Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.28

Question 1. Find $\int\sqrt{3+2x-x^2}dx$

Solution:

$\int\sqrt{3+2x+x^2}dx=\int\sqrt{4-(x-1)^2}dx\\$
Let considered x – 1 = t,
so that dx = dt
Thus, $\int\sqrt{3+2x+x^2}dx\\=\int\sqrt{4-t^2}dt\\ =\frac{1}{2}t\sqrt{4-t^2}+\frac{4}{2}sin^{-1}\left(\frac{t}{2}\right)+C\\ =\frac{1}{2}(x-1)\sqrt{3+2x-x^2}+2sin^{-1}\left(\frac{x-1}{2}\right)+C$

Question 2. Evaluate $\int\sqrt{x^2+x+1}dx$

Solution:

Let I = $\int\sqrt{x^2+x+1}dx\\ =\int\sqrt{x^2+x+\frac{1}{4}+\frac{3}{4}}dx\\ =\int\sqrt{\left(x+\frac{1}{2}^2\right)+\left(\frac{\sqrt{3}}{2}\right)^2}dx\\ =\frac{\left(x+\frac{1}{2}\right)}{2}\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}+\frac{\left(\frac{\sqrt{3}}{2}\right)^2}{2}.log\left|\left(x+\frac{1}{2}\right)+\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right|+c\\ =\left(\frac{2x+1}{4}\right)\sqrt{x^2+x+1}+\frac{3}{8}log\left|\left(\frac{2x+1}{2}\right)+\frac{1}{2}\sqrt{x^2+x+1}\right|+c\\ I=\left(\frac{2x+1}{4}\right)\sqrt{x^2+x+1}+\frac{3}{8}log\left|{2x+1}+\sqrt{x^2+x+1}\right|+c\\$

Question 3. Evaluate $\int\sqrt{x-x^2}dx$

Solution:

I = $\int\sqrt{x-x^2}dx\\ \int\sqrt{\frac{1}{4}-\frac{1}{4}+x-x^2}dx\ \ \ \ \ \ -(Add\ and\ subtract \frac{1}{4})\\ \\ \int\sqrt{\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}-x\right)^2}dx\\ = -\left(\frac{1-2x}{4}\right)\sqrt{\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}-x\right)^2}-\frac{\left(\frac{1}{2}\right)^2}{2}sin^{-1}\left(\frac{\frac{1-x}{2}}{\frac{1}{2}}\right)+c\\$
Hence, $I=\left(\frac{2x-1}{4}\right)\sqrt{x-x^2}+\frac{1}{8}sin^{-1}(2x-1)+c$

Question 4. Evaluate $\int\sqrt{1+x-2x^2}dx$

Solution:

Let I = $\int\sqrt{1+x-2x^2}dx\\ =\sqrt{2}\int\sqrt{\frac{1}{2}+\frac{x}{2}-x^2}dx\\ =\sqrt{2}\int\sqrt{\frac{9}{16}+-\left(\frac{1}{16}-\frac{x}{2}+x\right)^2}dx\\ =\sqrt{2}\int\sqrt{\left(\frac{3}{4}\right)^2-\left(x-\frac{1}{4}\right)^2}dx\\ =\sqrt{2}\left\{\frac{\left(x-\frac{1}{4}\right)}{2}\sqrt{\frac{1}{2}+\frac{x}{2}-x^2}+\frac{9}{32}sin^{-1}\left(\frac{x-\frac{1}{4}}{\frac{3}{4}}\right)\right\}+c\\$
Therefore, I = $\frac{1}{8}(4x-1)\sqrt{1+x-2x^2}+\frac{9\sqrt{2}}{32}sin^{-1}\left(\frac{4x-1}{3}\right)+c$

Question 5. $\int cosx\sqrt{4-sin^2x}dx$

Solution:

I = $\int cosx\sqrt{4-sin^2x}dx$
Let us considered sinx = t
So, on differentiating, we get
cosx dx = dt
I = $\int\sqrt{4-t^2}dt\\ =\int\sqrt{2^2-t^2}dt\\ =\frac{t}{2}\sqrt{2^2-t^2}+\frac{4}{2}sin^{-1}\frac{t}{2}+c\\$
Therefore, I = $\frac{1}{2}sinx\sqrt{4-sin^2x}+2sin^{-1}\left(\frac{sinx}{2}\right)+c$

Question 6. Evaluate $\int e^x\sqrt{e^{2x}+1}dx$

Solution:

I = $\int e^x\sqrt{e^{2x}+1}dx$
Let us considered e^x = t
So, on differentiating, we get
e^xdx = dt
Therefore, I = $\int\sqrt{t^2+1^2}dt\\ =\frac{t}{2}\sqrt{t^2+1^2}+\frac{1}{2}log\left|t+\sqrt{t^2+1}\right|+c$
Hence, I = $\frac{e^x}{2}\sqrt{e^{2x}+1}+\frac{1}{2}log\left|e^x+\sqrt{e^{2x}+1}\right|+c$

Question 7. Evaluate $\int\sqrt{9-x^2}dx$

Solution:

