RD Sharma Class 12 Exercise 18.10 Indefinite Integrals Solutions Maths

RD Sharma Class 12 Exercise 18.10 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:05 PM IST

RD Sharma books are some of the oldest and renowned maths books of all time in India. They are the most detailed and exam-oriented books for maths. RD Sharma books are widely used all over the country by CBSE schools for studies and exam paper setting.

RD Sharma Class 12th Exercise 18.10 contains the chapter ‘Indefinite Integrals.’ This is a small exercise consisting of only ten questions, from which eight are Level 1, and two are Level 2. The level one questions are simple and easy to solve similarly, level two questions have less complexity but are somewhat lengthy and are based on integrating the functions w.r.t to x.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.10
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.10

Indefinite Integrals Exercise 18.10 Question 1

Answer: $\frac{2}{7}(x+2)^{\frac{7}{2}}-\frac{8}{5}(x+2)^{\frac{5}{2}}+\frac{8}{3}(x+2)^{\frac{3}{2}}+c$
Hint: To solve this type of problem, we use substitution method. Let $x+2=t$ then solve the integral by general method.
Given: $\int x^{2} \sqrt{x+2} d x$
Solution:
Let $I=\int x^{2} \sqrt{x+2} d x$
Substitute $x+2=t \Rightarrow d x=d t$
$\begin{aligned} I &=\int(t-2)^{2} \sqrt{t} d t \qquad\qquad(\because x=t-2) \\ & \end{aligned}$
$=\int\left(t^{2}+4-4 t\right) t^{\frac{1}{2}} d t \qquad\qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \\$
$=\int\left(t^{2} \cdot t^{\frac{1}{2}}+4 t^{\frac{1}{2}}-4 t . t^{\frac{1}{2}}\right) d t \\$
$=\int\left(t^{\frac{5}{2}}+4 t^{\frac{1}{2}}-4 t^{\frac{3}{2}}\right) d t \qquad\qquad\left(\because a^{m} \cdot a^{n}=a^{m+n}\right)$
$\begin{array}{r} =\int t^{\frac{5}{2}} d t+4 \int t^{\frac{1}{2}} d t-4 \int t^{\frac{3}{2}} d t \\ \end{array}$
$\quad\left[\because \int\{f(x) \pm a g(x)\} d x=\int f(x) d x \pm a \int g(x) d x\right]$
$=\frac{t^{\frac{5}{2}+1}}{\frac{5}{2}+1}+4 \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}-4 \frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}+c \qquad\qquad\left[\because \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$
$\begin{aligned} &=\frac{t^{\frac{7}{2}}}{\frac{7}{2}}+4 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}-4 \frac{t^{\frac{5}{2}}}{\frac{5}{2}}+c \\ & \end{aligned}$
$=\frac{2}{7}(t)^{\frac{7}{2}}-4 \cdot \frac{2}{5}(t)^{\frac{5}{2}}+4 \frac{2}{3}(t)^{\frac{3}{2}}+c \\$
$=\frac{2}{7}(x+2)^{\frac{7}{2}}-\frac{8}{5}(x+2)^{\frac{5}{2}}+\frac{8}{3}(x+2)^{\frac{3}{2}}+c \qquad \qquad[\because x+2=t]$

Indefinite Integrals Exercise 18.10 Question 2

Answer: $\frac{2}{5}(x-1)^{\frac{5}{2}}+2(x-1)^{\frac{1}{2}}+\frac{4}{3}(x-1)^{\frac{3}{2}}+c$
Hint: Use substitution method to solve this type of integral
Given: $\int \frac{x^{2}}{\sqrt{x-1}} d x$
Solution:
Let $I=\int \frac{x^{2}}{\sqrt{x-1}} d x$
Substitute $x-1=t \Rightarrow d x=d t$ then
$\begin{aligned} I &=\int \frac{(t+1)^{2}}{\sqrt{t}} d t \qquad(\because x=t+1) \\ & \end{aligned}$
$=\int \frac{t^{2}+1+2 t}{\sqrt{t}} d t \qquad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \\$
$=\int\left(\frac{t^{2}}{\sqrt{t}}+\frac{1}{\sqrt{t}}+2 \frac{t}{\sqrt{t}}\right) d t \\$
$=\int\left(t^{2-\frac{1}{2}}+t^{-\frac{1}{2}}+2 t^{1-\frac{1}{2}}\right) d t$
$\begin{aligned} &=\int\left(t^{\frac{3}{2}}+t^{-\frac{1}{2}}+2 t^{\frac{1}{2}}\right) d t \\ & \end{aligned}$
$=\int t^{\frac{3}{2}} d t+\int t^{-\frac{1}{2}} d t+2 \int t^{\frac{1}{2}} d t$
$\begin{aligned} &=\frac{t^{\frac{3}{2}}{2}}{\frac{3}{2}+1}+\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+2 \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c \qquad\left[\because f^{n} d t=\frac{t^{n+1}}{n+1}+c\right] \\ & \end{aligned}$
$=\frac{t^{\frac{5}{2}}}{\frac{5}{2}}+\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+2 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c$
$\begin{aligned} &=\frac{2}{5}(t)^{\frac{5}{2}}+2(t)^{\frac{1}{2}}+2 \cdot \frac{2}{3}(t)^{\frac{3}{2}}+c \\ & \end{aligned}$
$=\frac{2}{5}(x-1)^{\frac{5}{2}}+2(x-1)^{\frac{1}{2}}+\frac{4}{3}(x-1)^{\frac{3}{2}}+c \qquad[\because x-1=t]$

