RD Sharma Class 12 Exercise 18.5 Indefinite Integrals Solutions Maths

RD Sharma Class 12 Exercise 18.5 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 11:56 AM IST

The RD Sharma books have widely gained popularity and are used by most students as a guide to clear their doubts. For example, students tend to have more doubts while they solve the homework sums in mathematics. And concepts like Indefinite Integrals are challenging to work out. In such cases, the RD Sharma Class 12th Exercise 18.5 solution book lends a helping hand.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.5

Indefinite Integrals Exercise18. 5 Question 1

Answer: $\frac{1}{6}(2 x+3)^{\frac{3}{2}}-\frac{1}{2}(2 x+3)^{\frac{1}{2}}+C$
Hint: Use Integration by partial fraction
Given: $\int \frac{x+1}{2x+3}dx$
Solution:
$= \int \frac{x+1}{2x+3}dx$
$\begin{aligned} &=\frac{1}{2} \int \frac{2 x+2}{\sqrt{2 x+3}} d x \\ &=\frac{1}{2} \int \frac{2 x+3-1}{\sqrt{2 x+3}} d x \\ &=\frac{1}{2} \int \frac{2 x+3}{\sqrt{2 x+3}} d x-\frac{1}{2} \int \frac{2 x+3}{\sqrt{2 x+3}} d x \\ &=\frac{1}{2} \int \sqrt{2 x+3} d x-\int(2 x+3)^{\frac{-1}{2}} \end{aligned}$
$=\frac{1}{2} \times \frac{1}{2} \int \frac{(2 x+3)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{1}{2} \times \frac{1}{2} \int \frac{(2 x+3)^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}$ $\left[\therefore \int(a x+b)^{n} d x=\frac{\frac{1}{a}(a x+b)^{n+1}}{n+1}\right]$
$\begin{aligned} &=\frac{1}{2} \int \frac{(2 x+3)^{\frac{3}{2}}}{3}-\frac{1}{2} \times(2 x+3)^{\frac{1}{2}}+C \\ &=\frac{1}{6}(2 x+3)^{\frac{3}{2}}-\frac{1}{2}(2 x+3)^{\frac{1}{2}}+C \end{aligned}$

Indefinite Integrals Exercise18.5 Question 2

Answer: $\frac{2}{5}(x+2)^{\frac{5}{2}}-\frac{6}{3}(x+2)^{\frac{1}{2}}+C$
Hint: Let $u= x+2$
Given: $\int x\sqrt{x+2}dx$
Solution:
Let $u= x+2$
$\frac{du}{dx}= 1$
Applying the substitution we have:
$\begin{aligned} &I=\int(u-2) \sqrt{u} d u \\ &\therefore I=(u-2) u^{\frac{1}{2}} d u \\ &=\int u^{\frac{3}{2}}-2 u^{\frac{1}{2}} d u \end{aligned}$
$\begin{aligned} &\therefore I=\frac{u^{\frac{5}{2}}}{\frac{5}{2}}-\frac{2 u^{\frac{3}{2}}}{\frac{3}{2}}+C \\ &\therefore I=\frac{2}{5} u^{\frac{5}{2}}-\frac{4}{3} u^{\frac{3}{2}}+C \\ &\therefore I=\frac{2}{5}(x+2)^{\frac{5}{2}}-\frac{4}{3}(x+2)^{\frac{3}{2}}+C \end{aligned}$

Indefinite Integrals Exercise18.5 Question 3

Answer: $\frac{2}{3}(x+4)^{\frac{3}{2}}-10(x+4)^{\frac{1}{2}}+C$
Hint: Use Integration by partial function.
Given: $\int \frac{x-1}{\sqrt{x+4}} d x \\$
Solution:
$\begin{aligned} &=\int \frac{x-1}{\sqrt{x+4}} d x \\ &=\int \frac{x+4-5}{\sqrt{x+4}} d x \end{aligned}$
$\begin{aligned} &=\int \frac{x+4}{\sqrt{x+4}} d x-5 \int \frac{d x}{\sqrt{x+4}} \\ &=\int \sqrt{x+4} d x-5 \int(x+4)^{\frac{-1}{2}} \end{aligned}$
$\begin{aligned} &=\frac{(x+4)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-5 \times \frac{(x+4)^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} \\ &=\frac{2}{3}(x+4)^{\frac{3}{2}}-10(x+4)^{\frac{1}{2}}+C \end{aligned}$

