RD Sharma Class 12 Exercise 18.17 Indefinite Integrals Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 18.17 Indefinite Integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 18.17 Indefinite Integrals Solutions Maths - Download PDF Free Online

Kuldeep MauryaUpdated on 24 Jan 2022, 12:18 PM IST

RD Sharma class 12th exercise 18.17 is the top choice of students when it comes to NCERT solutions for maths. This book is a must-have for all students who want to improve their math skills and develop their chances of scoring high in board exams. Since in class 12 students have their board exams near, they should get hold of the RD Sharma class 12 chapter 18 exercise 18.17 solution to help them prepare for their upcoming exams. RD Sharma Solutions

The RD Sharma class 12 solutions Indefinite Integrals 18.17 is another great book from the maths solutions series which will be beneficial for all students who have opted for maths. The 18th chapter of the maths textbook will teach students

Graphs, reverse power rule, indefinite integrals, and common functions too. Exercise 18.17 has 9 questions that seek to build the general understanding of the chapter. The RD Sharma class 12th exercise 18.17 will help solve doubts on calculations and problem-solving techniques.

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.17

Indefinite Integrals Exercise Very Short Answer Question 2

Answer:

Answer:-

Hint:-To solve this problem, use special integration formula
Given:-

Solution:- Let

Put

Indefinite Integrals Exercise Very Short Answer Question 2 Maths Textbook Solutio
Edit Q

Indefinite Integrals Exercise Very Short Answer Question 3

Answer:-$\frac{1}{\sqrt{2}}\sin^{-1}\left ( \frac{x+1}{\sqrt{\frac{}{2}}} \right )+c$
Hint:-To solve this problem, use special integration formula.
Given:-$\int \frac{1}{\sqrt{5-4x-2x^{2}}}dx$
Solution:- Let $\int \frac{1}{\sqrt{5-4x-2x^{2}}}dx\Rightarrow \int \frac{1}{\sqrt{-\left \{ \left ( 2x^{2} +4x-5\right ) \right \}}}dx$
$\begin{gathered} =\int \frac{1}{\sqrt{-2\left\{x^{2}+2 x-\frac{5}{2}\right\}}} d x \Rightarrow \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{x^{2}+2 x \cdot 1+(1)^{2}-(1)^{2}-\frac{5}{2}\right\}}} d x \\ \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{(x+1)^{2}-1-\frac{5}{2}\right\}}} d x \Rightarrow \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{(x+1)^{2}-\left(1+\frac{5}{2}\right)\right\}}} d x \end{gathered}$
$\begin{aligned} &=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{(x+1)^{2}-\left(\frac{2+5}{2}\right)\right\}}} d x \Rightarrow \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{(x+1)^{2}-\frac{7}{2}\right\}}} d x\\ &=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\frac{7}{2}-(x+1)^{2}}} d x \Rightarrow \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(\sqrt{\frac{7}{2}}\right)^{2}-(x+1)^{2}}} d x\\ &=\frac{1}{\sqrt{2}} \int \frac{1}{\left.\sqrt{-\left\{(x+1)^{2}-\left(\frac{2+5}{2}\right)\right.}\right\}} d x \Rightarrow \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{(x+1)^{2}-\frac{7}{2}\right\}}} d x\\ &=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\frac{7}{2}-(x+1)^{2}}} d x \Rightarrow \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(\frac{\sqrt{7}}{2}\right)^{2}-(x+1)^{2}}} d x \end{aligned}$
Put $x+1=t\Rightarrow dx=dt$ then
$I=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(\frac{\sqrt{7}}{2}\right)^{2}-t^{2}}} d t \Rightarrow \frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{t}{\frac{\sqrt{7}}{2}}\right)+c \quad \quad \quad \quad \quad\left.\because \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c\right]\\ = \frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{x+1}{\frac{\sqrt{7}}{2}}\right)+c \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\quad{[\because t=x+1]} \quad \begin{aligned} \end{aligned}$