Indefinite Integrals Exercise 18.10 Question 3

Answer: $\frac{2}{135}(3 x+4)^{\frac{5}{2}}-\frac{16}{81}(3 x+4)^{\frac{3}{2}}+\frac{32}{27}(3 x+4)^{\frac{1}{2}}+c$
Hint: Use substitution method to solve this type of integral
Given: $\int \frac{x^{2}}{\sqrt{3 x+4}} d x$
Solution:
$I=\int \frac{x^{2}}{\sqrt{3 x+4}} d x$
Put $3 x+4=t \Rightarrow 3 d x=d t \Rightarrow d x=\frac{d t}{3}$ then
$\begin{aligned} &I=\int \frac{\frac{1}{3^{2}}(t-4)^{2}}{\sqrt{t}} \frac{d t}{3} \qquad\left(\because x=\frac{t-4}{3}\right) \\ & \end{aligned}$
$\Rightarrow I=\frac{1}{3 \times 3^{2}} \int \frac{t^{2}+16-8 t}{\sqrt{t}} d t \qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$\begin{aligned} &\Rightarrow I=\frac{1}{27} \int\left\{\frac{t^{2}}{\sqrt{t}}+16 \frac{1}{\sqrt{t}}-8 \cdot \frac{t}{\sqrt{t}}\right\} d t \\ & \end{aligned}$
$\Rightarrow I=\frac{1}{27} \int\left\{t^{2-\frac{1}{2}}+16 t^{-\frac{1}{2}}-8 t^{1-\frac{1}{2}}\right\} d t$
$\begin{aligned} &\Rightarrow I=\frac{1}{27} \int\left(t^{\frac{3}{2}}-8 t^{\frac{1}{2}}+16 t^{-\frac{1}{2}}\right) d t \\ & \end{aligned}$
$\Rightarrow I=\frac{1}{27}\left[\int t^{\frac{3}{2}} d t-8 \int t^{\frac{1}{2}} d t+16 \int t^{-\frac{1}{2}} d t\right]$
$\Rightarrow I=\frac{1}{27}\left[\frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}-8 \cdot \frac{t^{\frac{1}{2}}{2}+1} + 16 \cdot \frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\right] \qquad\left[\because f t^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$
$\begin{aligned} &\Rightarrow I=\frac{1}{27}\left[\frac{t^{\frac{5}{2}}}{\frac{5}{2}}-8 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+16 \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\right] \\ & \end{aligned}$
$\Rightarrow I=\frac{1}{27}\left[\frac{2}{5}(t)^{\frac{5}{2}}-8 \cdot \frac{2}{3}(t)^{\frac{3}{2}}+16.2(t)^{\frac{1}{2}}+c\right] \\$
$\Rightarrow I=\frac{2}{135}(3 x+4)^{\frac{5}{2}}-\frac{16}{81}(3 x+4)^{\frac{3}{2}}+\frac{32}{27}(3 x+4)^{\frac{1}{2}}+c$

