RD Sharma Class 12 Exercise 18.28 Indefinite Integrals Solutions Maths

RD Sharma Class 12 Exercise 18.28 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 10:17 AM IST

Class 12 RD Sharma chapter 18 exercise 18.28 solution is an accomplishment for each understudy. In any case, students can't get their optimal results without thorough practice. Therefore, the one book that has been recommended to students for preparing and alluding to for RD Sharma class 12 solutions chapter 18 exercise 18.28 arrangement. RD Sharma solutions To have phenomenal practice, you need solicitations, and RD Sharma Class 12 Solutions Indefinite Integrals Ex 18.28 meets a splendid fundamental of maths questions. RD Sharma Maths scrutinizing material is one of the most astounding course books for class 12 maths unequivocally for students planning for genuine tests.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.28

Indefinite Integrals Excercise 18.28 Question 1

Answer:-
$\frac{1}{2}(x-1) \sqrt{3+2 x-x^{2}}+2 \sin ^{-1}\left(\frac{x-1}{2}\right)+c$
Hint:-
Adding and subtracting with 1.
Given:-
$\int \sqrt{3+2 x-x^{2}} d x$
Solution:-
$\begin{aligned} &=\int \sqrt{4+2 x-x^{2}-1} d x \\\\ &=\int \sqrt{(2)^{2}-\left(x^{2}-2 x+1\right)} d x \\\\ &=\int \sqrt{(2)^{2}-(x-1)^{2}} d x \end{aligned}$
$\begin{aligned} &=\frac{1}{2}(x-1) \sqrt{3+2 x-x^{2}}+\frac{1}{2} \times(2)^{2} \sin ^{-1}\left(\frac{x-1}{2}\right)+c \\\\ &=\frac{1}{2}(x-1) \sqrt{3+2 x-x^{2}}+2 \sin ^{-1}\left(\frac{x-1}{2}\right)+c \end{aligned}$

Using the formula

$\int \sqrt{a^{2}+x^{2}} d x=\frac{1}{2} x \sqrt{a^{2}+x^{2}}+\frac{1}{2} a^{2} \sin ^{-1}\left(\frac{x}{2}\right)+c$

Indefinite Integrals Excercise 18.28 Question 2

Answer:-
$\left(\frac{2 x+1}{4}\right) \sqrt{x^{2}+x+1}+\frac{3}{8} \log \left|(2 x+1)+\sqrt{x^{2}+x+1}\right|+c$
Hint:-
$\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c$
Given:-
$\int \sqrt{x^{2}+x+1} d x$
Solution:-
$\begin{aligned} &\int \sqrt{x^{2}+x+1} d x \\\\\ &=\int \sqrt{x^{2}+x+\frac{1}{4}+\frac{3}{4}} d x \\\\ &=\int \sqrt{x^{2}+x+\left(\frac{1}{5}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x \end{aligned}$
$\begin{aligned} &=\int \sqrt{\left(x+\frac{1}{5}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x \\\\ &=\frac{\left(x+\frac{1}{5}\right)}{2} \sqrt{\left(x+\frac{1}{5}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}+\frac{\left(\frac{\sqrt{3}}{2}\right)^{2}}{2} \log \left|\left(x+\frac{1}{5}\right)+\sqrt{\left(x+\frac{1}{5}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\right|+c \end{aligned}$

Using the formula

$\begin{aligned} &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \\\\ &=\left(\frac{2 x+1}{4}\right) \sqrt{x^{2}+x+1}+\frac{3}{8} \log \left|\left(x+\frac{1}{5}\right)+\sqrt{\left(x+\frac{1}{5}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\right|+c \end{aligned}$
$\begin{aligned} &=\left(\frac{2 x+1}{4}\right) \sqrt{x^{2}+x+1}+\frac{3}{8} \log \left|\left(\frac{2 x+1}{2}\right)+\frac{1}{2} \sqrt{x^{2}+x+1}\right|+c \\\\ &=\left(\frac{2 x+1}{4}\right) \sqrt{x^{2}+x+1}+\frac{3}{8} \log \left|(2 x+1)+\sqrt{x^{2}+x+1}\right|+c \end{aligned}$

