RD Sharma Solutions Class 12 Mathematics Chapter 18 FBQ

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:42 PM IST

The RD Sharma books are the most recommended solution guides for the class 12 students. Not every student can afford to pay separate tuition for challenging subjects like mathematics. Most of the students barely have any time to visit tuitions after their school hours. RD Sharma Solutions And when it comes to chapter 18, students find it harder to solve the Fill in the Blanks - FBQs. Therefore, the presence of RD Sharma Class 12th FBQ book is a must.

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RD Sharma Class 12 Solutions Chapter 18 FBQ Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:FBQ

Indefiite Integrals Excercise Fill in the Blanks Question 1

Answer:
$\sin x+\cos x+c$
Hint:
Make the function in complete square form and integrate it.
Solution:
Let
$\begin{aligned} &I=\int \sqrt{1-\sin 2 x} d x \\ &\Rightarrow I=\int \sqrt{\cos ^{2} x+\sin ^{2} x-2 \sin x \cos x} d x \quad\left[\cos ^{2} x+\sin ^{2} x=1, \quad \sin 2 x=2 \sin x \cos x\right] \\ &\Rightarrow I=\int \sqrt{(\cos x-\sin x)^{2}} d x \end{aligned}$
$\begin{aligned} &I=\int(\cos x-\sin x) d x \\ &I=\int \cos x d x-\int \sin x d x \\ &I=\sin x+\cos x+c \end{aligned}$

Indefiite Integrals Excercise Fill in the Blanks Question 2

Answer:
$-\sin x-\cos x+c$
Hint:
Simplify it and integrate it.
Solution:
$\begin{aligned} &I=\int \sqrt{1-\sin 2 x} d x \\ &I=\int \sqrt{\cos ^{2} x+\sin ^{2} x-2 \sin x \cos x} d x \quad\quad\quad\quad\quad\quad\quad\left[\cos ^{2} x+\sin ^{2} x=1, \sin 2 x=2 \sin x \cos x\right] \\ &\Rightarrow I=\int \sqrt{(\sin x-\cos x)^{2}} d x \end{aligned}$
$\begin{aligned} &I=\int(\sin x-\cos x) d x \\ &I=\int \sin x d x-\int \cos x d x \\ &I=-\cos x-\sin x+c \end{aligned}$

Indefiite Integrals Excercise Fill in the Blanks Question 3

Answer:
$\Rightarrow k=-\frac{1}{4}$
Hint:
Solve this integration and then compare with R.H.S.
Solution:
$\begin{aligned} &I=\int \frac{1}{\tan x+\cot x} d x \\ &I=\int \frac{1}{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}} d x \\ &I=\int \frac{1}{\frac{\sin ^{2} x+\cos ^{2} x}{\cos x \sin x}} d x \end{aligned}$
$\begin{aligned} &I=\int \frac{\cos x \sin x}{\sin ^{2} x+\cos ^{2} x} d x \quad\quad\quad\quad\quad\quad\quad\quad\left[\cos ^{2} x+\sin ^{2} x=1\right] \\ &I=\int \sin x \cos x d x \end{aligned}$
Multiply and divide by 2
$\begin{aligned} &I=\int \frac{2 \sin x \cos x}{2} d x \\ &I=\int \frac{\sin 2 x}{2} d x \quad \quad[\sin 2 x=2 \sin x \cos x] \end{aligned}$
$\begin{aligned} &I=-\frac{\cos 2 x}{4}+c \\ &=k \cos 2 x+c \\ &\Rightarrow k=-\frac{1}{4} \end{aligned}$

Indefiite Integrals Excercise Fill in the Blanks Question 4

Answer:
$\frac{x^{3}}{3}+\frac{2^{x}}{\log 2}+c$
Hint:
Solve this integration with standard formulas.
Solution:
$\begin{aligned} &I=\int\left(e^{2 \log x}+e^{x \log 2}\right) d x \\ &I=\int\left(e^{\log x^{2}}+e^{\log 2^{x}}\right) d x \end{aligned}$
$I=\int\left(x^{2}+2^{x}\right) d x \quad\quad\quad\quad\quad\quad {\left[e^{\log a}=a\right]} \\\\$
$I=\frac{x^{3}}{3}+\frac{2^{x}}{\log 2}+c \quad \quad \quad \quad \quad \quad \quad {\left[\int a^{x} d x=\frac{a^{x}}{\log a}+c\right]}$