Indefinite Integrals exercise 18.10 question 4

Answer: $2 \log |(x-1)|-\frac{1}{x-1}+c$
Hint: Use substitution method to solve this type of integral
Given: $\int \frac{2 x-1}{(x-1)^{2}} d x$
Solution:
Let $I=\int \frac{2 x-1}{(x-1)^{2}} d x$
Put $x-1=t \Rightarrow d x=d t$ then
$I=\int \frac{2(t+1)-1}{(t)^{2}} d t$
$\begin{aligned} &\Rightarrow I=\int \frac{2 t+2-1}{t^{2}} d t=\int \frac{2 t+1}{t^{2}} d t \\ & \end{aligned}$
$\Rightarrow I=\int \frac{2 t}{t^{2}}+\frac{1}{t^{2}} d t=\int\left\{2 t^{1-2}+t^{-2}\right\} d t \\$
$\Rightarrow I=\int\left(2 t^{-1}+t^{-2}\right) d t=\int\left\{\frac{2}{t}+t^{-2}\right\} d t \\$
$\Rightarrow I=2 \int \frac{1}{t} d t+\int t^{-2} d t$
$\Rightarrow I=2 \log |t|+\frac{t^{-2+1}}{-2+1}+c \qquad\left[\int \frac{1}{t} d t=\log |t|+c \& \int f^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$
$\begin{aligned} &\Rightarrow I=2 \log |t|+\frac{t^{-1}}{-1}+c \\ & \end{aligned}$
$\Rightarrow I=2 \log |(x-1)|-(x-1)^{-1}+c \\$
$\therefore I=2 \log |(x-1)|-\frac{1}{x-1}+c$

Indefinite Integrals exercise 18.10 question 5

Answer: $\frac{4}{7}(x+2)^{\frac{7}{2}}-\frac{16}{5}(x+2)^{\frac{5}{2}}+\frac{22}{3}(x+2)^{\frac{3}{2}}+c$
Hint: Use substitution method to solve this type of integral
Given: $\int\left(2 x^{2}+3\right) \sqrt{x+2} d x$
Solution: Let $I=\int\left(2 x^{2}+3\right) \sqrt{x+2} d x$
Put $x+2=t \Rightarrow d x=d t$
$\begin{aligned} &I=\int\left\{2(t-2)^{2}+3\right\} \sqrt{t} d t \qquad(\because x=t-2) \\ & \end{aligned}$
$\Rightarrow I=\int\left\{2\left(t^{2}-4 t+4\right)+3\right\} \sqrt{t} d t \qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$\begin{aligned} &\Rightarrow I=\int\left\{2 t^{2}-8 t+8+3\right\} \sqrt{t} d t \\ & \end{aligned}$
$\Rightarrow I=\int\left(2 t^{2}-8 t+11\right) \sqrt{t} d t \\$
$\Rightarrow I=\int\left(2 t^{2} \sqrt{t}-8 t \cdot \sqrt{t}+11 \sqrt{t}\right) d t$
$\begin{aligned} &\Rightarrow I=\int\left(2 t^{2+\frac{1}{2}}-8 t^{1+\frac{1}{2}}+11 t^{\frac{1}{2}}\right) d t \\ & \end{aligned}$
$\Rightarrow I=\left[2 \int t^{\frac{5}{2}} d t-8 \int t^{\frac{3}{2}} d t+11 \int t^{\frac{1}{2}} d t\right]$
$\Rightarrow I=\left[2 \cdot \frac{t^{\frac{5}{2}+1}}{\frac{5}{2}+1}-8 \cdot \frac{t^{\frac{3}{2}}{ } \frac{1}{2}+1} + 11 \cdot \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\right] \qquad\left[\because \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$
$\begin{aligned} &\Rightarrow I=\left[2 \cdot \frac{t^{\frac{7}{2}}}{\frac{7}{2}}-8 \cdot \frac{t^{\frac{5}{2}}}{\frac{5}{2}}+11 \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\right] \\ & \end{aligned}$
$\Rightarrow I=2 \cdot \frac{2}{7} t^{\frac{7}{2}}-8 \cdot \frac{2}{5} t^{\frac{5}{2}}+11 \cdot \frac{2}{3} t^{\frac{3}{2}}+c \\$
$\Rightarrow I=\frac{4}{7}(x+2)^{\frac{7}{2}}-\frac{16}{5}(x+2)^{\frac{5}{2}}+\frac{22}{3}(x+2)^{\frac{3}{2}}+c$