Indefinite Integrals Excercise 18.28 Question 3

Answer:-
$\frac{1}{2}(2 x-1) \sqrt{x-x^{2}}+\frac{1}{8} \sin ^{-1}(2 x-1)+c$
Hint:-
Add and subtract $\frac{1}{4}$
Given:-
$\int \sqrt{x-x^{2}} d x$
Solution:-
$\begin{aligned} &\int \sqrt{x-x^{2}} d x \\\\ &=\int \sqrt{\frac{1}{4}-\frac{1}{4}+x-x^{2}} d x \end{aligned}$

$\begin{aligned} &=\int \sqrt{\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}-x\right)^{2}} d x \\\\ &=\left(\frac{1-2 x}{4}\right) \sqrt{\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}-x\right)^{2}}-\frac{\left(\frac{1}{2}\right)^{2}}{2} \sin ^{-1} \frac{\left(\frac{1-2 x}{4}\right)}{\frac{1}{2}}+c \end{aligned}$
Using the formula
$\begin{aligned} &\int \sqrt{a^{2}-x^{2}}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c \\\\ &=\left(\frac{2 x-1}{4}\right) \sqrt{x-x^{2}}+\frac{1}{8} \sin ^{-1}(2 x-1)+c \end{aligned}$

Indefinite Integrals Excercise 18.28 Question 4

Answer:-
$\frac{1}{8}(4 x-1) \sqrt{1+x-2 x^{2}}+\frac{9 \sqrt{2}}{32} \sin ^{-1}(2 x-1)+c$

Multiplying by $\sqrt{2}$ .

Given:-
$\int \sqrt{1+x-2 x^{2}} d x$
Solution:-
$\int \sqrt{1+x-2 x^{2}} d x$
By multiplying by $\sqrt{2}$
$\begin{aligned} &=\sqrt{2} \int \sqrt{\frac{1}{2}+\frac{x}{2}-x^{2}} d x \\\\ &=\sqrt{2} \int \sqrt{\frac{9}{16}-\left(\frac{1}{16}-\frac{x}{2}+x^{2}\right)} d x \\\\ &=\sqrt{2} \int \sqrt{\left(\frac{3}{4}\right)^{2}-\left(x-\frac{1}{4}\right)^{2}} d x \end{aligned}$

$\begin{aligned} &=\sqrt{2}\left\{\left(\frac{\left(x-\frac{1}{4}\right)}{2}\right) \sqrt{\frac{1}{2}+\frac{x}{2}-x^{2}}+\frac{9}{32} \sin ^{-1}\left(\frac{x-\frac{1}{4}}{3 / 4}\right)^{2}\right\}+c \\\\ &=\frac{1}{8}(4 x-1) \sqrt{1+x-2 x^{2}}+\frac{9 \sqrt{2}}{32} \sin ^{-1}\left(\frac{4 x-1}{3}\right)+c \end{aligned}$
Using the formula
$\int \sqrt{a^{2}-x^{2}}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c$

Indefinite Integrals Exercise 18.28 Question 5

Answer:-
$\frac{1}{2} \sin x \sqrt{4-\sin ^{2} x}+2 \sin ^{-1} \frac{\sin x}{2}+c$
Hint:-
Taking $\sin x = t$
Given:-
$\int \cos x \sqrt{4-\sin ^{2} x} d x$
Solution:-
By taking $\sin x = t$
$\cos x d x=d t$
Hence
$\begin{aligned} &=\int \sqrt{4-t^{2}} d t \\\\ &=\int \sqrt{(2)^{2}-(t)^{2}} d t \\\\ &=\frac{1}{2} t \sqrt{4-t^{2}}+\frac{1}{2}(2)^{2} \sin ^{-1} \frac{t}{2}+c \end{aligned}$

Using the formula
$\begin{aligned} &\int \sqrt{a^{2}-x^{2}}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c \\\\ &=\frac{1}{2} \sin x \sqrt{4-\sin ^{2} x}+2 \sin ^{-1} \frac{\sin x}{2}+c \end{aligned}$