Indefinite Integrals Exercise 18.17 Question 4

Answer$\frac{1}{\sqrt{3}}\log \left | \left ( x+\frac{5}{6} \right ) +\sqrt{x^{2}+\frac{5x}{3}}+\frac{7}{3}\right |+c$
Hint:-To solve this problem, use special integration formula.
Given:-$\int \frac{1}{\sqrt{3x^{2}+5x+7}}dx$
Solution:-
Let
$\int \frac{1}{\sqrt{3x^{2}+5x+7}}dx\Rightarrow \int \frac{1}{\sqrt{3\left \{ x^{2}+\frac{5x}{3}+\frac{7}{3} \right \}}}dx$
$\frac{1}{\sqrt{3}}\int \frac{1}{\sqrt{x^{2}+2.x.\frac{5}{6}+\left ( \frac{5}{6} \right )^{2}-\left ( \frac{5}{6} \right )^{2}+\frac{7}{3}}}dx$
$\frac{1}{\sqrt{3}}\int \frac{1}{\sqrt{\left ( x+\frac{5}{6} \right )^{2}-\frac{25}{36}+\frac{7}{3}}}dx$
$\frac{1}{\sqrt{3}}\int \frac{1}{\sqrt{\left ( x+\frac{5}{6} \right )^{2}-\left ( \frac{25-84}{36} \right )}}dx$
$\frac{1}{\sqrt{3}}\int \frac{1}{\sqrt{\left ( x+\frac{5}{6} \right )^{2}+\frac{59}{36}}}dx$
Put
$x+\frac{5}{6}=t\Rightarrow dx=dt$
$\begin{gathered} I=\frac{1}{\sqrt{3}} \int \frac{1}{\sqrt{t^{2}+\left(\frac{\sqrt{59}}{6}\right)^{2}}} d x \\ =\frac{1}{\sqrt{3}} \log \left|t+\sqrt{t^{2}+\left(\frac{\sqrt{59}}{6}\right)^{2}}\right|+c \\ {\left[\because \int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c\right]} \end{gathered}$
$\begin{aligned} &=\frac{1}{\sqrt{3}} \log \left(x+\frac{5}{6}\right)+\sqrt{\left(x+\frac{5}{6}\right)^{2}+\frac{59}{36}} \mid+c \\ &=\frac{1}{\sqrt{3}} \log \left(x+\frac{5}{6}\right)+\sqrt{(x)^{2}+\left(\frac{5}{6}\right)^{2}+2 \cdot \frac{5}{6} x+\frac{59}{36} \mid+c} \quad\quad\quad\quad\quad\quad\quad\left[\because t=x+\frac{5}{6}\right] \\ &=\frac{1}{\sqrt{3}} \log \left(x+\frac{5}{6}\right)+\sqrt{x^{2}+\frac{5 x}{3}+\frac{25}{36}+\frac{59}{36}} \mid+c \end{aligned}$
$\begin{aligned} &=\frac{1}{\sqrt{3}} \log \left(x+\frac{5}{6}\right)+\sqrt{x^{2}+\frac{5 x}{3}+\frac{84}{36}} \mid+c \\ &=\frac{1}{\sqrt{3}} \log \left(x+\frac{5}{6}\right)+\sqrt{x^{2}+\frac{5 x}{3}+\frac{7}{3}} \mid+c \end{aligned}$

