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RD Sharma Class 12 Exercise 18.22 Indefinite integrals Solutions Maths - Download PDF Free Online
The Class 12 RD Sharma chapter 18 exercise 18.22 solution is the best solution for the chapter of Indefinite integrals for the students of class 12th. Students of any board can take these solutions for help if they want to score high in the exams. RD Sharma Solutions The RD Sharma class 12th exercise 18.22 is highly recommended for solving the complex chapters like Indefinite integrals, not only teachers but students as well prefer these solutions for thorough practice.
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- RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
- Indefinite Integrals Excercise:18.22
- Below are mentioned few benefits of the RD Sharma class 12 solution chapter 18 exercise 18.22 :-
RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Chapter 18 - Indefinite Integrals - Ex-18.1
Indefinite Integrals Excercise:18.22
Indefinite Integrals Exercise 18.22 Question 1.
Answer: $\frac{1}{6} \tan ^{-1}\left(\frac{3 \tan x}{2}\right)+c$Given: $\int \frac{1}{4 \cos ^{2} x+9 \sin ^{2} x} d x$
Hint: Use substitution method
Solution: $\int \frac{1}{4 \cos ^{2} x+9 \sin ^{2} x} d x$
Dividing numerator and denominator by $\cos ^{2} x$
$\begin{aligned} &=\int \frac{\frac{1}{\cos ^{2} x}}{\frac{4 \cos ^{2} x}{\cos ^{2} x}+\frac{9 \sin ^{2} x}{\cos ^{2} x}} d x \\ &=\int \frac{\sec ^{2} x d x}{4+9 \tan ^{2} x} \end{aligned}$
$Now,Let$
$\begin{aligned} &\tan x=t \\ &\sec ^{2} x d x=d t \end{aligned}$ (Differentiating w.r.t to x)
$\text { So, } \int \frac{d t}{4+9 t^{2}}$
$=\frac{1}{9} \int \frac{d t}{\frac{4}{9}+t^{2}}$
$=\frac{1}{9} \times \frac{3}{2}\left[\tan ^{-1}\left(\frac{t}{\frac{2}{3}}\right)\right]+c$ $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$
$=\frac{1}{6} \tan ^{-1}\left(\frac{3 \tan x}{2}\right)+c$ $[\tan x=t]$
Indefinite Integrals Exercise 18.22 Question 2.
Answer: $\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{2 \tan x}{\sqrt{5}}\right)+c$Given:$\int \frac{1}{4 \sin ^{2} x+5 \cos ^{2} x} d x$
Hint: Divide the numerator and denominator by $\cos ^{2} x$ and then use substitution method
Solution:
$\int \frac{1}{4 \sin ^{2} x+5 \cos ^{2} x} d x$
Dividing the numerator and denominator by $\cos ^{2} x$
$\Rightarrow \int \frac{\frac{1}{\cos ^{2} x}}{\frac{4 \sin ^{2} x}{\cos ^{2} x}+5 \frac{\cos ^{2} x}{\cos ^{2} x}} d x$
$=\int \frac{\sec ^{2} x}{4 \tan ^{2} x+5} d x$
Let $\begin{gathered} t=\tan x \\ d t=\sec ^{2} x d x \end{gathered}$ (Differentiating w.r.t to x)
Now, $\int \frac{1}{4 t^{2}+5} d t$
$=\frac{1}{4} \int \frac{1}{t^{2}+\frac{5}{4}} d t$ $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$
$=\frac{1}{4} \times \frac{2}{\sqrt{5}} \tan ^{-1}\left(\frac{t}{\frac{\sqrt{5}}{2}}\right)+c$
$=\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{2 \tan x}{\sqrt{5}}\right)+c$
Indefinite Integrals Exercise 18.22 Question 3.
Answer: $\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x+1}{\sqrt{3}}\right)+c$Given: $\int \frac{2}{2+\sin 2 x} d x$
Hint: You must know about substitution method
Solution: $\int \frac{2}{2+\sin 2 x} d x$
$\begin{aligned} &=\int \frac{2}{2+2 \sin x \cos x} d x \\ &=\int \frac{1}{1+\sin x \cos x} d x \end{aligned}$
Dividing the numerator and denominator by $\cos ^{2} x$
$\begin{aligned} &=\int \frac{\sec ^{2} x}{\sec ^{2} x+\tan x} d x \\ &=\int \frac{\sec ^{2} x}{1+\tan ^{2} x+\tan x} d x \end{aligned}$
Let
$\begin{aligned} &\tan x=t \\ &\sec ^{2} d x=d t \end{aligned}$ (Differentiating w.r.t x)
Now, $\int \frac{1}{1+t^{2}+t} d t$
$=\int \frac{1}{t^{2}+2 t \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d t$
$=\int \frac{1}{\left(t+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d t$
$=\frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+c$ $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$
$\begin{aligned} &=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t+1}{\sqrt{3}}\right)+c \\ &=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x+1}{\sqrt{3}}\right)+c \end{aligned}$
Indefinite Integrals Exercise 18.22 Question 4.