Indefinite Integrals Exercise 18.28 Question 6

Answer:-
$\frac{1}{2} e^{x} \sqrt{e^{2 x}+1}+\frac{1}{2} \log \left\{e^{x}+\sqrt{e^{2 x}+1}\right\}+c$
Hint:-
Let $e^{x}=z$
Given:-
$\int e^{x} \sqrt{e^{2 x}+1} d x$
Solution:-
By taking $e^{x}=z$
$e^{x} d x=z d x$
Hence
$\begin{aligned} &=\int \sqrt{z^{2}+1} d z \\\\ &=\int \sqrt{(z)^{2}+(1)^{2}} d z \\\\ &=\frac{1}{2} z \sqrt{z^{2}+1}+\frac{1}{2}(1)^{2} \log \left\{z+\sqrt{z^{2}+1}\right\}+c \end{aligned}$

Using the formula
$\begin{aligned} &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \\\\ &=\frac{1}{2} e^{x} \sqrt{e^{2 x}+1}+\frac{1}{2} \log \left\{e^{x}+\sqrt{e^{2 x}+1}\right\}+c \end{aligned}$

Indefinite Integrals Exercise 18.28 Question 7

Answer:-
$\frac{1}{2} x \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1} \frac{x}{3}+c$
Hint:-
Using the formula
$\int \sqrt{a^{2}-x^{2}}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c$
Given:-
$\int \sqrt{9-x^{2}} d x$
Solution:-
By using the formula
$\begin{aligned} &=\int \sqrt{(3)^{2}-(x)^{2}} d x \\\\ &=\frac{1}{2} x \sqrt{9-x^{2}}+\frac{1}{2}(3)^{2} \sin ^{-1} \frac{x}{3}+c \end{aligned}$
Using the formula
$\begin{aligned} &\int \sqrt{a^{2}-x^{2}}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c \\\\ &=\frac{1}{2} x \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1} \frac{x}{3}+c \end{aligned}$

Indefinite Integrals Exercise 18.28 Question 8

Answer:-
$2 x \sqrt{x^{2}+\frac{25}{16}}+\frac{25}{8} \log \left[x+\sqrt{x^{2}+\frac{25}{16}}\right]+c$
Hint:-
Taking common 4 and then use the formula.
Given:-
$\int \sqrt{16 x^{2}+25} d x$
Solution:-
$\begin{aligned} &=4 \int \sqrt{x^{2}+\frac{25}{16}} d x \\\\ &=4 \int \sqrt{(x)^{2}+\left(\frac{5}{4}\right)^{2}} d x \\\\ &=4\left[\frac{1}{2} x \sqrt{(x)^{2}+\left(\frac{5}{4}\right)^{2}}+\frac{1}{2} \times \frac{25}{16} \log \left[x+\sqrt{x^{2}+\frac{25}{16}}\right]+c\right] \end{aligned}$

Using the formula
$\begin{aligned} &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \\\\ &=2 x \sqrt{x^{2}+\frac{25}{16}}+\frac{25}{8} \log \left[x+\sqrt{x^{2}+\frac{25}{16}}\right]+c \end{aligned}$

Indefinite Integrals Exercise 18.28 Question 9

Answer:-
$2 x \sqrt{x^{2}-\frac{5}{4}}+\frac{5}{2} \log \left|x+\sqrt{x^{2}-\frac{5}{4}}\right|+c$
Hint:-
Taking common 2 and then use the formula.
Given:-
$\int \sqrt{4 x^{2}-5} d x$
Solution:-
$\begin{aligned} &=4 \int \sqrt{x^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}} d x \\\\ &=4\left[\frac{1}{2} x \sqrt{(x)^{2}-\frac{5}{4}}-\frac{1}{2} \times \frac{5}{4} \log \left[x+\sqrt{x^{2}-\frac{5}{4}}\right]+c\right] \end{aligned}$

By using the formula
$\begin{aligned} &\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+c \\\\ &=2 x \sqrt{x^{2}-\frac{5}{4}}+\frac{5}{2} \log \left|x+\sqrt{x^{2}-\frac{5}{4}}\right|+c \end{aligned}$