Indefinite Integrals Exercise18.5 Question 4

Answer: $\frac{2}{135}(9 x+20)(3 x+5)^{\frac{3}{2}}+C$
Hint: Let $t= 3x+5$
Given: $\int \left ( x+2 \right )\sqrt{3x+5}dx$
Solution:
Substitute $3x+5=t$
$\begin{aligned} &\Rightarrow x=\frac{t-5}{3} \Rightarrow 3 d x=d t \\ &\Rightarrow d x=\frac{d t}{3} \end{aligned}$
$\begin{aligned} &\therefore I=\int\left(\frac{t-5}{3}+2\right) \sqrt{t} \frac{d t}{3} \\ &\therefore I=\frac{1}{3} \int\left(\frac{t-5+6}{3}\right) \sqrt{t} d t \end{aligned}$

On taking 3 as common and multiply, we get

$=\frac{1}{9} \int\left(t^{\frac{3}{2}}+t^{\frac{1}{2}}\right) d t$
On integrating we get
$=\frac{1}{9}\left[\frac{t^{\frac{3}{2}}+1}{\frac{3}{2}+1}+\frac{t^{\frac{1}{2}}+1}{\frac{1}{2}+1}\right]+C$
On simplifying
$=\frac{1}{9}\left[\frac{2}{5} t^{\frac{5}{2}}+\frac{2}{3} t^{\frac{3}{2}}\right]+C$
On substituting the value of t
$\begin{aligned} &=\frac{1}{9}\left[\frac{2}{5}(3 x+5)^{\frac{5}{2}}+\frac{2}{3}(3 x+5)^{\frac{3}{2}}\right]+C \\ &=\frac{2}{9}\left[(3 x+5)^{\frac{3}{2}}\left\{\frac{3 x+5}{5}+\frac{1}{3}\right\}\right]+C \end{aligned}$
$\begin{aligned} &=\frac{2}{9}\left[(3 x+5)^{\frac{3}{2}}\left\{\frac{9 x+15+5}{15}\right\}\right]+C \\ &=\frac{2}{9}\left[(3 x+5)^{\frac{3}{2}}\left\{\frac{9 x+20}{15}\right\}\right]+C \\ &=\frac{2}{135}(3 x+5)^{\frac{3}{2}}(9 x+20)+C \end{aligned}$

Indefinite Integrals Exercise18.5 Question 5

Answer: $\frac{2}{27}(6 x+1) \sqrt{3 x+2}+C$
Hint: Use Integration by partial function.
Given: $\int \frac{2x+1}{\sqrt{3x+2}}dx$
Solution: On multiplying and dividing by 3 in the equation, we get
$=\frac{1}{3} \int\left(\frac{6 x+3}{\sqrt{3 x+2}}\right) d x$
The equation can be written as
$=\frac{1}{3} \int\left(\frac{6 x+4-1}{\sqrt{3 x+2}}\right) d x$
So taking 2 common from and subtracting
$=\frac{1}{3} \int\left(\frac{2(3 x+2)}{\sqrt{3 x+2}}-\frac{1}{\sqrt{3 x+2}}\right) d x$
On simplifying
$=\frac{1}{3} \int( 2 \sqrt{3 x+2}-\frac{1}{\sqrt{3 x+2}}) d x$
By splitting the integral
$=\frac{1}{3} \int 2(3 x+2)^{\frac{1}{2}} d x-\frac{1}{3}\int(3 x+2)^{-\frac{1}{2}} d x$
So integrating we get
$\begin{aligned} &=\frac{1}{3}\left[2\left\{\frac{(3 x+2)^{\frac{1}{2}+1}}{3\left(\frac{1}{2}+1\right)}\right\}-\frac{(3 x+2)^{-\frac{1}{2}+1}}{3\left(-\frac{1}{2}+1\right)}\right]+C \\ &=\frac{1}{3}\left[\frac{4}{9}(3 x+2)^{\frac{3}{2}}-\frac{2}{3}(3 x+2)^{\frac{1}{2}}\right]+C \end{aligned}$

By Simplifying, we get
$\begin{aligned} &=\frac{4}{27}(3 x+2)^{\frac{3}{2}}-\frac{2}{9}(3 x+2)^{\frac{1}{2}}+C \\ &=\sqrt{3 x+2}\left[\frac{4}{27}(3 x+2)-\frac{2}{9}\right]+C \\ &=\sqrt{3 x+2}\left[\frac{4(3 x+2)-6}{27}\right]+C \\ &=\sqrt{3 x+2}\left[\frac{12 x+8-6}{27}\right]+C \\ &=\frac{2}{27}(6 x+1) \sqrt{3 x+2}+C \end{aligned}$