Answer: $\frac{1}{2 \sqrt{3}} \log \left|\frac{1+\sqrt{3} \tan x}{1-\sqrt{3} \tan x}\right|+c$Given: $\int \frac{\cos x}{\cos 3 x} d x$
Hint: Use the formula of $\cos 3 x$
Solution:
$\int \frac{\cos x}{\cos 3 x} d x$
$=\int \frac{\cos x}{4 \cos ^{3} x-3 \cos x} d x$
$\left(\because \cos 3 x=4 \cos ^{3} x-3 \cos x\right)$
$=\int \frac{1}{4 \cos ^{2} x-3} d x$
Dividing numerator and denominator by $\cos ^{2} x$
$\begin{aligned} &=\int \frac{\sec ^{2} x}{4-3 \sec ^{2} x} d x \\ &=\int \frac{\sec ^{2} x}{4-3\left(1+\tan ^{2} x\right)} d x \end{aligned}$
$\left(\because \sec ^{2} x=1+\tan ^{2} x\right)$
$=\int \frac{\sec ^{2} x}{1-3 \tan ^{2} x} d x$
Let
$\begin{aligned} &\tan x=t \\ &\sec ^{2} x d x=d t \end{aligned}$ (Differentiating w.r.t x)
$\begin{aligned} &=\int \frac{1}{1-3 t^{2}} d t \\ &=\frac{1}{3} \int \frac{1}{\left(\frac{1}{\sqrt{3}}\right)^{2}-t^{2}} d t \end{aligned}$
$=\frac{1}{3} \times \frac{1}{2} \times \frac{1}{\frac{1}{\sqrt{3}}} \log \left|\frac{\frac{1}{\sqrt{3}}+t}{\frac{1}{\sqrt{3}}-t}\right|+c$ $\left[\int \frac{d t}{a^{2}-t^{2}}=\frac{1}{2 a} \log \left|\frac{a+t}{a-t}\right|+c\right]$
$=\frac{1}{6} \times \sqrt{3} \log \left|\frac{1+\sqrt{3} t}{1-\sqrt{3} t}\right|+c$
$=\frac{1}{2 \sqrt{3}} \log \left|\frac{1+\sqrt{3} \tan x}{1-\sqrt{3} \tan x}\right|+c$
Indefinite Integrals Exercise 18.22 Question 5.
Answer: $\frac{1}{2} \tan ^{-1}(2 \tan x)+c$Given: $\int \frac{1}{1+3 \sin ^{2} x} d x$
Hint: Use the formula of $\sec ^{2} x$ and then apply substitution method
Solution: $\int \frac{1}{1+3 \sin ^{2} x} d x$
Divide numerator and denominator by $\cos ^{2} x$
$=\int \frac{\frac{1}{\cos ^{2} x}}{\frac{1}{\cos ^{2} x}+\frac{3 \sin ^{2} x}{\cos ^{2} x}} d x$
$=\int \frac{\sec ^{2} x}{\sec ^{2} x+3 \tan ^{2} x} d x$
$=\int \frac{\sec ^{2} x}{1+\tan ^{2} x+3 \tan ^{2} x} d x$
$\left(\sec ^{2} x=1+\tan ^{2} x\right)$
$=\int \frac{\sec ^{2} x}{1+4 \tan ^{2} x} d x$
$Let$
$\mathrm{t}=\tan x$ (Differentiating w.r.t x)
$d t=\sec ^{2} x d x$
$=\int \frac{1}{1+4 t^{2}} d t$
$=\frac{1}{4} \int \frac{1}{\left(\frac{1}{2}\right)^{2}+t^{2}} d t$
$=\frac{1}{4} \times \frac{1}{\frac{1}{2}} \tan ^{-1}\left(\frac{t}{\frac{1}{2}}\right)+c$ $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$
$=\frac{1}{2} \tan ^{-1}(2 \tan x)+c$
Indefinite Integrals Exercise 18.22 Question 6.