Indefinite Integrals Exercise 18.28 Question 10

Answer:-
$\frac{4 x+3}{8} \sqrt{2 x^{2}+3 x+4}+\frac{23 \sqrt{2}}{32} \log \left|x+\frac{3}{4} \sqrt{x^{2}-\frac{3}{2}+2}\right|+c$
Hint:-
Multiplying by $2\sqrt{2}$
Given:-
$\int \sqrt{2 x^{2}+3 x+4} d x$
Solution:-
$\begin{aligned} &=\frac{1}{2 \sqrt{2}} \int \sqrt{16 x^{2}+24 x+32} d x \\\\ &=\frac{1}{2 \sqrt{2}} \int \sqrt{(4 x)^{2}+2.4 x \cdot 3+(3)^{2}+23} d x \\\\ &=\frac{1}{2 \sqrt{2}} \int \sqrt{(4 x+3)^{2}+\sqrt{23}^{2}} d x \end{aligned}$

By using the formula

$\begin{aligned} &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \\\\ &=\frac{1}{2 \sqrt{2}}\left[\frac{1}{2} \times \frac{1}{4} \times(4 x+3) \sqrt{(4 x+3)^{2}}+\frac{\sqrt{23}^{2}}{2} \times \frac{1}{4} \log \left[(4 x+3)+\sqrt{(4 x+3)^{2}+\sqrt{23}}\right]+c\right] \end{aligned}$
$\begin{aligned} &=\frac{4 x+3}{8} \sqrt{2 x^{2}+3 x+4}+\frac{23 \sqrt{2}}{32} \log \left|(4 x+3)+\sqrt{2 x^{2}+3 x+4}\right|+c \\\\ &=\frac{4 x+3}{8} \sqrt{2 x^{2}+3 x+4}+\frac{23 \sqrt{2}}{32} \log \left|\left(x+\frac{3}{4}\right) \sqrt{x^{2}-\frac{3}{2}+2}\right|+\frac{23 \sqrt{2}}{32} \log 4+c \\\\ &=\frac{4 x+3}{8} \sqrt{2 x^{2}+3 x+4}+\frac{23 \sqrt{2}}{32} \log \left|x+\frac{3}{4} \sqrt{x^{2}-\frac{3}{2}+2}\right|+c \end{aligned}$

Indefinite Integrals Exercise 18.28 Question 11

Answer:-
The answer is
$\frac{1}{4}(2 x+1) \sqrt{3-2 x-2 x^{2}}+\frac{7 \sqrt{2}}{8} \sin ^{-1} \frac{2 x+1}{\sqrt{7}}+c$
Hints:-
$\begin{aligned} &\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c \\\\ &\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+c \\\\ &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{aligned}$

Given:-

$\int \sqrt{3-2 x-2 x^{2}} d x$
Solution:-
Let,
$\begin{aligned} &I=\int \sqrt{3-2 x-2 x^{2}} d x \\\\ &\therefore \int \sqrt{3-2\left(x^{2}+2\left(\frac{1}{2}\right) x\right)} d x=\int \sqrt{3-2\left(x^{2}+2\left(\frac{1}{2}\right) x+\left(\frac{1}{2}\right)^{2}\right)+2\left(\frac{1}{2}\right)^{2}} d x \end{aligned}$
We have
$I=\int \sqrt{\frac{7}{4}-2\left(x+\frac{1}{2}\right)^{2}} d x=\int \sqrt{2} \sqrt{\left(\frac{\sqrt{7}}{2}\right)^{2}-\left(x+\frac{1}{2}\right)^{2}} d x$
As I match with the form
$\begin{aligned} &\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c \\\\ &I=\sqrt{2}\left\{\frac{x+\frac{1}{2}}{2} \sqrt{\left(\frac{\sqrt{7}}{2}\right)^{2}-\left(x+\frac{1}{2}\right)^{2}}\right\}+\frac{\frac{7}{4}}{2} \sin ^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{7}}{2}}\right)+c \end{aligned}$
$I=\frac{1}{4}(2 x+1) \sqrt{2\left\{\left(\frac{\sqrt{7}}{2}\right)^{2}-\left(x+\frac{1}{2}\right)^{2}\right\}}+\frac{7 \sqrt{2}}{8} \sin ^{-1}\left(\frac{2 x+1}{\sqrt{7}}\right)+c$
$I=\frac{1}{4}(2 x+1) \sqrt{3-2 x-2 x^{2}}+\frac{7 \sqrt{2}}{8} \sin ^{-1}\left(\frac{2 x+1}{\sqrt{7}}\right)+c$