Indefiite Integrals Excercise Fill in the Blanks Question 5

Answer: $k= \frac{1}{7}$
Hint: Use simple integration
Solution:
$\begin{aligned} &I=\int \frac{\sin ^{6} x}{\cos ^{8} x} d x \\ &I=\int \frac{\sin ^{6} x}{\cos ^{6} x \cdot \cos ^{2} x} d x \\ &I=\int \tan ^{6} x \cdot \sec ^{2} x d x \end{aligned}$

Let

$\tan x = y$ …(i)
$\begin{aligned} &\sec ^{2} x d x=d y \\ &\Rightarrow I=\int y^{6} d y \\ &=\frac{y^{7}}{7}+c \end{aligned}$

$I=\frac{\tan ^{7} x}{7}+c$ from equation(i)

As given
$\begin{aligned} &\int \frac{\sin ^{6} x}{\cos ^{8} x} d x=k \tan ^{7} x+c \\ & \end{aligned}$
$\frac{\tan ^{7} x}{7}+c=k \tan ^{7} x+c$
Comparing both sides
$k= \frac{1}{7}$

Indefiite Integrals Excercise Fill in the Blanks Question 6

Answer:
$\frac{e^{x}}{x+4}+c$
Hint:
Simplify the function and then integrate it.
Solution:
$\begin{aligned} &I=\int \frac{x+3}{(x+4)^{2}} e^{x} d x \\ &I=\int \frac{x+4-1}{(x+4)^{2}} e^{x} d x \\ &I=I=\int\left[\frac{1}{x+4}-\frac{1}{(x+4)^{2}}\right] e^{x} d x \end{aligned}$

Let $f(x)=\frac{1}{x+4}$

then ${f}'\left (x \right )=-\frac{1}{\left ( x+4 \right )^{2}}$
$\begin{aligned} &\Rightarrow I=\int\left[f(x)+f^{\prime}(x)\right] e^{x} d x \\ &\Rightarrow I= \int\left[f(x)+f^{\prime}(x)\right] e^{x} d x \\ &\Rightarrow I=f(x) e^{x}+c \\ &I=\frac{e^{x}}{x+4}+c \end{aligned}$

Indefiite Integrals Excercise Fill in the Blanks Question 7

Answer:
$\frac{1}{2} e^{2 \sin x}+c$
Hint:
Simplify the function and then integrate it.
Solution:
$I=\int \cos x e^{2 \sin x} d x$
Let $\sin x=y$ …(i)
$\begin{aligned} &\Rightarrow \cos x d x=d y \\ &\Rightarrow I=\int e^{2 y} d y \\ &I=\frac{e^{2 y}}{2}+c \end{aligned}$

$I=\frac{e^{2 \sin x}}{2}+c$ …from equation (i)

Indefiite Integrals Excercise Fill in the Blanks Question 8

Answer:
$\frac{e^{x}}{x+1}+c$
Hint:
Simplify the function and then integrate it with the help of formula.
Solution:
$\begin{aligned} &I=\int \frac{x e^{x}}{(x+1)^{2}} d x \\ &I=\int e^{x}\left[\frac{x+1-1}{(x+1)^{2}}\right] d x \\ &I=\int e^{x}\left[\frac{1}{x+1}-\frac{1}{(x+1)^{2}}\right] d x \end{aligned}$

As we know that, $\int\left[f(x)+f^{\prime}(x)\right] e^{x} d x=f(x) e^{x}+c$

$I=e^{x} \frac{1}{x+1}+c$

Indefiite Integrals Excercise Fill in the Blanks Question 9

Answer:
$-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+c$
Hint:
Simplify the function and then integrate it by using formula.
Solution:
$I=\int \frac{\sin x}{3+4 \cos ^{2} x} d x$
Let $\cos x=y$
$\begin{aligned} &\Rightarrow-\sin x d x=d y \\ &\Rightarrow I=-\int \frac{d y}{3+4 y^{2}} \end{aligned}$

