Students of class 12 are very well aware that the chapter of Indefinite integrals is very vast portion to cover in less time and therefore RD Sharma class 12 solution of Indefinite integrals exercise 18.24 gives an important amount of knowledge about the few basic concepts and also the high level concepts of the chapter. The RD Sharma class 12th exercise 18.24 will help the student to get a firm hold on this chapter and also help in ace the chapter to keep ahead of everyone and improve the performance gradually resulting in scoring very high marks in the board exams.
Also Read - RD Sharma Solutions For Class 9 to 12 Maths
Chapter 18 - Indefinite Integrals - Ex-18.1
Indefinite Integrals exercise 18.24 question 1 sub question (i)
Answer: $\frac{1}{2} x+\frac{1}{2} \log |\sin x-\cos x|+c$Indefinite Integrals exercise 18.24 question 1 sub question (ii)
Answer: $\frac{1}{2} x-\frac{1}{2} \log |\cos x-\sin x|+c$Indefinite Integrals exercise 18.24 question 1 sub question (iii)
Answer: $\mathrm{I}=2 \mathrm{x}-3 \tan ^{-1}\left(\tan \frac{x}{2}+1\right)+\mathrm{c}$Indefinite Integrals exercise 18.24 question 1 sub question (iv)
Answer: $\frac{p}{p^{2}+q^{2}} x+\frac{q}{p^{2}+q^{2}} \log |p \cos x+q \sin x|+c$Hint: use the formula in which ,
Put Numerator = λ denominator+ μ (derivative of denominator)
Given: $\int \frac{1}{p+q \tan x}$
Explanation:
$\text { Let, } l=\int \frac{1}{p+q \tan x} \mathrm{dx}$
$\begin{aligned} &=\int \frac{d x}{p+q \frac{\sin x}{\cos x}} \\ &=\int \frac{\cos x}{p \cos x+q \sin x} d x \end{aligned}$
$\text { Let, } \operatorname{cos} x=\lambda(p \cos x+q \sin x)+\mu(-p \sin x+q \cos x)$
Equating coefficients of Cos x and Sin x, we get.
$\begin{aligned} &p \lambda+q \mu=1 \\ &q \lambda-p \mu=0 \end{aligned}$
Solving these, we get.
$\begin{aligned} &\frac{\lambda}{0-p}=\frac{\mu}{-q-0}=\frac{1}{-p^{2}-q^{2}} \\ &\frac{\lambda}{-p}=\frac{\mu}{-q}=\frac{1}{-\left(p^{2}+q^{2}\right)} \end{aligned}$
$\lambda=\frac{p}{p^{2}+q^{2}} \quad, \quad \mu=\frac{q}{p^{2}+q^{2}}$
$\operatorname{cos} x=\frac{p}{p^{2}+q^{2}}(p \cos x+q \sin x)+\frac{q}{p 2+q 2}(-p \sin x+q \cos x)$
$\mathrm{I}=\int \frac{\frac{p}{p_{2}+q_{2}}(\mathrm{p} \cos \mathrm{x}+\mathrm{q} \sin \mathrm{x})+\frac{q}{p_{2}+q_{2}}(-\mathrm{p} \operatorname{sin} \mathrm{x}+\mathrm{q} \cos \mathrm{x})}{p \cos x+q \sin x} d x$
$=\frac{p}{p 2+q 2} \int d x+\frac{q}{p 2+q 2} \int\left[\frac{-p \sin x+q \cos x}{p \operatorname{cos} x+q \sin x}\right] d x$
$I=\frac{p}{p 2+q 2} x+\frac{q}{p 2+q 2} \log (p \cos x+q \sin x)+c$
Indefinite Integrals exercise 18.24 question 1 sub question (v)
Answer: $2 x+\log |2 \cos x+\sin x+3|+C$Indefinite Integrals exercise 18.24 question 1 sub question (vi)