Indefinite Integrals Exercise18.5 Question 6

Answer: $\frac{2}{49}(7 x+17) \sqrt{7 x+9}+C$
Hint: Try to take out common factors
Given: $\int \frac{(3 x+5)}{\sqrt{7 x+9}} d x \\$
Solution: On multiplying and dividing by 3 in the equation, we get
$\begin{aligned} &=\int \frac{(3 x+5)}{\sqrt{7 x+9}} d x \\ &=3 \int \frac{x+\frac{5}{3}}{\sqrt{7 x+9}} d x \end{aligned}$

$\begin{aligned} &I=\frac{3}{7} \int \frac{7 x+\frac{35}{3}}{\sqrt{7 x+9}} d x \\ &I=\frac{3}{7} \int \frac{7 x+9-9+\frac{35}{3}}{\sqrt{7 x+9}} d x \\ &I=\frac{3}{7} \int \frac{7 x+9+\frac{8}{3}}{\sqrt{7 x+9}} d x \end{aligned}$

$\begin{aligned} &I=\frac{3}{7} \int \frac{7 x+9}{\sqrt{7 x+9}} d x+\int \frac{8}{7} \cdot \frac{1}{\sqrt{7 x+9}} d x \\ &I=\frac{3}{7} \int \sqrt{7 x+9} d x+\int \frac{8}{7} \cdot \frac{1}{\sqrt{7 x+9}} d x \end{aligned}$
$I=\frac{2}{49}(7x+9)\sqrt{7x+9}+\frac{16}{49}\sqrt{7x+9}$

Indefinite Integrals Exercise18.5 Question 7

Answer: $\frac{2}{3}(x+4)^{\frac{3}{2}}-8(x+4)^{\frac{1}{2}}+C$
Hint: Let $x+4=t$
Given: $\int \frac{x}{\sqrt{x+4}}dx$
Solution:
Let $x+4=t , x=t+4$
$\begin{aligned} &d x=d t \\ &=\int \frac{t-4}{\sqrt{t}} d t \\ &=\int \frac{t}{\sqrt{t}} d t-\int \frac{4}{\sqrt{t}} d t \end{aligned}$
$\begin{aligned} &=\frac{2}{3}(t)^{\frac{3}{2}}-4 \cdot 2(t)^{\frac{1}{2}}+C \\ &=\frac{2}{3}(x+4)^{\frac{3}{2}}-8(x+4)^{\frac{1}{2}}+C \end{aligned}$

Indefinite Integrals Exercise18.5 Question 8

Answer: $\frac{4}{3}(1+3 x)^{\frac{1}{2}}-\frac{2}{9}(1+3 x)^{\frac{3}{2}}-\frac{2}{3}(1+3 x)^{\frac{1}{2}}+C$
Hint: Let $1+3x=t$
Given: $\int \frac{\left ( 2-3x \right )}{\sqrt{1+3x}}dx$
Solution:
$=\int \frac{2}{\sqrt{1+3 x}} d x-\int \frac{3 x}{\sqrt{1+3 x}} d x$
Let $1+3x=t , 3x=t-1$
$dx=\frac{dt}{3}$
$\begin{aligned} &=2 \int(1+3 x)^{\frac{-1}{2}} d x-\frac{1}{3} \int \frac{t-1}{\sqrt{t}} d x \\ &=2\left[\frac{(1+3 x)^{\frac{-1}{2}+1}}{-\frac{1}{2}+1} \times \frac{1}{3}\right]-\frac{1}{3}\left[\int \sqrt{t} d t-\int(t)^{\frac{-1}{2}}\right] \end{aligned}$
$\begin{aligned} &=\frac{4}{3}(1+3 x)^{\frac{1}{2}}-\frac{1}{3}\left[2 \times \frac{(t)^{\frac{3}{2}}}{3}-2(t)^{\frac{1}{2}}\right]+C \\ &=\frac{4}{3}(1+3 x)^{\frac{1}{2}}-\frac{2}{9}(1+3 x)^{\frac{3}{2}}-\frac{2}{3}(1+3 x)^{\frac{1}{2}}+C \end{aligned}$