Answer: $\frac{1}{\sqrt{15}} \tan ^{-1}\left(\frac{\sqrt{3} \tan x}{\sqrt{5}}\right)+c$Given: $\int \frac{1}{3+2 \cos ^{2} x} d x$
Hint: Divide Numerator and Denominator by $\cos ^{2} x$ and then apply substitution method.
Solution: $\int \frac{1}{3+2 \cos ^{2} x} d x$
On dividing Numerator and Denominator by $\cos ^{2} x$
$=\int \frac{\sec ^{2} x}{3 \sec ^{2} x+2}$
$=\int \frac{\sec ^{2} x}{3\left(1+\tan ^{2} x\right)+2} d x$
$\left(\sec ^{2} x=1+\tan ^{2} x\right)$
$=\int \frac{\sec ^{2} x}{5+3 \tan ^{2} x} d x$
$Let$
$t=\tan \: \: x$ (Differentiating w.r.t x)
$d t=\sec ^{2} x d x$
$Now$,$\int \frac{1}{5+3 t^{2}} d t$
$=\frac{1}{3} \int \frac{1}{\left(\left(\sqrt{\frac{5}{3}}\right)^{2}+t^{2}\right)} d t$
$=\frac{1}{3} \times \frac{\sqrt{3}}{\sqrt{5}} \tan ^{-1}\left(\frac{t}{\sqrt{\frac{5}{3}}}\right)+c$ $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$
$=\frac{1}{\sqrt{15}} \tan ^{-1}\left(\frac{\sqrt{3} \tan x}{\sqrt{5}}\right)+c$
Indefinite Integrals Exercise 18.22 Question 7.
Answer: $\frac{1}{5} \log \left|\frac{\tan x-2}{2 \tan x+1}\right|+c$Given: $\int \frac{1}{(\sin x-2 \cos x)(2 \sin x+\cos x)} d x$
Hint: Apply substitution method and divide Numerator and Denominator by $\cos ^{2} x$
Solution:
$\int \frac{1}{(\sin x-2 \cos x)(2 \sin x+\cos x)} d x$
$=\int \frac{1}{2 \sin ^{2} x+\sin x \cos x-4 \sin x \cos x-2 \cos ^{2} x} d x$
$=\int \frac{1}{2 \sin ^{2} x-3 \sin x \cos x-2 \cos ^{2} x} d x$
On dividing Numerator and Denominator by $\cos ^{2} x$
$=\int \frac{\sec ^{2} x}{2 \tan ^{2} x-3 \tan x-2} d x$
$Let$
$\tan x=t$ (Differentiate w.r.t x)
$\sec ^{2} d x=d t$
$Now,$$\int \frac{1}{2 t^{2}-3 t-2} d t$
$=\frac{1}{2} \int \frac{1}{t^{2}-\frac{3}{2} t-1} d t$
$=\frac{1}{2} \int \frac{1}{t^{2}-2 t\left(\frac{3}{4}\right)+\left(\frac{3}{4}\right)^{2}-\left(\frac{3}{4}\right)^{2}-1} d t$
$=\frac{1}{2} \int \frac{1}{\left(t-\frac{3}{4}\right)^{2}-\left(\frac{5}{4}\right)^{2}} d t$
$=\frac{1}{2} \times \frac{1}{2 \times \frac{5}{4}} \log \left|\frac{t-\frac{3}{4}-\frac{5}{4}}{t-\frac{3}{4}+\frac{5}{4}}\right|+c$ $\left[\int \frac{d t}{t^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{t-a}{t+a}\right|+c\right]$
$\begin{aligned} &=\frac{1}{2} \times \frac{2}{5} \log \left|\frac{2t-4}{2 t+1}\right|+c \\ &=\frac{1}{5} \log \left|\frac{2\tan x-4}{2 \tan x+1}\right|+c \end{aligned}$
Indefinite Integrals Exercise 18.22 Question 8.