I = $\int\sqrt{3^2-x^2}$
We already have,
$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{a}+c$
Therefore, I = $\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}sin^{-1}\frac{x}{3}+c$

Question 8. Evaluate $\int\sqrt{16x^2+25}dx$

Solution:

Let us assume I = $\int\sqrt{16x^2+25}dx$
$=\int\sqrt{(4x)^2+5^2}dx\\ =4\int\sqrt{x^2+\left(\frac{5}{4}\right)^2}dx\\ =4\left\{\frac{x}{2}\sqrt{x^2+\left(\frac{5}{4}\right)^2}+{\frac{\left(\frac{5}{4}\right)^2}{2}}log\left|x+\sqrt{x^2+\left(\frac{5}{4}\right)^2}\right|\right\}+c\\$
Therefore, I = $2x\sqrt{x^2+\frac{25}{16}}+\frac{25}{8}log\left|x+\sqrt{x^2+\frac{25}{16}}\right|+c\\$

Question 9. Evaluate $\int\sqrt{4x^2-5}dx$

Solution:

Let us assume I = $\int\sqrt{4x^2-5}dx$
$=2\int\sqrt{x^2-\left(\frac{\sqrt5}{2}\right)}dx\\ =2\left\{\frac{x}{2}\sqrt{x^2-\frac{5}{4}}-\frac{5}{8}log\left|x+\sqrt{x^2-\frac{5}{4}}\right|+c\right\}$
Therefore, I = $x\sqrt{x^2-\frac{5}{4}}-\frac{5}{4}log\left|x+\sqrt{x^2-\frac{5}{4}}\right|+c$

Question 10. Evaluate $\int\sqrt{2x^2+3x+4}dx$

Solution:

Let us assume I = $\int\sqrt{2x^2+3x+4}dx$
$=\sqrt{2}\int\sqrt{x^2+\frac{3}{2}x+2}dx\\ =\sqrt{2}\int\sqrt{x^2+\frac{3}{2}x+\frac{9}{16}+\frac{23}{16}}dx\\ =\sqrt{2}\int\sqrt{\left(x+\frac{3}{4}\right)^2+\left(\frac{\sqrt{23}}{4}\right)^2}dx\\ =\sqrt{2}\left\{\frac{\left(x+\frac{3}{4}\right)}{2}\sqrt{x^2+\frac{3}{2}x+2}+\frac{23}{32}.log\left|\left(x+\frac{3}{4}\right)+\sqrt{x^2+\frac{3}{2}x+2}\right|+c\right\}$
Therefore, I = $\frac{4x+3}{8}\sqrt{2x^2+3x+4}+\frac{23\sqrt{2}}{32}.log\left|\left(x+\frac{3}{4}\right)+\sqrt{x^2+\frac{3}{2}x+2}\right|+c$

Question 11. Evaluate $\int\sqrt{3-2x-2x^2}dx$

Solution:

Let us assume I = $\int\sqrt{3-2x-2x^2}dx$
$=\sqrt{2}\int\sqrt{\frac{3}{2}-x-x^2}dx\\ =\sqrt{2}\int\sqrt{\frac{7}{4}-\left(\frac{1}{4}+x+x^2\right)}dx\ \ \ \ -(Add\ and\ subtract\ \frac{1}{4})\\ =\sqrt{2}\int\sqrt{\left(\frac{\sqrt{7}}{2}\right)^2-\left(x+\frac{1}{2}\right)^2}dx\\ =\sqrt{2}\left\{\frac{x+\frac{1}{2}}{2}\sqrt{\frac{3}{2}-x+x^2}+\frac{7}{8}sin^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{7}}{2}}\right)+c\right\}$
Therefore, I = $\frac{2x+1}{4}\sqrt{3-2x-2x^2}+\frac{7\sqrt{2}}{8}sin^{-1}\left(\frac{2x+1}{\sqrt{7}}\right)+c$

Question 12. Evaluate $\int x\sqrt{x^4+1}dx$

Solution:

Let us assume x² = t
On differentiating we get
2x dx = dt
Therefore, I = $\frac{1}{2}\int\sqrt{t^2+1^2}dt\\ =\frac{1}{2}\left\{\frac{t}{2}\sqrt{t^2+1}+\frac{1}{2}log\left|t+\sqrt{t^2+1}\right|\right\}+c\\$
Hence, I = $\frac{1}{2}\left\{\frac{x^2}{2}\sqrt{x^4+1}+\frac{1}{2}log\left|x^2+\sqrt{x^4+1}\right|\right\}+c\\$

Question 13. Evaluate $\int x^2\sqrt{a^6-x^6}dx$

Solution:

I = $\int x^2\sqrt{a^6-x^6}dx$
Let us considered x³ = t
So, on differentiating, we get
3x²dx = dt
Therefore, I = $\frac{1}{3}\int\sqrt{a^6-t^2}dt\\ = \frac{1}{3}\left\{\frac{t}{2}\sqrt{a^6-t^2}+\frac{a^6}{2}sin^{-1}\left(\frac{t}{a^3}\right)\right\}+c\\$
Hence, I = $\frac{x^3}{6}\sqrt{a^6-x^6}+\frac{a^6}{6}sin^{-1}\left(\frac{x^3}{a^3}\right)+c$