Indefinite Integrals Exercise 18.28 Question 12

Answer:-
The answer is
$I=\frac{x^{2}}{4} \sqrt{x^{4}+1}+\frac{1}{4} \log \left|x+x^{2} \sqrt{x^{4}+1}\right|+c$
Hints:-
$\begin{aligned} &\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c \\\\ &\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+c \\\\ &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{aligned}$
Given:-

$I=\int x \sqrt{x^{4}+1} d x$
Solution:-
Let,
$I=\int x \sqrt{x^{4}+1} d x$
Let, $x^{2}= t$
Differentiating both sides,
$\begin{aligned} &\Rightarrow 2 x d x=d t \\\\ &\Rightarrow x d x=\frac{1}{2} d t \end{aligned}$
Substituting $x^{2}$ with t, we have
$\begin{aligned} &I=\frac{1}{2} \int \sqrt{t^{2}+1} d t \\\\ &I=\frac{1}{2} \int \sqrt{t^{2}+1^{2}} d t \end{aligned}$
As I match with the form

$\begin{aligned} &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \\\\ &I=\frac{1}{2}\left\{\frac{t}{2} \sqrt{t^{2}+1}+\frac{1}{2} \log \left|t+\sqrt{t^{2}+1}\right|\right\}+c \\\\ &I=\frac{t}{4} \sqrt{t^{2}+1}+\frac{1}{4} \log \left|t+\sqrt{t^{2}+1}\right|+c \end{aligned}$
Putting the value of t back.
$\begin{aligned} &I=\frac{x^{2}}{4} \sqrt{\left(x^{2}\right)^{2}+1}+\frac{1}{4} \log \left|x^{2}+\sqrt{\left(x^{2}\right)^{2}+1}\right|+c \\\\ &I=\frac{x^{2}}{4} \sqrt{x^{4}+1}+\frac{1}{4} \log \left|x+x^{2} \sqrt{x^{4}+1}\right|+c \end{aligned}$

Indefinite Integrals exercise 18.28 question 13

Answer:-
$\frac{x^{3}}{6} \sqrt{a^{6}-x^{6}}+\frac{a^{6}}{6} \sin ^{-1}\left(\frac{x^{3}}{a^{3}}\right)+c$
Hints:-
$\begin{aligned} &\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c \\\\ &\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+c \\\\ &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{aligned}$

Given:-

$\int x^{2} \sqrt{a^{6}-x^{6}} d x$
Solution:-
Let,
$\begin{aligned} &I=\int x^{2} \sqrt{a^{6}-x^{6}} d x \\\\ &I=\int x^{2} \sqrt{a^{6}-\left(x^{3}\right)^{2}} d x \end{aligned}$
Let, $x^{3}= t$
Differentiating both sides,
$\begin{aligned} &\Rightarrow 3 x^{2} d x=d t \\\\ &\Rightarrow x^{2} d x=\frac{1}{3} d t \end{aligned}$

Substituting $x^{3}$ with t, we have
$\begin{aligned} &I=\frac{1}{3} \int \sqrt{\left(a^{3}\right)^{2}-t^{2}} d t \\\\ &I=\frac{1}{2} \int \sqrt{\left(a^{3}\right)^{2}-t^{2}} d t \end{aligned}$

As I match with the form
$\begin{aligned} &\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c \\\\ &\therefore I=\frac{1}{3}\left\{\frac{t}{2} \sqrt{a^{6}-t^{2}}+\frac{a^{6}}{2} \sin ^{-1}\left(\frac{t}{a^{3}}\right)\right\}+c \\\\ &I=\frac{t}{4} \sqrt{t^{2}+1}+\frac{1}{4} \log \left|t+\sqrt{t^{2}+1}\right|+c \end{aligned}$

Putting the value of t i.e.t=x3

$I=\frac{x^{3}}{6} \sqrt{a^{6}-x^{6}}+\frac{a^{6}}{6} \sin ^{-1}\left(\frac{x^{3}}{a^{3}}\right)+c$

