RD Sharma Class 12 Exercise 18.24 Indefinite integrals Solutions Maths

RD Sharma Class 12 Exercise 18.24 Indefinite integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:27 PM IST

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.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise: 18.24

Indefinite Integrals exercise 18.24 question 1 sub question (i)

Answer: $\frac{1}{2} x+\frac{1}{2} \log |\sin x-\cos x|+c$
Hint: $\sin x-\cos x=t \Rightarrow(\cos x+\sin x) d x=d t$
Given: $\int \frac{1}{1-\operatorname{Cot} x} d x$
Explanation:
$\begin{aligned} &\text { Let } I=\int \frac{1}{1-\operatorname{Cot} x} \mathrm{dx} \\ &\end{aligned}$
$\begin{aligned} &=\int \frac{1}{1-\frac{\cos x}{\sin x}} \mathrm{dx} \\ &=\int \frac{\sin x}{\sin x-\cos x} \mathrm{dx} \end{aligned}$
$\begin{aligned} &=\frac{1}{2} \int \frac{2 \sin x}{\sin x-\cos x} \mathrm{dx} \\ &=\frac{1}{2} \int \frac{(\sin x-\cos x)+(\sin x+\cos x)}{(\sin x-\cos x)} \mathrm{dx} \end{aligned}$
$=\frac{1}{2} \int 1 d x+\frac{1}{2} \int \frac{\sin x+\cos x}{\sin x-\cos x} \mathrm{dx}$
$\text { Let, } \quad \sin x-\cos x=t \Rightarrow(\cos x+\sin x) d x=d t$
$\begin{aligned} &\mathrm{I}=\frac{x}{2}+\frac{1}{2} \int \frac{d t}{t} \\ &=\frac{x}{2}+\frac{1}{2} \log |\mathrm{t}|+\mathrm{c} \\ &=\frac{x}{2}+\frac{1}{2} \log |\operatorname{Sin} x-\operatorname{Cos} x|+c \end{aligned}$

Indefinite Integrals exercise 18.24 question 1 sub question (ii)

Answer: $\frac{1}{2} x-\frac{1}{2} \log |\cos x-\sin x|+c$
Hint: $\text { Put, } \cos x-\sin x=t \Rightarrow(-\sin x-\cos x) d x=d t$
Given: $\int \frac{1}{1-\tan x} \mathrm{dx}$
Explanation:
$\text { Let I}=\int \frac{1}{1-\tan x} \mathrm{dx}$
$\begin{aligned} &=\int \frac{1}{1-\frac{\sin x}{\cos x}} \mathrm{dx} \\ &=\int \frac{\cos x}{\cos x-\sin x} \mathrm{dx} \end{aligned}$
$\begin{aligned} &=\frac{1}{2} \int \frac{2 \cos x}{\cos x-\sin x} \mathrm{~d} \mathrm{x} \\ &=\frac{1}{2} \int \frac{(\cos x-\sin x)+(\cos x+\sin x)}{(\cos x-\sin x)} \mathrm{dx} \end{aligned}$
$\begin{aligned} &=\frac{1}{2} \int 1 d x+\frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} \mathrm{~d} x \\ &=\frac{x}{2}+\frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} \mathrm{~d} x \end{aligned}$
$\text { Put, } \operatorname{cos} x-\sin x=t \Rightarrow(-\sin x-\cos x) d x=d t$
$\begin{aligned} &\mathrm{I}=\frac{x}{2}+\frac{1}{2} \int \frac{-(d t)}{t} \\ &=\frac{x}{2}-\frac{1}{2} \log |\mathrm{t}|+\mathrm{c} \\ &=\frac{x}{2}-\frac{1}{2} \log |\operatorname{cos} \mathrm{x}-\operatorname{sin} \mathrm{x}|+\mathrm{c} \end{aligned}$

Indefinite Integrals exercise 18.24 question 1 sub question (iii)