Indefinite Integrals exercise 18.10 question 6

Answer: $(x+1)+\log |(x+1)|+\frac{1}{x+1}+c$
Hint: Use substitution method to solve this type of integral
Given: $\int \frac{x^{2}+3 x+1}{(x+1)^{2}} d x$
Solution: let $I=\int \frac{x^{2}+3 x+1}{(x+1)^{2}} d x$
Substitute $x+1=t \Rightarrow d x=d t$ then
$\Rightarrow I=\int \frac{(t-1)^{2}+3(t-1)+1}{t^{2}} d t \qquad(\because x=t-2)$
$\begin{aligned} &\Rightarrow I=\int \frac{t^{2}-2 t+1+3 t-3+1}{t^{2}} d t \qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \\ & \end{aligned}$
$\Rightarrow I=\int \frac{t^{2}+t-1}{t^{2}} d t \\$
$\Rightarrow I=\int \frac{t^{2}}{t^{2}}+\frac{t}{t^{2}}-\frac{1}{t^{2}} d t$
$\begin{aligned} &\Rightarrow I=\int\left(1+\frac{1}{t}-t^{-2}\right) d t \\ & \end{aligned}$
$\Rightarrow I=\int t^{0} d t+\int \frac{1}{t} d t-\int t^{-2} d t$
$\Rightarrow I=\frac{t^{0+1}}{0+1}+\log |t|-\frac{t^{-2+1}}{-2+1}+c \qquad\left[\int \frac{1}{t} d t=\log |t|+c \& \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$
$\begin{aligned} &\Rightarrow I=t+\log |t|-\frac{t^{-1}}{-1}+c \\ & \end{aligned}$
$\Rightarrow I=t+\log |t|+\frac{1}{t}+c \\$
$\therefore I=(x+1)+\log |(x+1)|+\frac{1}{x+1}+c$

Indefinite Integrals Excercise 18.10 Question 7

Answer: $-\frac{2}{5}(1-x)^{\frac{5}{2}}+\frac{4}{3}(1-x)^{\frac{3}{2}}+-2(1-x)^{\frac{1}{2}}+c$
Hint: Use substitution method to solve this type of integral
Given: $\int \frac{x^{2}}{\sqrt{1-x}} d x$
Solution: Let $I=\int \frac{x^{2}}{\sqrt{1-x}} d x$
Substitute $1-x=t \Rightarrow d x=-d t$ then
$\begin{aligned} &I=\int \frac{(1-t)^{2}}{\sqrt{t}}(-d t) \qquad(\because x=1-t) \\ & \end{aligned}$
$\Rightarrow I=-\int\left(\frac{1+t^{2}-2 t}{\sqrt{t}}\right) d t \qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$\begin{aligned} &\Rightarrow I=-\int\left\{\frac{1}{\sqrt{t}}+\frac{t^{2}}{\sqrt{t}}-\frac{2 t}{\sqrt{t}}\right\} d t \\ & \end{aligned}$
$\Rightarrow I=-\int\left(t^{\frac{-1}{2}}+t^{2-\frac{1}{2}}-2 t^{1-\frac{1}{2}}\right) d t \\$
$\Rightarrow I=-\left[\int t^{-\frac{1}{2}} d t+\int t^{\frac{3}{2}} d t-2 \int t^{\frac{1}{2}} d t\right.$
$\Rightarrow I=-\left[\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}-2 \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\right] \qquad\left[\because \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$
$\begin{aligned} &\Rightarrow I=-\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\frac{t^{\frac{5}{2}}}{\frac{5}{2}}-2 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\right] \\ & \end{aligned}$
$\Rightarrow I=-\left[2 . t^{\frac{1}{2}}+\frac{2}{5} t^{\frac{5}{2}}-2 \cdot \frac{2}{3} t^{\frac{3}{2}}\right]+c \\$
$\Rightarrow I=-2(1-x)^{\frac{1}{2}}-\frac{2}{5}(1-x)^{\frac{5}{2}}+\frac{4}{3}(1-x)^{\frac{3}{2}}+c \qquad[\because 1-x=t]$
$\therefore I=-\frac{2}{5}(1-x)^{\frac{5}{2}}+\frac{4}{3}(1-x)^{\frac{3}{2}}+-2(1-x)^{\frac{1}{2}}+c$

Indefinite Integrals Excercise 18.10 Question 7

Indefinite Integrals Excercise 18.10 Question 8

Answer: $\frac{(1-x)^{25}}{25}-\frac{(1-x)^{24}}{24}+c$
Hint: Use substitution method to solve this type of integral
Given: $\int x(1-x)^{23} d x$
Solution: let $I=\int x(1-x)^{23} d x$
Substitute $1-x=t \Rightarrow-d x=d t$ then
$\begin{aligned} &I=\int(1-t) \cdot t^{23}(-d t) \qquad(\because x=1-t) \\ & \end{aligned}$
$\Rightarrow I=-\int(1-t) \cdot t^{23} d t=-\int t^{23}-t . t^{23} d t \\$
$\Rightarrow I=-\int\left(t^{23}-t^{24}\right) d t=-\int t^{23} d t+\int t^{24} d t$
$\begin{aligned} &\Rightarrow I=-\frac{t^{23+1}}{23+1}+\frac{t^{24+1}}{24+1}+c \qquad\left[\because \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right] \\ & \end{aligned}$
$\Rightarrow I=\frac{t^{25}}{25}-\frac{t^{24}}{24}+c \\$
$\therefore I=\frac{(1-x)^{25}}{25}-\frac{(1-x)^{24}}{24}+c$