Indefinite Integrals exercise 18.28 question 14

Answer:-
The answer is
$I=\frac{\log x}{2} \sqrt{(\log x)^{2}+16}+8 \log \left|\log x+\sqrt{(\log x)^{2}+16}\right|+c$
Hints:-
$\begin{aligned} &\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c \\\\ &\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+c \\\\ &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{aligned}$

Given:-

$I=\int \frac{\sqrt{(\log x)^{2}+16}}{x} d x$
Solution:-
Let,
$I=\int \frac{1}{x} \sqrt{16+(\log x)^{2}} d x$
Let, $\log x=t$
Differentiating both sides,
$\Rightarrow \frac{1}{x} d x=d t$
Substituting logx with t, we have
$\begin{aligned} &I=\int \sqrt{t^{2}+16} d t \\\\ &I=\int \sqrt{t^{2}+4^{2}} d t \end{aligned}$

As I match with the form
$\begin{aligned} &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \\\\ &I=\left\{\frac{t}{2} \sqrt{t^{2}+16}+\frac{16}{2} \log \left|t+\sqrt{t^{2}+16}\right|\right\}+c \end{aligned}$

Putting the value of t back.

$I=\frac{\log x}{2} \sqrt{(\log x)^{2}+16}+8 \log \left|\log x+\sqrt{(\log x)^{2}+16}\right|+c$

Indefinite Integrals exercise 18.28 question 15

Answer:-
$\frac{1}{2}(x-a) \sqrt{2 a x-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x-a}{a}\right)+c$
Hint:-
Using the formula
$\begin{aligned} &\int \sqrt{a^{2}-x^{2}}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+c \\\\ &\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+c \\\\ &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{aligned}$

Given:-

$\int \sqrt{2 a x-x^{2}} d x$
Solution:-
Let,
$\begin{aligned} & I=\int \sqrt{2 a x-x^{2}} d x \\ &\therefore I=\int \sqrt{-\left(x^{2}-2 a x\right)} d x=\int \sqrt{(a)^{2}-\left(x^{2}-2 a x+a^{2}\right)} d x \end{aligned}$
We have
$I=\int \sqrt{(a)^{2}-(x-a)^{2}} d x$
As I match with the form
$\begin{aligned} &\int \sqrt{a^{2}-x^{2}}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+c \\\\ &I=\frac{x-a}{2} \sqrt{a^{2}-(x-a)^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x-a}{a}\right)+c \\\\ &I=\frac{1}{2}(x-a) \sqrt{2 a x-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x-a}{a}\right)+c \end{aligned}$

Indefinite Integrals Exercise 18.28 Question 16

Answer:-
$I=\frac{x}{2} \sqrt{3-x^{2}}+\frac{3}{2} \sin ^{-1}\left(\frac{x}{\sqrt{3}}\right)+c$
Hint:-
Using the formula
$\begin{aligned} &\int \sqrt{a^{2}-x^{2}}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+c \\\\ &\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+c \\\\ &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{aligned}$

Given:-

$\int \sqrt{3-x^{2}} d x$
Solution:-
$\begin{aligned} &\text { Let, } I=\int \sqrt{3-x^{2}} d x \\ &\therefore I=\int \sqrt{3-x^{2}} d x=\int \sqrt{(\sqrt{3})^{2}-x^{2}} d x \end{aligned}$
As I match with the form
$\begin{aligned} &\int \sqrt{a^{2}-x^{2}}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+c \\\\ &I=\frac{x}{2} \sqrt{3-x^{2}}+\frac{3}{2} \sin ^{-1}\left(\frac{x}{\sqrt{3}}\right)+c \end{aligned}$

Indefinite Integrals Exercise 18.28 Question 17

Answer:-
$\frac{x-1}{2} \sqrt{x^{2}-2 x}-\frac{1}{2} \log \left|x-1+\sqrt{x^{2}-2 x}\right|+c$
Hint:-
$\begin{aligned} &\int \sqrt{a^{2}-x^{2}}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+c \\\\ &\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+c \\\\ &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{aligned}$