$\begin{aligned} &\Rightarrow I=-\frac{1}{4} \int \frac{d y}{y^{2}+\frac{3}{4}} \\\\ & \end{aligned}$
$\Rightarrow I=-\frac{1}{4} \int \frac{d y}{y^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}$
$\begin{aligned} &I=-\frac{1}{4}\left(\frac{2}{\sqrt{3}}\right) \tan ^{-1}\left(\frac{y}{\frac{\sqrt{3}}{2}}\right)+c \\ &I=-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+c \end{aligned}$

Indefiite Integrals Excercise Fill in the Blanks Question 10

Answer:
$x+\log \left|\frac{x-1}{x+1}\right|+c$
Hint:
To solve, this we use partial fraction.
Solution:
$\begin{aligned} &I=\int \frac{x^{2}+1}{x^{2}-1} d x \\ &I=\int \frac{(x-1)^{2}+2 x}{x^{2}-1} d x \quad\quad\quad\quad\quad\quad\left[(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \end{aligned}$
$\begin{aligned} &I=\int \frac{(x-1)^{2}}{x^{2}-1} d x+\int \frac{2 x}{x^{2}-1} d x \\ &I=\int \frac{x-1}{x+1} d x+\int \frac{2 x}{x^{2}-1} d x \end{aligned}$

$\begin{aligned} &I=\int \frac{x+1-2}{x+1} d x+\int \frac{2 x}{x^{2}-1} d x \\ &I=\int d x-2 \int \frac{1}{x+1} d x+\int \frac{2 x}{x^{2}-1} d x \\\ &I=x-2 \log |x+1|+\log \left|x^{2}-1\right|+c \end{aligned}$
$\begin{aligned} &I=x+\log \left|\frac{x^{2}-1}{(x+1)^{2}}\right|+c \\ &I=x+\log \left|\frac{x-1}{x+1}\right|+c \end{aligned}$

Indefiite Integrals Excercise Fill in the Blanks Question 11

Answer:
$\frac{x^{3}}{3}+c$
Hint:
Simplify it and integrate it.
Solution:
$I=\int \frac{e^{5 \log _{\varepsilon} x}-e^{4 \log _{e} x}}{e^{3 \log _{e} x}-e^{2 \log _{e} x}} d x$
$I=\int \frac{e^{\log _{e} x^{5}}-e^{\log _{\varepsilon} x^{4}}}{e^{\log _{e} x^{3}}-e^{\log _{\varepsilon} x^{2}}} d x$
$I=\int \frac{x^{5}-x^{4}}{x^{3}-x^{2}} d x$
$\begin{aligned} &I=\int \frac{x^{4}(x-1)}{x^{2}(x-1)} d x \\ &I=\int x^{2} d x \\ &I=\frac{x^{3}}{3}+c \end{aligned}$

Indefiite Integrals Excercise Fill in the Blanks Question 12

Answer:
$\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+c$
Hint:
Simplify it by factorizing the numerator and then integrate it.
Solution:
$\begin{aligned} &I=\int \frac{x^{4}+x^{2}+1}{x^{2}-x+1} d x \\ &I=\int \frac{\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)}{x^{2}-x+1} d x \\ &I=\int\left(x^{2}+x+1\right) d x \\ &I=\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+c \end{aligned}$

Indefiite Integrals Excercise Fill in the Blanks Question 13

Answer:
$\frac{-1}{\sin x+\cos x}+c$
Hint:
Simplify it and then integrate it.
Solution:
$\begin{aligned} &I=\int \frac{\cos x-\sin x}{1+\sin 2 x} d x \\ &I=\int \frac{\cos x-\sin x}{\cos ^{2} x+\sin ^{2} x+2 \sin x \cos x} d x \quad\quad\quad\quad\quad\quad\quad\quad\left[\cos ^{2} x+\sin ^{2} x=1, \quad \sin 2 x=2 \sin x \cos x\right] \\ &I=\int \frac{\cos x-\sin x}{(\cos x+\sin x)^{2}} d x \end{aligned}$