Indefinite Integrals Excercise 18.10 Question 9

Answer: $2 \sqrt{x}-4 x^{\frac{1}{4}}+4 \log \left|1+x^{\frac{1}{4}}\right|+c$
Hint: Use substitution method to solve this type of integral
Given: $\int \frac{1}{\sqrt{x}+\sqrt[4]{x}} d x$
Solution:
Let $I=\int \frac{1}{\sqrt{x}+\sqrt[4]{x}} d x$
In the given integral, the exponent of $x \text { are } \frac{1}{2} \& \frac{1}{4}$ and the LCM of denominators is 4
Substitute $x=t^{4} \Rightarrow d x=4 t^{3} d t$ then
$\begin{aligned} &\Rightarrow I=\int \frac{1}{t^{2}+t}\left(4 t^{3}\right) d t \qquad\left(\because x=t^{4}\right) \\ & \end{aligned}$
$\Rightarrow I=4 \int \frac{t^{3}}{t^{2}+t} d t \\$
$\Rightarrow I=\int \frac{t^{3}+t^{2}-t^{2}}{t^{2}+t} d t\qquad\left[\text { we can write } t^{3}=t^{3}+t^{2}-t^{2}\right]$
$\begin{aligned} &\Rightarrow I=4 \int \frac{t\left(t^{2}+t\right)-t^{2}}{\left(t^{2}+t\right)} d t \\ & \end{aligned}$
$\Rightarrow I=4 \int\left(t-\frac{t^{2}}{t^{2}+t}\right) d t \\$
$\Rightarrow I=4 \int\left\{t-\frac{t^{2}+t}{t^{2}+t}-\frac{t}{t^{2}+t}\right\} d t$
$\begin{aligned} &\Rightarrow I=4 \int\left\{t-1-\frac{t}{t(t+1)}\right\} d t=4 \int\left\{t-1-\frac{1}{(t+1)}\right\} d t \\ & \end{aligned}$
$\Rightarrow I=4 \int t d t-4 \int t^{0} d t-4 \int \frac{1}{t+1} d t$
$\begin{aligned} &\Rightarrow I=4 \frac{t^{1+1}}{1+1}-4 \log |t+1|-4 t+c \\ & \end{aligned}$
$\Rightarrow I=4 \frac{t^{2}}{2}-4 t-4 \log |t+1|+c \\$
$\therefore I=2 \sqrt{x}-4 x^{\frac{1}{4}}+4 \log \left|1+x^{\frac{1}{4}}\right|+c$

Indefinite Integrals Excercise 18.10 Question 10

Answer: $3 x^{\frac{1}{3}}+3 \log \left|x^{\frac{1}{3}}-1\right|+c$
Hint: Use substitution method to solve this type of integral
Given: $\int \frac{1}{x^{\frac{1}{3}}\left(x^{\frac{1}{3}}-1\right)} d x$
Solution: let $I=\int \frac{1}{x^{\frac{1}{3}}\left(x^{\frac{1}{3}}-1\right)} d x$
Substitute $x=t^{3} \Rightarrow d x=3 t^{2} d t$ then
$\begin{aligned} I &=I=\int \frac{1}{t(t-1)} 3 t^{2} d t \quad\left(\because x^{\frac{1}{3}}=t\right) \\ & \end{aligned}$
$=3 \int \frac{t^{2}}{t(t-1)} d t=3 \int \frac{t}{t-1} d t \\$
$=3 \int \frac{t+1-1}{t-1} d t=3 \int\left(\frac{t-1}{t-1}+\frac{1}{t-1}\right)dt$
$\begin{aligned} &=3 \frac{t^{0+1}}{0+1}+3 \log |t-1|+c \qquad\left[\int \frac{1}{t} d t=\log |t|+c \& \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right] \\ & \end{aligned}$
$=3 t+3 \log |t-1|+c \\$
$\therefore I=3 x^{\frac{1}{3}}+3 \log \left|x^{\frac{1}{3}}-1\right|+c$

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