^2+1}dx       is log√2.

Question 6. \int_{0}^{\infty} \frac{1}{a^2+b^2x^2}dx

Solution:

We have,

I = \int_{0}^{\infty} \frac{1}{a^2+b^2x^2}dx

I = \int_{0}^{\infty} \frac{1}{b^2}\left(\frac{1}{\frac{a^2}{b^2}+x^2}\right)dx

I = \frac{1}{b^2}\int_{0}^{\infty}\frac{1}{\frac{a^2}{b^2}+x^2}dx

I = \frac{1}{b^2}\left[\frac{b}{a}tan^{-1}\frac{bx}{a}\right]^{\infty}_{0}

I = \frac{1}{ab}\left[tan^{-1}\frac{bx}{a}\right]^{\infty}_{0}

I = 1/ab[tan^-1∞ - tan^-10] 

I = 1/ab[π/2 - 0]  

I = 1/ab[π/2]  

I = π/2ab   

Therefore, the value of \int_{0}^{\infty} \frac{1}{a^2+b^2x^2}dx       is π/2ab.

Question 7. \int_{-1}^{1} \frac{1}{1+x^2}dx

Solution:

We have,

I = \int_{-1}^{1} \frac{1}{1+x^2}dx

I = \left[tan^{-1}x\right]_{-1}^{1}

I = [tan^-11 - tan^-1(-1)]  

I = [π/4 - (-π/4)]   

I = [π/4 + π/4]  

I = 2π/4 

I = π/2  

Therefore, the value of \int_{-1}^{1} \frac{1}{1+x^2}dx       is π/2.

Question 8. \int_{0}^{\infty} e^{-x}dx

Solution:

We have,

I = \int_{0}^{\infty} e^{-x}dx

I = \left[-e^{-x}\right]^{\infty}_{0}

I = -e^-∞ - (-e⁰) 

I = − 0 + 1

I = 1

Therefore, the value of \int_{0}^{\infty} e^{-x}dx       is 1.

Question 9. \int_{0}^{1} \frac{x}{x+1}dx

Solution:

We have,

I = \int_{0}^{1} \frac{x}{x+1}dx

I = \int_{0}^{1} \frac{(x+1)-1}{x+1}dx

I = \int_{0}^{1} \frac{x+1}{x+1}dx-\int_{0}^{1}\frac{1}{x+1}dx

I = \int_{0}^{1} 1dx-\int_{0}^{1}\frac{1}{x+1}dx

I = \left[x\right]^1_0-\left[log(x+1)\right]^1_0

I = [1 − 0] − [log(1 + 1) − log(0 + 1)]

I = 1 − [log2 − log1]

I = 1 - log2/1 

I = 1 − log 2

I = log e − log 2

I = loge/2 

Therefore, the value of \int_{0}^{1} \frac{x}{x+1}dx        is loge/2.

Question 10. \int_{0}^{\frac{\pi}{2}} (sinx+cosx)dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} (sinx+cosx)dx

I = \int_{0}^{\frac{\pi}{2}} (sinx)dx+\int_{0}^{\frac{\pi}{2}}(cosx)dx

I = \left[-cosx\right]_{0}^{\frac{\pi}{2}}+\left[sinx\right]_{0}^{\frac{\pi}{2}}

I = [-cosπ/2 + cos0] + [sinπ/2 - sin0] 

I = [−0 + 1] + 1

I = 1 + 1

I = 2

Therefore, the value of \int_{0}^{\frac{\pi}{2}} (sinx+cosx)dx      is 2.

Question 11. \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} cotxdx

Solution:

We have,

I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} cotxdx

I = \left[log(sinx)\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}

I = log(sinπ/2) - log(sinπ/4)

I = log1 - log1/√2 

I = log\frac{1}{\frac{1}{\sqrt{2}}}

I = log√2  

Therefore, the value of \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} cotxdx     is log√2.

Question 12. \int_{0}^{\frac{\pi}{4}} secxdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{4}} secxdx

I = \left[log(secx+tanx)\right]^{\frac{\pi}{4}}_0

I = log(secπ/4 + tanπ/4 - log(sec0 + tan0) 

I = log(√2 + 1) - log(1 + 0) 

I = log(\frac{\sqrt{2}+1}{1})

I = log(√2 + 1) 

Therefore, the value of \int_{0}^{\frac{\pi}{4}} secxdx     is log(√2 + 1).