Answer: $\mathrm{I}=2 \mathrm{x}-3 \tan ^{-1}\left(\tan \frac{x}{2}+1\right)+\mathrm{c}$
Hint: Substitute $\tan \frac{x}{2}=\mathrm{t}$
Given: $\int \frac{3+2 \cos x+4 \sin x}{2 \sin x+\cos x+3} d x$
Explanation:
$\text { Let, } I=\int \frac{3+4 \sin x+2 \cos x}{3+2 \sin x+\cos x} d x$
$=\int \frac{6+4 \sin x+2 \cos x-3}{3+2 \sin x+\cos x} \mathrm{~d} \mathrm{x}$
$=\int \frac{6+4 \sin x+2 \cos x}{3+2 \sin x+\cos x} \mathrm{dx}-\int \frac{-3}{3+2 \sin x+\cos x} \mathrm{dx}$
$=\int \frac{2(3+2 \sin x+\cos x)}{(3+2 \sin x+\cos x)} d x-\int \frac{-3}{3+2 \sin x+\cos x} d x$
$=\int 2 d x-3 \int \frac{1}{3+2 \sin x+\cos x} d x$
substituting , $\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \text { and } \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$
$=2 x-3 \int \frac{1}{3+2 \times\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)+\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$
$=2 x-3 \int \frac{1+\tan ^{2} \frac{x}{2}}{3\left(1+\tan ^{2} \frac{x}{2}\right)+4 \tan \frac{x}{2}+1-\tan ^{2} \frac{x}{2}} d x$
$=2 x-3 \int \frac{\sec ^{2} \frac{x}{2}}{2 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}+4} d x$
Substitute, $\tan \frac{x}{2}=\mathrm{t}$
$\Rightarrow \operatorname{Sec}^{2} \frac{x}{2}\cdot \frac{1}{2} \mathrm{dx}=\mathrm{dt}$
$\begin{aligned} &I=2 x-3 \int \frac{2}{2 t^{2}+4 t+4} d t \\ &I=2 x-3 \int \frac{2}{2\left(t^{2}+2 t+2\right)} d t \\ &I=2 x-3 \int \frac{1}{(t+1)^{2}+1} d t \end{aligned}$
$\mathrm{I}=2 \mathrm{x}-3 \tan ^{-1}(\mathrm{t}+1)+\mathrm{c} \quad\left[\because \int \frac{1}{1+x^{2}} d x=\tan ^{-1} x+c\right]$
$\mathrm{I}=2 \mathrm{x}-3 \tan ^{-1}\left(\tan \frac{x}{2}+1\right)+c$

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: $2 x+\log |2 \cos x+\sin x+3|+C$
Hint: Using formula, put numerator = λ(denominator)+µ(derivative of denominator)
Given: $\int \frac{5 \cos x+6}{2 \cos x+\sin x+3} d x$
Explanation:
$\text { Let, } 1=\int \frac{5 \cos x+6}{2 \cos x+\sin x+3} d x$
$\text { Let, } 5 \cos x+6=A(2 \cos x+\sin x+3)+B(-2 \sin x+\cos x)$ $............(i)$
$\Rightarrow 5 \cos x+6=(A-2 B) \operatorname{Sin} x+(2 A+B) \operatorname{Cos} x+3 A$
Comparing co-efficient of like terms,
$\begin{aligned} &\mathrm{A}-2 \mathrm{~B}=0\quad \ldots \text { (ii) }\\ &2 \mathrm{~A}+\mathrm{B}=5 \quad \ldots \text { (iii) }\\ &3 A=6 \; \; \; \; \; \; \; \; \quad \ldots \text { (iv) } \end{aligned}$
From eqn (iv) A = 2
Put value of A in equation (ii)
$2-2B = 0 \Rightarrow B =1$
We get B = 1 and A = 2
By putting values of A, B and C in equation $(ii)$
$\therefore I=\int\left[\frac{2(2 \operatorname{cos} x+\sin x+3)+(-2 \sin x+\cos x)}{2 \cos x+\sin x+3}\right] d x$
$=2 \int \mathrm{dx}+\int \frac{(-2 \sin x+\cos x)}{(2 \cos x+\sin x+3)} \mathrm{d} \mathrm{x}$
Putting $2 \operatorname{cos} x+\sin x+3=t$
$\Rightarrow(-2 \sin x+\cos x) d x=d t$
$\therefore I=2 \int \mathrm{dx}+\int \frac{1}{\mathrm{t}} \mathrm{d} \mathrm{t}$
$\begin{aligned} &=2 x+\log |t|+C \\ &\text { Put } t=2 \cos x+\sin x+3 \\ &\therefore I=2 x+\log |2 \cos x+\sin x+3|+c \end{aligned}$

