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

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

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

Kuldeep MauryaUpdated on 24 Jan 2022, 10:17 AM IST

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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

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