Question 13. \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} cosecxdx

Solution:

We have,

I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} cosecxdx

I = \left[log|cosecx-cotx|\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} 

I = [log|cosecπ/4 - cotπ/4|] - [log|cosecπ/6 - cotπ/6|]

I = [log|√2 - 1|] - [log|2 - √3|] 

I = log(\frac{\sqrt{2}-1}{2-\sqrt{3}})

Therefore, the value of \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} cosecxdx     is log(\frac{\sqrt{2}-1}{2-\sqrt{3}})     .

Question 14. \int_{0}^{1} \frac{1-x}{1+x}dx

Solution: 

We have, 

I = \int_{0}^{1} \frac{1-x}{1+x}dx

Let x = cos 2t, so we have,

=> dx = –2 sin 2t dt

Now, the lower limit is,

=> x = 0

=> cos 2t = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is,

=> x = 1

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I = \int_{\frac{\pi}{4}}^{0} \frac{1-cos2t}{1+cos2t}(-2sin2t)dt

I = \int_{\frac{\pi}{4}}^{0}\frac{2sin^2t}{2cos^2t}(-2sin2t)dt

I = \int^{\frac{\pi}{4}}_{0}\frac{sin^2t}{cos^2t}(2sin2t)dt

I = \int^{\frac{\pi}{4}}_{0}\frac{sin^2t}{cos^2t}(4sintcost)dt

I = \int^{\frac{\pi}{4}}_{0}\frac{4sin^3t}{cost}dt

Let cos t = z, so we have,

=> – sin t dt = dz

=> sin t dt = – dz

Now, the lower limit is,

=> t = 0

=> z = cos t

=> z = cos 0

=> z = 1

Also, the upper limit is,

=> t = π/4

=> z = cos t

=> z = cos π/4

=> z = 1/√2

So, the equation becomes,

I = \int^{\frac{\pi}{4}}_{0}\frac{4sin^2t(sint)}{cost}dt

I = \int^{\frac{1}{\sqrt{2}}}_{1}\frac{-4(1-z^2)}{z}dz

I = -4\int^{\frac{1}{\sqrt{2}}}_{1}\frac{1-z^2}{z}dz

I = -4\int^{\frac{1}{\sqrt{2}}}_{1}(\frac{1}{z}-\frac{z^2}{z})dz

I = -4\int^{\frac{1}{\sqrt{2}}}_{1}(\frac{1}{z}-z)dz

I = -4\left[logz-\frac{z^2}{2}\right]^{\frac{1}{\sqrt{2}}}_1

I = -4[(log1/√2 - 1/2(2)) - (log1 - 1/2)]

I = -4[(log1/√2 - 1/4) - (0 - 1/2)]

I = -4[log1/√2 - 1/4 - 0 + 1/2]

I = -4[-log√2 + 1/4]

I = 4log√2 - 1 

I = 4 × 1/2log2 - 1

I = 2log2 - 1 

Therefore, the value of \int_{0}^{1} \frac{1-x}{1+x}dx     is 2log2 - 1.

Question 15. \int_{0}^{\pi} \frac{1}{1+sinx}dx

Solution: 

We have,

I = \int_{0}^{\pi} \frac{1}{1+sinx}dx

I = \int_{0}^{\pi} \frac{1-sinx}{(1+sinx)(1-sinx)}dx

I = \int_{0}^{\pi} \frac{1-sinx}{1-sin^2x}dx

I = \int_{0}^{\pi} \frac{1-sinx}{cos^2x}dx

I = \int_{0}^{\pi}\frac{1}{cos^2x}-\frac{sinx}{cos^2x}dx

I = \int_{0}^{\pi}sec^2x-\frac{sinx}{cosx(cosx)}dx

I = \int_{0}^{\pi}(sec^2x-tanxsecx)dx

I = \int_{0}^{\pi}sec^2xdx-\int_{0}^{\pi}tanxsecxdx

I = \left[tanx\right]^{\pi}_0-\left[secx\right]^{\pi}_0

I = [tan π – tan0] – [sec π – sec 0]

I = [0 – 0] – [–1 – 1]

I = 0 – (–2)

I = 2

Therefore, the value of \int_{0}^{\pi} \frac{1}{1+sinx}dx     is 2.

