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.
Indefinite Integrals Excercise:18.17
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$ $\left [ \because t=x-\frac{3}{2} \right ]$The class 12 RD Sharma chapter 18 exercise 18.17 solution will be quite handy for students who like to practice at home and test their knowledge about the concerns they learn at school. Chapter 18 of the maths book is a lengthy and complex chapter with many sample questions. Hence, the answers in the RD Sharma class 12 solutions chapter 18 ex 18.17 will be of immense help to them.
RD Sharma class 12 solutions chapter 18 ex 18.17 contains accurate answers which are crafted by skilled professionals in mathematics. These answers employ some easy and modern calculations which will help students solve questions faster and more accurately. The answers in the RD Sharma class 12th exercise 18.17 can be extremely useful in testing their knowledge and marking their performance.
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