Indefinite Integrals Exercise18.5 Question 9

Answer: $\frac{1}{3}(2 x-1)^{\frac{3}{2}}(3 x+4)+C$
Hint: Let $5x+3=\lambda \left ( 2x-1\right )+\mu$
Given: $\int \left ( 5x+3 \right )\sqrt{2x-1}dx$
Solution: Comparing the coefficient, we get
$\begin{aligned} &\Rightarrow 2 \lambda=5 \Rightarrow \lambda=\frac{5}{2} \Rightarrow-\lambda+\mu=3 \\ &\Rightarrow \frac{-5}{2}+\mu=3 \Rightarrow \mu=3+\frac{5}{2} \Rightarrow \mu=\frac{11}{2} \\ &=\int[\lambda(2 x-1)+\mu] \sqrt{2 x-1} d x \Rightarrow \int\left[\frac{5}{2}(2 x-1)+\frac{11}{2}\right] \sqrt{2 x-1} d x \end{aligned}$
$\begin{aligned} &\therefore I=\int(5 x+3) \sqrt{2 x-1} d x \Rightarrow \frac{5}{2} \int(2 x-1)^{1+\frac{1}{2}} d x+\frac{11}{2} \int(2 x-1)^{\frac{1}{2}} d x \\ &\Rightarrow \frac{5}{2} \int(2 x-1)^{\frac{3}{2}}+\frac{11}{2} \int(2 x-1)^{\frac{1}{2}} d x \\ &\Rightarrow \frac{5}{2} \frac{(2 x-1)^{\frac{3}{2}+1}}{2\left[\frac{3}{2}+1\right]}+\frac{11}{2} \frac{(2 x-1)^{\frac{1}{2}+1}}{2\left[\frac{1}{2}+1\right]}+C \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{5}{2} \frac{(2 x-1)^{\frac{5}{2}}}{2\left[\frac{5}{2}\right]}+\frac{11}{2} \frac{(2 x-1)^{\frac{3}{2}}}{2\left[\frac{3}{2}\right]}+C \Rightarrow \frac{5}{2} \frac{(2 x-1)^{\frac{5}{2}}}{5}+\frac{11}{2} \frac{(2 x-1)^{\frac{3}{2}}}{3}+C \\ &\Rightarrow \frac{1}{2}(2 x-1)^{\frac{5}{2}}+\frac{11}{2}(2 x-1)^{\frac{3}{2}}+C \Rightarrow \frac{1}{6}(2 x-1)^{\frac{3}{2}}[3(2 x-1)+11]+C \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{1}{6}(2 x-1)^{\frac{3}{2}}[6 x-3+11]+C \Rightarrow \frac{1}{6}(2 x-1)^{\frac{3}{2}}[6 x+8]+C \\ &\Rightarrow \frac{1}{6}(2 x-1)^{\frac{3}{2}}[2(3 x+4)]+C \Rightarrow \frac{2}{6}(2 x-1)^{\frac{3}{2}}[3 x+4]+C \\ &\Rightarrow \frac{1}{3}(2 x-1)^{\frac{3}{2}}(3 x+4)+C \end{aligned}$

Indefinite Integrals Exercise18.5 Question 10

Answer: $\frac{1}{a-b}\left\{\frac{2}{5}(x+a)^{\frac{5}{2}}-\frac{2 a}{3}(x+a)^{\frac{3}{2}}+\frac{2}{5}(x+b)^{\frac{5}{2}}-\frac{2 b}{3}(x+b)^{\frac{3}{2}}\right\}+C$
Hint: Use: $(a-b)(a b)=a^{2}-b^{2}$
Given: $\int \frac{x}{\sqrt{x+a}-\sqrt{x+b}}$
Solution:
$\begin{aligned} &=\int \frac{x}{\sqrt{x+a}-\sqrt{x+b}} \Rightarrow \frac{x}{\sqrt{x+a}-\sqrt{x+b}} \Rightarrow \frac{\sqrt{x+a}+\sqrt{x+b}}{\sqrt{x+a}+\sqrt{x+b}} \\ &=\frac{x(\sqrt{x+a}+\sqrt{x+b})}{(\sqrt{x+a})^{2}-(\sqrt{x+b})^{2}} \Rightarrow \frac{x \sqrt{x+a}+x \sqrt{x+b}}{x+a-(x+b)} \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{1}{a-b}\left\{(x+a)^{\frac{3}{2}}-a(x+a)^{\frac{1}{2}}+(x+b)^{\frac{3}{2}}-b(x+b)^{\frac{1}{2}}\right\} \\ &\Rightarrow \int \frac{1}{\sqrt{x+a}-\sqrt{x+b}} d x=\int \frac{1}{(a-b)}\left\{(x+a)^{\frac{3}{2}}-a(x+a)^{\frac{1}{2}}+(x+b)^{\frac{3}{2}}-b(x+b)^{\frac{1}{2}}\right\} \end{aligned}$
$\begin{aligned} &\Rightarrow \int \frac{1}{\sqrt{x+a}-\sqrt{x+b}} d x=\int \frac{1}{(a-b)}\left\{(x+a)^{\frac{3}{2}}-a(x+a)^{\frac{1}{2}}+(x+b)^{\frac{3}{2}}-b(x+b)^{\frac{1}{2}}\right\} \\ &\Rightarrow \frac{1}{a-b}\left\{(x+a)^{\frac{3}{2}} d x-\int a(x+a)^{\frac{1}{2}} d x+\int(x+b)^{\frac{3}{2}} d x-\int b(x+b)^{\frac{1}{2}} d x\right\} \end{aligned}$ $\begin{aligned} &\Rightarrow \frac{1}{a-b}\left\{\frac{1}{1+\frac{3}{2}}(x+a)^{\frac{3}{2}}-a \frac{1}{1+\frac{1}{2}}(x+a)^{\frac{1}{2}}+\frac{1}{1+\frac{3}{2}}(x+b)^{\frac{3}{2}}-b \frac{1}{1+\frac{1}{2}}(x+b)^{\frac{1}{2}}\right\} \\ &=\frac{1}{a-b}\left\{\frac{2}{5}(x+a)^{\frac{5}{2}}-\frac{2 a}{3}(x+a)^{\frac{3}{2}}+\frac{2}{5}(x+b)^{\frac{5}{2}}-\frac{2 b}{3}(x+b)^{\frac{3}{2}}\right\}+C \end{aligned}$