Given:-

$\int \sqrt{x^{2}-2 x} d x$
Solution:-
Let, $I=\int \sqrt{x^{2}-2 x} d x$
We have
$\begin{aligned} &I=\int \sqrt{x^{2}-2 x} d x \\ &I=\int \sqrt{x^{2}-2 x+1^{2}-1^{2}} d x \\ &I=\int \sqrt{(x-1)^{2}-(1)^{2}} d x \end{aligned}$

As I match with the form
$\begin{aligned} &\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+c \\\\ &I=\frac{x-1}{2} \sqrt{(x-1)^{2}-1}-\frac{1}{2} \log \left|x-1+\sqrt{(x-1)^{2}-1}\right|+c \\\\ &I=\frac{x-1}{2} \sqrt{x^{2}-2 x}-\frac{1}{2} \log \left|x-1+\sqrt{x^{2}-2 x}\right|+c \end{aligned}$

Indefinite Integrals Exercise 18 point 28 Question 18

Answer:-
$I=\frac{1}{2}(x-1) \sqrt{2 x-x^{2}}+\frac{1}{2} \sin ^{-1}((x-1))+c$
Hint:-
Using the formula
$\begin{aligned} &\int \sqrt{a^{2}-x^{2}}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+c \\\\ &\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+c \\\\ &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{aligned}$

Given:-

$\int \sqrt{2 x-x^{2}} d x$
Solution:-
$\begin{aligned} &\text { Let, } I=\int \sqrt{2 x-x^{2}} d x \\\\ &\therefore I=\int \sqrt{-\left(x^{2}-2(1) x\right)} d x \end{aligned}$
$\begin{aligned} &I=\int \sqrt{1^{2}-(x-a)^{2}} d x \\\\ &I=\int \sqrt{(1)^{2}-(x-1)^{2}} d x \end{aligned}$
As I match with the form $\begin{aligned} &\int \sqrt{a^{2}-x^{2}}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+c \\\\ &I=\frac{x-1}{2} \sqrt{(1)^{2}-(x-1)^{2}}+\frac{1^{2}}{2} \sin ^{-1}\left(\frac{x-1}{1}\right)+c \\\\ &I=\frac{1}{2}(x-1) \sqrt{2 x-x^{2}}+\frac{1}{2} \sin ^{-1}(x-1)+c \end{aligned}$

Rd Sharma class 12 chapter 18 exercise 18.28 answers that are clear and reviewed. It further helps students with getting a handle on formulae and tending to methods. Rd Sharma Class 12th exercise 18.28 has around 18 inquiries, including its subparts, and it joins themes like: -

Questions Related to antiderivative
Joining of mathematical capacities
Combination of outstanding capacities
Random issues
Mathematical translation of endless vital
Correlation among separation and combination
Methods of combination

Benefits of picking RD Sharma Mathematics Solutions from Career360 include:

Career360, you will need to get all of the solutions, so no convincing motivation to go somewhere else.
You can similarly benchmark your display premise with these solutions.
These solutions are available free of cost in career360.
Extraordinary yet arranged showing of the subjects.
A point by point clarification of thoughts and formulae
Assists understudies to rehearse with no issue
You can likewise benchmark your exhibition dependent on these solutions.

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download E-book

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

1. Does Career360 provide RD Sharma class 12 chapter 18 exercise 18.28 solution?

Career360 gives point-by-point and accurate solutions for RD Sharma class 12 chapter 18 ex 18.28. Subject matter experts provide RD Sharma Class 12 Solutions to work with a smooth and clear awareness of thoughts.

2. Are the RD Sharma Class 12 Maths Solutions Chapter 18 agreeable for CBSE students?

It's essentially suggested that class 12 RD Sharma chapter 18 exercise 18.28 plan students pick the RD Sharma Class 12 Solutions from Career360 as reference material for solutions.

3. How do students get the RD Sharma class 12th exercise 18.28 Solutions?

Students can get to the RD Sharma Solutions for Class 12 Maths open as a downloadable PDF at the Career360 webpage. In this RD Sharma class, 12 solutions chapter 18 ex 18.28 helps ace subject advisers to help students in their solutions.

4. How many requests are there in class 12 RD Sharma chapter 18?

There are 18 requests in Class 12 RD Sharma chapter 18 exercise 18.28.