Answer: $\frac{18 x}{25}+\frac{1}{25} \log |3 \sin x+4 \cos x|+c$Indefinite Integrals exercise 18.24 question 1 sub question (vii)
Answer: $\frac{3}{25} x-\frac{4}{25} \log |3 \sin x+4 \cos x|+C$Indefinite Integrals exercise 18.24 question 1 sub question (viii)
Answer: $\frac{18}{25} x+\frac{1}{25} \log |3 \operatorname{sin} x+4 \cos x|+C$Indefinite Integrals exercise 18.24 question 1 sub question (ix)
Answer: $\frac{4}{25} x+\frac{3}{25} \log |4 \cos x+3 \operatorname{Sin} x|+C$Hint: use the formula in which ,
Put Numerator = λ denominator+ μ (derivative of denominator)
Given: $\int \frac{1}{4+3 \tan x} d x$
Explanation:
$\text { Let } \mathrm{I}=\int \frac{1}{4+3 \tan x} \mathrm{dx}$
$\begin{aligned} &=\int \frac{1}{4+3\left(\frac{\sin x}{\cos x}\right)} \mathrm{dx} \\ &=\int \frac{\cos x}{4 \cos x+3 \sin x} \mathrm{dx} \end{aligned}$
$\cos x=\lambda(4 \cos x+3 \sin x)+\mu(-4 \sin x+3 \cos x)$
Equating co- efficient of Cos x and Sin x, We get,
$\begin{aligned} &4 \lambda+3 \mu-1=0 \ \\ &3 \lambda-4 \mu-0=0 \end{aligned}$
Solving these, we get;
$\begin{aligned} &\frac{\lambda}{0-4}=\frac{\mu}{-3+0}=\frac{1}{-16-9} \quad \lambda=\frac{4}{25} \quad, \quad \mu=\frac{3}{25} \\ &\therefore \cos x=\frac{4}{25}(4 \cos x+3 \sin x)+\frac{3}{25}(-4 \sin x+3 \cos x) \end{aligned}$
$\therefore \mathrm{I}=\int \frac{\frac{4}{25}(4 \cos x+3 \sin x)+\frac{3}{25}(-4 \sin x+3 \cos x)}{4 \cos x+3 \sin x} \mathrm{dx}$
$=\int \frac{4}{25} d x+\frac{3}{25} \cdot \int \frac{-4 \sin x+3 \cos x}{4 \cos x+3 \sin x} \mathrm{~d} \mathrm{x}$
put $4cos x + 3 sin x = t$ in second integral and differentiate both sides,
we get, $(-4 \sin x+3 \cos x) d x=d t$
$I=\int \frac{4}{25} d x+\frac{3}{25} \cdot \int \frac{1}{t} \mathrm{dt}$
$=\frac{4}{25} x+\frac{3}{25} \log |t|+C$
$\text { put } t=4 \cos x+3 \sin x$
$I=\frac{4}{25} x+\frac{3}{25} \log |4 \cos x+3 \operatorname{sin} x|+C$
Indefinite Integrals exercise 18.24 question 1 sub question (x)
Answer:$I=\log |3 \operatorname{Cos} x+2 \operatorname{Sin} x|+2 x+C$Indefinite Integrals exercise 18.24 question 1 sub question (xi)
Answer: $I=\frac{9}{41} \log |4 \operatorname{Cos} x+5 \sin x|+\frac{40 x}{41}+C$The Class 12 RD Sharma chapter 18 exercise 18.24 solution of Indefinite integrals of class 12 consists of 11 questions which deals with various topics like,
The formulae of fundamental integration
The exponential functions of integration
Special integrals
Special integration by parts
Important integral theorem
Benefits for using the RD Sharma class 12 solution chapter 18 exercise 18.24 :-
It has been observed that the RD Sharma class 12th exercise 18.24 contains questions that have frequently been asked in the board exams, so a thorough practice of the solution can help you score better.
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