Question 16. \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1}{1+sinx}dx

Solution: 

We have,

I = \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1}{1+sinx}dx

I = \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1-sinx}{(1+sinx)(1-sinx)}dx

I = \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1-sinx}{1-sin^2x}dx

I = \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1-sinx}{cos^2x}dx

I = \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\frac{1}{cos^2x}-\frac{sinx}{cos^2x}dx

I = \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}sec^2x-\frac{sinx}{cosx(cosx)}dx

I = \int_{\frac{-\pi}{4}}^{\frac{-\pi}{4}}(sec^2x-tanxsecx)dx

I = \int_{\frac{-\pi}{4}}^{\frac{-\pi}{4}}sec^2xdx-\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}tanxsecxdx

I = \left[tanx\right]^{\frac{\pi}{4}}_{\frac{-\pi}{4}}-\left[secx\right]^{\frac{\pi}{4}}_{\frac{-\pi}{4}}

I = \left[tanx\right]^{\frac{\pi}{4}}_{\frac{-\pi}{4}}-\left[secx\right]^{\frac{\pi}{4}}_{\frac{-\pi}{4}}

I = [tan π/4 – tan(–π/4)] – [sec π/4 – sec (–π/4)]

I = [1 – (–1)] – [sec π/4 – sec (π/4)]

I = 2 – 0

I = 2

Therefore, the value of \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1}{1+sinx}dx     is 2.

Question 17. \int_{0}^{\frac{\pi}{2}} cos^2xdx

Solution: 

We have,

I = \int_{0}^{\frac{\pi}{2}} cos^2xdx

I = \int_{0}^{\frac{\pi}{2}} (\frac{1+cos2x}{2})dx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} (1+cos2x)dx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}1dx+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}cos2xdx

I = \frac{1}{2}\left[x\right]^{\frac{\pi}{2}}_0+\frac{1}{2}\left[\frac{sin2x}{2}\right]^{\frac{\pi}{2}}_0

I = \frac{1}{2}\left[x\right]^{\frac{\pi}{2}}_0+\frac{1}{4}\left[sin2x\right]^{\frac{\pi}{2}}_0

I = 1/2[π/2 - 0] + 1/4[sinπ - sin0]

I = 1/2[π/2] + 1/4[0 - 0]

I = π/4 

Therefore, the value of \int_{0}^{\frac{\pi}{2}} cos^2xdx      is π/4.

Question 18. \int_{0}^{\frac{\pi}{2}} cos^3xdx

Solution: 

We have,

I = \int_{0}^{\frac{\pi}{2}} cos^3xdx

I = \int_{0}^{\frac{\pi}{2}} \frac{cos3x+3cosx}{4}dx

I = \frac{1}{4}\int_{0}^{\frac{\pi}{2}} (cos3x+3cosx)dx

I = \frac{1}{4}\int_{0}^{\frac{\pi}{2}} cos3xdx+\frac{3}{4}\int_{0}^{\frac{\pi}{2}}cosxdx

I = \frac{1}{4}\left[\frac{sin3x}{3}\right]^{\frac{\pi}{2}}_0+\frac{3}{4}[sinx]^{\frac{\pi}{2}}_{0}

I = \frac{1}{12}\left[sin3x\right]^{\frac{\pi}{2}}_0+\frac{3}{4}[sinx]^{\frac{\pi}{2}}_{0}

I = 1/12 [-1 - 0] + 3/4[1 - 0]

I = 3/4 - 1/12

I = (9 - 1)/12

I = 8/12

I = 2/3

Therefore, the value of \int_{0}^{\frac{\pi}{2}} cos^3xdx      is 2/3.