Indefinite Integrals Exercise 18.17 Question 5

Answer$\sin^{-1}\left ( \frac{2x-\alpha-\beta}{\beta -\alpha } \right )+c$
Hint:-To solve this problem, use special integration formula
Given:-$\int \frac{1}{\sqrt{\left ( 1-\alpha \right )\left ( \beta -x \right )}}dx,\left ( \beta > \alpha \right )$
Solution:-
Let $\begin{aligned} I=& \int \frac{1}{\sqrt{(x-\alpha)(\beta-x)}} d x=\int \frac{1}{\sqrt{x \beta-x^{2}-\alpha \beta+\alpha x}} d x \\ &=\int \frac{1}{\sqrt{-\left\{x^{2}-x(\alpha+\beta)+\alpha \beta\right\}}} d x \\ &=\int \frac{1}{\sqrt{-\left\{x^{2}-2 x \cdot \frac{\alpha+\beta}{2}+\left(\frac{\alpha+\beta}{2}\right)^{2}-\left(\frac{\alpha+\beta}{2}\right)^{2}+\alpha \beta\right\}}} d x \end{aligned}$
$\begin{aligned} &=\int \frac{1}{\sqrt{-\left[\left\{x-\left(\frac{\alpha+\beta}{2}\right)^{2}\right\}-\left\{\frac{\alpha^{2}+\beta^{2}+2 \alpha \beta}{4}-\alpha \beta\right\}\right]}} d x \\ &=\int \frac{1}{\sqrt{-\left[\left\{x-\left(\frac{\alpha+\beta}{2}\right)\right\}^{2}-\left\{\frac{\alpha^{2}+\beta^{2}+2 \alpha \beta-4 \alpha \beta}{4}\right\}\right]}} d x \\ &=\int \frac{1}{\sqrt{-\left[\left\{x-\left(\frac{\alpha+\beta}{2}\right)\right\}^{2}-\left\{\frac{\alpha^{2}+\beta^{2}-2 \alpha \beta}{4}\right\}\right]}} d x \end{aligned}$
$\begin{aligned} &=\int \frac{1}{\sqrt{-\left\{\left\{x-\left(\frac{\alpha+\beta}{2}\right)\right\}^{2}-\left(\frac{\alpha-\beta}{2}\right)^{2}\right]}} \Rightarrow \int \frac{1}{\sqrt{-\left[\left\{x-\left(\frac{\alpha+\beta}{2}\right)\right\}^{2}-\left(\frac{\beta-\alpha}{2}\right)^{2}\right]} d x}\quad\quad\quad \quad(\because \beta>\alpha) \\ &=\int \frac{1}{\sqrt{\left(\frac{\beta-\alpha}{2}\right)^{2}-\left\{x-\left(\frac{\alpha+\beta}{2}\right)\right\}^{2}}} d x \end{aligned}$
$\begin{aligned} &x-\left(\frac{\alpha+\beta}{2}\right)=t \Rightarrow d x=d t \text { then } \\ &\mathrm{I}=\int \frac{1}{\sqrt{\left(\frac{\beta-\alpha}{2}\right)^{2}-t^{2}}} d t=\sin ^{-1}\left(\frac{t}{\frac{\beta-\alpha}{2}}\right)+c \quad\quad\quad\quad\quad\left[\because \int \frac{1}{a^{2}-x^{2}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c\right] \end{aligned}$
$\begin{aligned} &=\sin ^{-1}\left(\frac{x-\left(\frac{\alpha+\beta}{2}\right)}{\frac{\beta-\alpha}{2}}\right)+c \Rightarrow \sin ^{-1}\left(\frac{2 x-(\alpha+\beta)}{\frac{2}{\frac{\beta-\alpha}{2}}}\right)+c\quad\quad\quad\quad\quad \quad\left[\because t=x-\left(\frac{\alpha+\beta}{2}\right)\right] \\ &=\sin ^{-1}\left(\frac{2 x-\alpha-\beta}{\beta-\alpha}\right)+c \end{aligned}$

Indefinite Integrals Exercise 18.17 Question 6

Answer:-$\frac{1}{\sqrt{2}}\sin^{-1}\left ( \frac{4x+3}{\sqrt{65}} \right )+c$
Hint:-to solve this problem, use special integration formula
Given:-$\int \frac{1}{\sqrt{7-3 x-2 x^{2}}} d x$
Solution:-
Let
$\begin{aligned} &I=\int \frac{1}{\sqrt{7-3 x-2 x^{2}}} d x \Rightarrow \int \frac{1}{\sqrt{-2\left\{x^{2}+\frac{3}{2} x-7 / 2\right\}}}dx \\ &=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{x^{2}+2 x \cdot \frac{3}{4}+\left(\frac{3}{4}\right)^{2}-\left(\frac{3}{4}\right)^{2}-7 / 2\right\}}}dx \\ &=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{\left(x+\frac{3}{4}\right)^{2}-\frac{9}{16}-\frac{7}{2}\right\}}} d x \end{aligned}$
$\begin{aligned} &= \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{\left(x+\frac{3}{4}\right)^{2}-\frac{9+56}{16}\right\}}} d x \\ &= \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{\left(x+\frac{3}{4}\right)^{2}-\frac{65}{16}\right\}}} d x \Rightarrow \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{+\frac{65}{16}-\left(x+\frac{3}{4}\right)^{2}}} d x \\ &= \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(\frac{\sqrt{65}}{4}\right)^{2}-\left(x+\frac{3}{4}\right)^{2}}} d x \end{aligned}$
$\begin{aligned} &\text { put } \mathrm{x}+\frac{3}{4}=t \Rightarrow d x=d t \text { then } \\ &I=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(\frac{\sqrt{65}}{4}\right)^{2}-t^{2}}} d t \Rightarrow \frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{t}{\left(\frac{\sqrt{65}}{4}\right)}\right)+c \quad\quad\quad\quad\left[\because \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c\right] \end{aligned}$
$\begin{aligned} &=\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{x+3 / 4}{\frac{\sqrt{65}}{4}}\right)+c &\quad\quad\quad\quad{[\because t=x+3 / 4]}\\ &=\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{4 x+3}{\frac{4}{\sqrt{65}}}{4}\right) +c \Rightarrow \frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{4 x+3}{\sqrt{65}}\right)+c\\ \\ \end{aligned}$