Answer: $\tan ^{-1}\left(\tan ^{2} x\right)+c$Given: $\int \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} d x$
Hint: Divide Numerator and denominator by $\cos ^{4} x$
Solution:
$\int \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} d x$ $(\sin 2 x=2 \sin x \cos x)$
$=\int \frac{2 \sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x$
On dividing Numerator and denominator by $\cos ^{4} x$
$=\int \frac{2 \tan x \cdot \sec ^{2} x d x}{\tan ^{4} x+1}$
$Let$
$\begin{aligned} &\tan ^{2} x=t \\ &2 \tan x \sec ^{2} x d x=d t \end{aligned}$ (Differentiate w.r.t x)
$Now,$$\int \frac{1}{t^{2}+1} d t$
$=\tan ^{-1} t+c$ $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$
$=\tan ^{-1}\left(\tan ^{2} x\right)+c$
Indefinite Integrals Exercise 18.22 Question 10.
Answer: $\frac{1}{2} \log \left|\frac{\tan x}{\tan x+2}\right|+c$Given: $\int \frac{1}{\sin ^{2} x+\sin 2 x} d x$
Hint: Use the formula of sin 2x and then apply substitution method
Solution: $\int \frac{1}{\sin ^{2} x+\sin 2 x} d x$
$=\int \frac{1}{\sin ^{2} x+2 \sin x \cos x} d x$ $(\sin 2 x=2 \sin x \cos x)$
On dividing numerator and denominator by $\cos ^{2} x$, we get
$=\int \frac{\sec ^{2} x}{\tan ^{2} x+2 \tan x} d x$
$Let$
$t=\tan x$
$d t=\sec ^{2} x d x$ (Differentiate w.r.t x)
$Now,$ $\int \frac{1}{t^{2}+2 t} d t$
$\begin{aligned} &=\int \frac{1}{t^{2}+2 t+(1)^{2}-(1)^{2}} d t \\ &=\int \frac{1}{(t+1)^{2}-(1)^{2}} d t \end{aligned}$$\left[\int \frac{d t}{t^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{t-a}{t+a}\right|+c\right]$
$=\frac{1}{2} \log \left|\frac{t+1-1}{t+1+1}\right|+c$
$\begin{aligned} &=\frac{1}{2} \log \left|\frac{t}{t+2}\right|+c \\ &=\frac{1}{2} \log \left|\frac{\tan x}{\tan x+2}\right|+c \end{aligned}$
Indefinite Integrals Exercise 18.22 Question 11.
Answer: $\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan x)+c$Given: $\int \frac{1}{\cos 2 x+3 \sin ^{2} x} d x$
Hint: Divide numerator and denominator by $\cos ^{2} x$ and then use substitution method
Solution:
$\begin{aligned} &\int \frac{1}{\cos 2 x+3 \sin ^{2} x} d x \\ &\left(\cos 2 x=\cos ^{2} x-\sin ^{2} x\right) \\ &=\int \frac{1}{\cos ^{2} x-\sin ^{2} x+3 \sin ^{2} x} d x \\ &=\int \frac{1}{\cos ^{2} x+2 \sin ^{2} x} d x \end{aligned}$
On dividing numerator and denominator by $\cos ^{2} x$, we get
$=\int \frac{\sec ^{2} x}{1+2 \tan ^{2} x} d x$
$Let$
$\begin{aligned} &\tan x=t \\ &\sec ^{2} x d x=d t \end{aligned}$ (Differentiate w.r.t x)
$Now,$$\int \frac{1}{1+2 t^{2}} d t$
$=\frac{1}{2} \int \frac{1}{\left(\frac{1}{\sqrt{2}}\right)^{2}+t^{2}} d t$ $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$
$=\frac{1}{2} \times \frac{\sqrt{2}}{1} \tan ^{-1}\left(\frac{t}{\frac{1}{\sqrt{2}}}\right)+c$$=\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan x)+c$
RD Sharma class 12 solution of Indefinite Integrals exercise 18.22 contains 11 questions, covered the following topics that are mentioned below:-
Integration of trigonometric and inverse trigonometric functions
Integration of exponential functions
Below are mentioned few benefits of the RD Sharma class 12 solution chapter 18 exercise 18.22 :-
The RD Sharma class 12th exercise 18.22 solution is crafted by great experts that are famous in the field of mathematics and thus it gives a vast proportion of knowledge to students in a way that no school can provide.
The syllabus of NCERT is quite vast and difficult, therefore students face trouble while solving questions from it and that is why RD Sharma class 12 chapter 18 exercise 18.22 is recommended for solved questions provided in it.
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RD Sharma class 12th exercise 18.22 gives a brief idea about the concepts and explains it in easy, moderate and tough level which will prepare you to solve any type of question asked in the exam.
RD Sharma Chapter-wise Solutions
- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable
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