Question 19. \int_{0}^{\frac{\pi}{6}} cosxcos2xdx

Solution: 

We have,

I = \int_{0}^{\frac{\pi}{6}} cosxcos2xdx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{6}} 2cosxcos2xdx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{6}} (cos3x + cosx)dx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{6}}cos3xdx + \frac{1}{2}\int_{0}^{\frac{\pi}{6}}cosxdx

I = \frac{1}{2}\left[\frac{sin3x}{3}\right]^{\frac{\pi}{6}}_0 + \frac{1}{2}\left[sinx\right]^{\frac{\pi}{6}}_0

I = \frac{1}{6}\left[sin3x\right]^{\frac{\pi}{6}}_0 + \frac{1}{2}\left[sinx\right]^{\frac{\pi}{6}}_0

I = 1/6[sinπ/2 - sin0] + 1/2[sinπ/6 - sin0]

I = 1/6[1 - 0] + 1/2[1/2 - 0]

I = 1/6 + 1/4

I = (4 + 6)/24 

I = 10/24 

I = 5/12 

Therefore, the value of \int_{0}^{\frac{\pi}{6}} cosxcos2xdx      is 5/12.

Question 20. \int_{0}^{\frac{\pi}{2}} sinxsin2xdx

Solution: 

We have,

I = \int_{0}^{\frac{\pi}{2}} sinxsin2xdx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} 2sinxsin2xdx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} (cosx - cos3x)dx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}cosxdx - \frac{1}{2}\int_{0}^{\frac{\pi}{2}}cos3xdx

I = \frac{1}{2}\left[sinx\right]^{\frac{\pi}{2}}_0-\frac{1}{2}\left[\frac{sin3x}{3}\right]^{\frac{\pi}{2}}_0

I = \frac{1}{2}\left[sinx\right]^{\frac{\pi}{2}}_0-\frac{1}{6}[sin3x]^{\frac{\pi}{2}}_0

I = 1/2[sinπ/2 - sin0] - 1/6[sin3π/2 - sin0]

I = 1/2[1 - 0] - 1/6[-1 - 0]

I = 1/2 - 1/6(-1)

I = 1/2 + 1/6 

I = (6 + 2)/12

I = 8/12 

I = 2/3

Therefore, the value of \int_{0}^{\frac{\pi}{2}} sinxsin2xdx      is 2/3.

Question 21. \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (tanx+cotx)^2dx

Solution: 

We have,

I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (tanx+cotx)^2dx

I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (\frac{sinx}{cosx}+\frac{cosx}{sinx})^2dx

I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (\frac{sin^2x+cos^2x}{cosxsinx})^2dx

I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (\frac{1}{cosxsinx})^2dx

I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\frac{1}{cos^2xsin^2x}dx

I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\frac{4}{4cos^2xsin^2x}dx

I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\frac{4}{(sin2x)^2}dx

I = 4\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}cosec^22xdx

I = 4\left[\frac{-cot2x}{2}\right]_{\frac{\pi}{3}}^{\frac{\pi}{4}}

I = 2\left[-cot2x\right]_{\frac{\pi}{3}}^{\frac{\pi}{4}}

I = 2[-cotπ/2 + cot2π/3]

I = 2[-1/√3 - 0]

I = -2/√3 

Therefore, the value of \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (tanx+cotx)^2dx      is -2/√3.

Question 22. \int_{0}^{\frac{\pi}{2}} cos^4xdx

Solution: 

We have,

I = \int_{0}^{\frac{\pi}{2}} cos^4xdx

I = \int_{0}^{\frac{\pi}{2}} (cos^2x)^2dx

I = \int_{0}^{\frac{\pi}{2}} (\frac{1+cos2x}{2})^2dx

I = \frac{1}{4}\int_{0}^{\frac{\pi}{2}} (1+cos2x)^2dx

I = \frac{1}{4}\int_{0}^{\frac{\pi}{2}} (1+cos^22x+2cos2x)dx

I = \frac{1}{4}\int_{0}^{\frac{\pi}{2}} (1+\frac{1+cos4x}{2}+2cos2x)dx

I = \frac{1}{4}\left[x+\frac{x}{2}+\frac{sin4x}{8}+sin2x\right]^{\frac{\pi}{2}}_0

I = 1/4[π/2 + π/4 + 0 + 0 - 0 - 0 - 0 - 0]

I = 1/4[3π/4] 

I = 3π/16 

Therefore, the value of \int_{0}^{\frac{\pi}{2}} cos^4xdx     is 3π/16.

Summary

Exercise 20.1 in RD Sharma's Class 12 textbook focuses on definite integrals. This section typically covers:

• The concept of definite integrals
• Properties of definite integrals
• Evaluation of definite integrals using fundamental theorems of calculus
• Integration techniques applied to definite integrals