Let $\cos x+\sin x=y$ …(i)
$\begin{aligned} &(-\sin x+\cos x) d x=d y \\ &\Rightarrow I=\int \frac{d y}{y^{2}} \\ &I=\frac{-1}{y}+c \\ &I=\frac{-1}{\sin x+\cos x}+c \end{aligned}$

Indefiite Integrals Excercise Fill in the Blanks Question 14

Answer:
$-\cos e^{x}+c$
Hint:
Simplify the function and then integrate it.
Solution:
$I=\int e^{x} \sin e^{x} d x$
Let $e^{x} =y$ …(i)
$\begin{aligned} &\Rightarrow e^{x} d x=d y \\ &\Rightarrow I=\int \sin y d y \end{aligned}$
$\begin{aligned} &I=-\cos y+c \\ &I=-\cos e^{x}+c \end{aligned}$

Indefiite Integrals Excercise Fill in the Blanks Question 15

Answer:
$\frac{-1}{6} e^{-3 x^{2}}+c$
Hint:
Simplify it and integrate it.
Solution:
$I=\int \frac{x}{e^{3 x^{2}}} d x$
Let $x^{2} =y$ …(i)
$\begin{aligned} &\Rightarrow 2 x d x=d y \\ &x d x=\frac{d y}{2} \\ &\Rightarrow I=\frac{1}{2} \int \frac{d y}{e^{3 y}} \end{aligned}$
$\begin{aligned} &I=\frac{1}{2} \int e^{-3 y} d y \\ &I=\frac{1}{2}\left(\frac{e^{-3 y}}{-3}\right)+c \end{aligned}$

$I=-\frac{e^{-3 x^{2}}}{6}+c$ using…(i)

Indefiite Integrals Excercise Fill in the Blanks Question 16

Answer:
$x^{x}+c$
Hint:
By putting $x^{x}= y$ , Simplify it and integrate it.
Solution:
$I=\int x^{x}(1+\log x) d x$
Let $x^{x}= y$ …(i)
Applying logarithm on both sides
$\begin{aligned} &\log x^{x}=\log y \\ &x \log x=\log y \end{aligned}$

diff. both sides

$\left(x \cdot \frac{1}{x}+\log x\right) d x=\frac{1}{y} d y$
$(1+\log x) d x=\frac{d y}{y}$ …(ii)
$\begin{aligned} &\Rightarrow I=\int y \frac{d y}{y} \\ &I=\int d y \\ &I=y+c \end{aligned}$

$I=x^{x}+c$ …from equi (i)

Indefiite Integrals Excercise Fill in the Blanks Question 17

Answer:
$\log \left|e^{x}+1\right|+c$
Hint:
Simplify it and integrate it.
Solution:
$I=\int \frac{e^{x}}{e^{x}+1} d x$
Let $e^{x}+1 =y$ …(i)
$\begin{aligned} &\Rightarrow e^{x} d x=d y \\ &\Rightarrow I=\int \frac{d y}{y} \end{aligned}$
$\begin{aligned} &I=\log |y|+c \\ &I=\log \left|e^{x}+1\right|+c \end{aligned}$ …from equi (i)

Indefiite Integrals Excercise Fill in the Blanks Question 18

Answer:
$\frac{1}{20}\left(x^{2}+6 x+10\right)^{10}+c$
Hint:
Simplify it and integrate it.
Solution:
$I=\int(x+3)\left(x^{2}+6 x+10\right)^{9} d x$
Let $x^{2}+6x+10 =y$ …(i)
$\begin{aligned} &(2 x+6) d x=d y \\ &(x+3) d x=\frac{d y}{2} \end{aligned}$
$\begin{aligned} &\Rightarrow I=\frac{1}{2} \int y^{9} d y \\ &I=\frac{1}{2}\left(\frac{y^{10}}{10}\right)+c \\ &I=\frac{1}{20}\left(x^{2}+6 x+10\right)^{10}+c \end{aligned}$