Question 14. Evaluate $\int\sqrt{\frac{16+(logx)^2}{x}}dx$

Solution:

I = $\int\sqrt{\frac{16+(logx)^2}{x}}dx$
Let us considered logx = t
So, on differentiating, we get
1/x dx = dt
Therefore, I = $\int\sqrt{16+t^2}dt\\ =\int\sqrt{4^2+t^2}dt\\ =\frac{t}{2}\sqrt{16+t^2}+\frac{16}{2}log\left|t+\sqrt{16+t^2}\right|$
Hence, I = $\frac{logx}{2}\sqrt{16+(logx)^2}+8log\left|logx+\sqrt{16+(logx)^2}\right|+c$

Question 15. Evaluate $\int\sqrt{2ax-x^2}dx$

Solution:

I = $\int\sqrt{2ax-x^2}dx$
$=\int\sqrt{a^2-(a^2-ax+x^2)}dx\ \ \ \ \ -(Add\ and\ subtract\ a^2)\\ =\int\sqrt{a^2-(a-x)^2}dx\\ =\int\sqrt{a^2-(x-a)^2}dx\\ =\frac{(x-a)}{2}\sqrt{2ax-x^2}+\frac{a^2}{2}sin^{-1}\left(\frac{x-a}{a}\right)+c$
Therefore, I = $\frac{1}{2}(x-a)\sqrt{2ax-x^2}+\frac{a^2}{2}sin^{-1}\left(\frac{x-a}{a}\right)+c$

Question 16. Evaluate $\int\sqrt{3-x^2}dx$

Solution:

Let I = $\int\sqrt{3-x^2}dx$
$=\int\sqrt{(\sqrt{3})^2-x^2}dx$
I = $\frac{x}{2}\sqrt{3-x^2}+\frac{3}{2}sin^{-1}\left(\frac{x}{\sqrt{3}}\right)+c$

My Personal Notes

Last Updated : 11 Feb, 2021

Related Articles

Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.28

Question 1. Find $\int\sqrt{3+2x-x^2}dx$

Question 2. Evaluate $\int\sqrt{x^2+x+1}dx$

Question 3. Evaluate $\int\sqrt{x-x^2}dx$

Question 4. Evaluate $\int\sqrt{1+x-2x^2}dx$

Question 5. $\int cosx\sqrt{4-sin^2x}dx$

Question 6. Evaluate $\int e^x\sqrt{e^{2x}+1}dx$

Question 7. Evaluate $\int\sqrt{9-x^2}dx$

Question 8. Evaluate $\int\sqrt{16x^2+25}dx$

Question 9. Evaluate $\int\sqrt{4x^2-5}dx$

Question 10. Evaluate $\int\sqrt{2x^2+3x+4}dx$

Question 11. Evaluate $\int\sqrt{3-2x-2x^2}dx$

Question 12. Evaluate $\int x\sqrt{x^4+1}dx$

Question 13. Evaluate $\int x^2\sqrt{a^6-x^6}dx$

Question 14. Evaluate $\int\sqrt{\frac{16+(logx)^2}{x}}dx$

Question 15. Evaluate $\int\sqrt{2ax-x^2}dx$

Question 16. Evaluate $\int\sqrt{3-x^2}dx$

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Related Articles

Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.28

Question 1. Find

Question 2. Evaluate

Question 3. Evaluate

Question 4. Evaluate

Question 5.

Question 6. Evaluate

Question 7. Evaluate

Question 8. Evaluate

Question 9. Evaluate

Question 10. Evaluate

Question 11. Evaluate

Question 12. Evaluate

Question 13. Evaluate

Question 14. Evaluate

Question 15. Evaluate

Question 16. Evaluate

Please Login to comment...

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Question 1. Find $\int\sqrt{3+2x-x^2}dx$

Question 2. Evaluate $\int\sqrt{x^2+x+1}dx$

Question 3. Evaluate $\int\sqrt{x-x^2}dx$

Question 4. Evaluate $\int\sqrt{1+x-2x^2}dx$

Question 5. $\int cosx\sqrt{4-sin^2x}dx$

Question 6. Evaluate $\int e^x\sqrt{e^{2x}+1}dx$

Question 7. Evaluate $\int\sqrt{9-x^2}dx$

Question 8. Evaluate $\int\sqrt{16x^2+25}dx$

Question 9. Evaluate $\int\sqrt{4x^2-5}dx$

Question 10. Evaluate $\int\sqrt{2x^2+3x+4}dx$

Question 11. Evaluate $\int\sqrt{3-2x-2x^2}dx$

Question 12. Evaluate $\int x\sqrt{x^4+1}dx$

Question 13. Evaluate $\int x^2\sqrt{a^6-x^6}dx$

Question 14. Evaluate $\int\sqrt{\frac{16+(logx)^2}{x}}dx$

Question 15. Evaluate $\int\sqrt{2ax-x^2}dx$

Question 16. Evaluate $\int\sqrt{3-x^2}dx$