The class 12 mathematics, chapter 18, Indefinite Integrals, is wholly based on solving integrations. RD Sharma Class 12 Solutions Chapter 18 ex 18.5, consists of 10 sums that need to be solved. There are nine questions in Level 1 and one question in Level 2 in this particular exercise. The only concept in this exercise is to evaluate the integrals. The RD Sharma Class 12th Exercise 18.5 solutions for all these sums are given in the RD Sharma Class 12 Chapter 18 Exercise 18.5. The students can use this resource material to solve the sums in the correct format.

These books follow the NCERT syllabus making it easier for the CBSE students to adapt it. Moreover, the RD Sharma Class 12th Exercise 18.5 book contains the solution for the questions given in the textbook; it also includes various additional sums for practice. This can help the students be ready to face challenging questions in the examinations.

If you find it hard to grasp the Indefinite Integrals concept, have the Class 12 RD Sharma Chapter 18 Exercise 18.5 Solution material with you for reference. This lets you solve the sums in the proper format without any confusion. Once you practice these sums, you will soon be able to perfect solving integrations in a short period. You can use this book to clear your doubts while doing homework, making assignments, and preparing for the exams.

The main advantage of using the RD Sharma Class 12 Solutions Indefinite Integrals is that you can access it from the top educational website, Career360, for free of cost. It also includes a download option that lets you download the solution PDF file into your device. In addition, many students have benefitted from using the RD Sharma Class 12th Exercise 18.5 solution material to make them well versed in the concept.

Students use the RD Sharma books to prepare for their exams; even teachers use them to pick questions for their exam question papers. Hence, by practicing with the RD Sharma Class 12 Solutions Chapter 18 Ex 18.5, you will be getting ready to face the questions in the examinations with complete confidence. So, download the RD Sharma solutions from the Career360 website and start preparing for your exams.

RD Sharma Chapter wise Solutions

Frequently Asked Questions (FAQs)

1. Where can I find the right solutions for the sums given in Class 12, mathematics chapter 18 portion?

The best solution guide for the students to refer to the sums given in chapter 18 is the RD Sharma Class 12th Exercise 18.5 solution book.

2. Does the RD Sharma book provide answers for the Level 2 questions?

The RD Sharma solution book consists of answers for the Level 1 and Level 2 questions. Thus, RD Sharma books are a one-stop solution for all the sums given in the textbook.

3. How can I download the RD Sharma solutions for Chapter 18, ex 18.5?

Visit the Career 360 website and search for the RD Sharma Class 12th Exercise 18.5 solution. You will be able to access the resource material; click the download option to download the PDF file into your device.

4. How many solved sums are given in exercise 18.1 in the RD Sharma mathematics guide?

There are ten questions in this exercise, and all the solutions are given in the Class 12 RD Sharma Chapter 18 Exercise 18.5 Solution material.

5. Where is the RD Sharma solution book available costless?

You can obtain the RD Sharma solution books as a freebie on the Career 360 website. The expert provided answers are given in it for the welfare of the students.