Indefinite Integrals Exercise 18.17Question 7

Answer:-$\sin^{-1}\left ( \frac{x+3}{5} \right )+c$
Hint: - To solve this problem, use special integration formula
Given:-$\int \frac{1}{\sqrt{16-6x-x^{2}}}dx$
Solution:-$Let\: I=\int \frac{1}{\sqrt{16-6x-x^{2}}}dx\Rightarrow \int \frac{1}{\sqrt{-\left \{ x^{2}+6x-16 \right \}}}dx$
$\begin{aligned} &=\int \frac{1}{\sqrt{-\left\{x^{2}+2 x .3+(3)^{2}-(3)^{2}-16\right\}}} d x \\ &=\int \frac{1}{\sqrt{-\left\{(x+3)^{2}-9-16\right\}}} d x=\int \frac{1}{\sqrt{-\left\{(x+3)^{2}-25\right\}}} d x \\ &\Rightarrow I=\int \frac{1}{\sqrt{25-(x+3)^{2}}} d x=\int \frac{1}{\sqrt{(5)^{2}-(x+3)^{2}}} d x \end{aligned}$
$\Gamma \begin{aligned} &\text { put } x+3=t \Rightarrow d x=d t \text { then }\quad\quad\quad\quad\quad\left[\because \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c\right]\\&I=\int \frac{1}{\sqrt{(5)^{2}-t^{2}}} d t=\sin ^{-1}\left(\frac{t}{5}\right)+c\\ &=\sin ^{-1}\left(\frac{x+3}{4}\right)+c \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad[\because t=x+3] \end{aligned}$

Indefinite Integrals Exercise 18.17 Question 8

Answer:-$\sin^{-1}\left ( \frac{x+3}{4} \right )+c$
Hint: - To solve this problem, use special integration formula
Given:-$\int \frac{1}{\sqrt{7-6x-x^{2}}}dx$
Solution:-
Let $I=\int \frac{1}{\sqrt{7-6x-x^{2}}}dx\Rightarrow \int \frac{1}{\sqrt{-\left \{ x^{2}+6x -7\right \}}}dx$
$\begin{aligned} &=\int \frac{1}{\sqrt{-\left\{x^{2}+2 x \cdot 3+(3)^{2}-(3)^{2}-7\right\}}} d x \\ &=\int \frac{1}{\sqrt{-\left\{(x+3)^{2}-9-7\right\}}} d x \Rightarrow \int \frac{1}{\sqrt{-\left\{(x+3)^{2}-16\right\}}} d x \\ &=\int \frac{1}{\sqrt{16-(x+3)^{2}}} d x \Rightarrow \int \frac{1}{\sqrt{(4)^{2}-(x+3)^{2}}} d x \\ &\text { put } \mathrm{x}+3=\mathrm{t} \Rightarrow \mathrm{dx}=\mathrm{dt} \text { then } \end{aligned}$
$\begin{aligned} &I=\int \frac{1}{\sqrt{(4)^{2}-t^{2}}} d t=\sin ^{-1}\left(\frac{t}{4}\right)+c \quad\quad\quad\quad\quad\left[\because \int \frac{1}{a^{2}-x^{2}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c\right] \\ &=\sin ^{-1}\left(\frac{x+3}{4}\right)+c \quad\quad\quad\quad\quad\quad\quad\quad\quad[\because t=x+3] \end{aligned}$

