RD Sharma Class 12 Ex 19.28 Solutions Chapter 19 Indefinite Integrals

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Here we provide RD Sharma Class 12 Ex 19.28 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.28 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.28
CategoryRD Sharma Solutions

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 ex = t

So, on differentiating, we get

exdx = 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

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

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