Indefiite Integrals Excercise Fill in the Blanks Question 19

Answer:
$\frac{1}{4}\left(\tan ^{-1} x\right)^{4}+c$
Hint:
Simplify it and solve it.
Solution:
$I=\int \frac{\left(\tan ^{-1} x\right)^{3}}{1+x^{2}} d x$
Let $tan^{-1}x =y$ …(i)
$\begin{aligned} &\frac{1}{1+x^{2}} d x=d y \\ &\Rightarrow I=\int y^{3} d y \\ &I=\frac{y^{4}}{4}+c \\ &I=\frac{\left(\tan ^{-1} x\right)^{4}}{4}+c \end{aligned}$ …from equi (i)

Indefiite Integrals Excercise Fill in the Blanks Question 20

Answer :
$e^{x}(1-\cot x)+c$
Hint:
Simplify it and integrate it.
Solution:
$\begin{aligned} &I=\int e^{x}\left(1-\cot x+\operatorname{cosec}^{2} x\right) d x\\ \end{aligned}$
$\begin{aligned} &\operatorname{Asf}(x)=1-\cot x\\ \end{aligned}$ …(i)
$\begin{aligned} &\Rightarrow f^{\prime}(x)=\operatorname{cosec}^{2} x \end{aligned}$
We will use the formula $I=\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x \Rightarrow I=e^{x} f(x)+c$
$\Rightarrow I=e^{x}(1-\cot x)+c$ …from equi (i)

The class 12 mathematics, chapter 18, Indefinite Integrals, is where students must be well-versed to answer the FBQs quickly. There are 20 FBQs in this chapter. The central concept in this portion is to find the result of the indefinite integration of trigonometric and logarithmic values, Evaluation of integrals by using trigonometric substitutions and Integration by parts. Small sums are given to be solved and found the integral value. Students must be well aware of the integration concept to attend this portion in the exams without any confusion. And to make it effortless, the RD Sharma Class 12 Chapter 18 FBQ book can be used.

FBQs must be completed within a short period. Only then would the students find time to recheck it and confirm their answers during examinations. It can be done using various tricks and formulas, which are given in the RD Sharma Class 12th FBQ book.

There are numerous practice questions given in the Class 12 RD Sharma Chapter 18 FBQ Solution resource material that will make the students gain a good amount of practice to work out the sums fast. And also, the RD Sharma Class 12 Solutions Indefinite Integrals FBQ will help you solve your homework, assignments and even prepare for the examinations. Using this resource material from day 1 of practice, you can easily score full marks in objective-type questions.

Looking at these benefits provided by the RD Sharma Class 12th FBQ solution book, do not mistake it to be highly priced. The best part of these books is that they are available at the Career360 websites for free of cost. Therefore, you can access every RD Sharma solution book for free at this reputed educational website and also download these resource materials for later use.

There is a high chance of questions being taken for the public exam from the RD Sharma books. Many students have benefitted by scoring good marks by using the RD Sharma Class 12 Solutions Chapter 18 FBQ as their reference material. Now it is your time to own a set of the best mathematics guides.

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

1. Where can I find the RD Sharma solution books for free?

The RD Sharma solution books can be found on the Career 360 website for free of cost.

2. Should I visit the career 360 website each time I need to refer to the solution book?

The Career 360 website gives you the option for downloading the solution books to your device for later reference.

3. Which solution book can the class 12 students use to clarify their doubts on chapter 18 FBQ?

The students can use the RD Sharma Class 12th FBQ solution material to clarify their doubts about the Blanks portion's Fill.

4. Are class 12 mathematics chapter 18 FBQs very hard?

Indefinite Integral is a challenging chapter in the syllabus. But nothing can be complicated with good practice and proper guidance. Use the RD Sharma solution books to gain good marks.

5. How many questions are there in the FBQ section in the chapter Indefinite Integrals?

There are 20 FBQ questions present in the Indefinite Integrals chapter. You can find the solutions for these questions in the RD Sharma Class 12th FBQ book.