Indefinite Integrals Exercise 18.17 Question 9

Answer:-$\frac{1}{\sqrt{5}}\log \left | \left ( x-\frac{1}{5} \right )+\sqrt{x^{2}-\frac{2x}{5}} \right |+c$
Hint: - To solve this problem, use special integration formula
Given:-$\int \frac{1}{\sqrt{5x^{2}-2x}}dx$
Solution:-
$Let\: I=\int \frac{1}{\sqrt{5x^{2}-2x}}dx=\int \frac{1}{\sqrt{5\left ( x^{2}-\frac{2}{5}x \right )}}dx$
$\begin{aligned} &\Rightarrow I=\frac{1}{\sqrt{5}} \int \frac{1}{\sqrt{x^{2}-2 x \cdot \frac{1}{5}+\left(\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}} d x \\ &\Rightarrow I=\frac{1}{\sqrt{5}} \int \frac{1}{\sqrt{\left(x-\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}} d x \end{aligned}$
$\begin{aligned} &\text { put } \mathrm{x}-\frac{1}{5}=t \Rightarrow d x=d t \text { then } \\ &\mathrm{I}=\frac{1}{\sqrt{5}} \int \frac{1}{\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}} d t \\ &I=\frac{1}{\sqrt{5}} \log \left|t+\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}\right|+c \quad\quad\quad\quad\quad\quad\left[\because \int \frac{1}{\sqrt{x^{2}-a^{2}}} d x=\log \mid x+\sqrt{x^{2}-a|}+c\right] \end{aligned}$
$\begin{aligned} &I=\frac{1}{\sqrt{5}} \log \left|\left(x-\frac{1}{5}\right)+\sqrt{\left(x-\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}\right|+c \quad\quad\quad\quad\left(\because t=x-\frac{1}{5}\right) \\ &\Rightarrow I=\frac{1}{\sqrt{5}} \log \left|\left(x-\frac{1}{5}\right)+\sqrt{x^{2}-\frac{2 x}{5}+\frac{1}{25}-} \frac{1}{25}\right|+c \\ &\Rightarrow I=\frac{1}{\sqrt{5}} \log \left|\left(x-\frac{1}{5}\right)+\sqrt{x^{2}-\frac{2 x}{5}}\right|+c \end{aligned}$

Indefinite Integrals Exercise 18.17 Question 2

Answer:-$\sin^{-1}\left ( \frac{2x-3}{\sqrt{41}} \right )+c$
Hint:-To solve this problem, use special integration formula
Given:-$\int \frac{1}{\sqrt{8+3x-x^{2}}}dx$
Solution:-Let $I=\int \frac{1}{\sqrt{8+3x-x^{2}}}dx=\int \frac{1}{\sqrt{-\left ( x^{2}-3x-8 \right )}}dx$
$\begin{aligned} &\Rightarrow I=\int \frac{1}{\sqrt{-\left\{x^{2}-2 \cdot x \cdot \frac{3}{2}+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}-8\right\}}} d x \\ &=\int \frac{1}{\sqrt{-\left\{\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}-8\right\}}} d x=\int \frac{1}{\sqrt{-\left\{\left(x-\frac{3}{2}\right)^{2}-\left(\frac{9+32}{4}\right)\right\}}} d x \end{aligned}$

$\begin{aligned} &=\int \frac{1}{\sqrt{-\left\{\left(x-\frac{3}{2}\right)^{2}-\frac{41}{4}\right\}}} d x=\int \frac{1}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}}} d x \\ &=\int \frac{1}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}}} d x \end{aligned}$
Put $x-\frac{3}{2}=t\Rightarrow dx=dt\: then$
$I=\int \frac{1}{\sqrt{\left ( \frac{\sqrt{41}}{2} \right )^{2}-t^{2}}}dt=\sin^{-1}\left (\frac{t}{\frac{\sqrt{41}}{2}} \right )+c$ $\left [ \because \int \frac{1}{\sqrt{a^{2}-x^{2}}}dx=\sin^{-1}\left ( \frac{x}{a} \right )+c \right ]$
$=\sin^{-1}\left ( \frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}} \right )+c=\sin^{-1}\left ( \frac{\frac{2x-3}{2}}{\frac{\sqrt{41}}{2}} \right )+c$ $\left [ \because t=x-\frac{3}{2} \right ]$

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