Indefinite Integrals exercise 18.24 question 1 sub question (vi)

Answer: $\frac{18 x}{25}+\frac{1}{25} \log |3 \sin x+4 \cos x|+c$
Hint: use the formula in which ,
Put Numerator = λ denominator+ μ (derivative of denominator)
Given: $\int \frac{2 \sin x+3 \cos x}{3 \sin x+4 \cos x} \mathrm{~d} x$
Explanation:
$\text { Let } I=\int \frac{2 \sin x+3 \cos x}{3 \sin x+4 \cos x} \mathrm{~d} \mathrm{x}$
$\text { Let } 2 \sin x+3 \cos x=\lambda(3 \sin x+4 \cos x)+\mu(3 \cos x-4 \sin x)$
Equating co-efficient of Cos x and Sin x, We get,
$\begin{array}{ll} 4 \lambda+3 \mu=3 & \text { i.e. } 4 \lambda+3 \mu-3=0 \\ 3 \lambda-4 \mu=2 & \text { i.e. } 3 \lambda-4 \mu-2=0 \end{array}$
Solving these, we get;
$\frac{\lambda}{-6-12}=\frac{\mu}{-9+8}=\frac{1}{-16-9}$
$\frac{\lambda}{-18}=\frac{\mu}{-1}=\frac{1}{-25}, \lambda=\frac{18}{25} \quad ; \quad \mu=\frac{1}{25}$
Therefore,
$2 \sin x+3 \cos x=\frac{18}{25}(3 \sin x+4 \cos x)+\frac{1}{25}(3 \cos x-4 \sin x)$
$\therefore I=\int \frac{\frac{18}{25}(3 \sin x+4 \cos x)+\frac{1}{25}(3 \cos x-4 \sin x)}{3 \sin x+4 \cos x} d x$

$=\frac{18}{25} \int 1 d x+\frac{1}{25} \int \frac{3 \cos x-4 \sin x}{3 \sin x+4 \cos x} d x$

Put $3sin x + 4cos x = t$ in second integral and differentiate both sides w.r.t $x$
We get , $(3 \cos x-4 \sin x) d x=d t$
$\begin{aligned} &I=\frac{18}{25} \int 1 d x+\frac{1}{25} \int \frac{1}{t} d t \\ &I=\frac{18}{25} x+\frac{1}{25} \log |t|+C \end{aligned}$
Put $t=3 \sin x+4 \cos x$
$I=\frac{18}{25} x+\frac{1}{25} \log |3 \sin x+4 \operatorname{Cos} x|+C$

Indefinite Integrals exercise 18.24 question 1 sub question (vii)

Answer: $\frac{3}{25} x-\frac{4}{25} \log |3 \sin x+4 \cos x|+C$
Hint: use the formula in which ,
Put Numerator = λ denominator+ μ (derivative of denominator)
Given: $\int \frac{1}{3+4 \cot x} \mathrm{~d} \mathrm{x}$
Explanation:
$\text { Let I} =\int \frac{1}{3+4 \operatorname{cot} x} \mathrm{dx}$
$\begin{aligned} &=\int \frac{1}{3+4\left(\frac{\cos x}{\sin x}\right)} d x \\ &=\int \frac{\sin x}{3 \sin x+4 \cos x} d x \end{aligned}$
$\operatorname{sin} x=\lambda(3 \sin x+4 \cos x)+\mu(3 \cos x-4 \sin x)$
Equating co- efficient of Cos x and Sin x, We get,
$\begin{aligned} &3 \lambda-4 \mu-1=0 \\ &4 \lambda+3 \mu-0=0 \end{aligned}$
Solving this, we get
$\frac{\lambda}{0+3}=\frac{\mu}{-4-0}=\frac{1}{9+16}, \lambda=\frac{3}{25}, \mu=\frac{-4}{25}$
Therefore,
$\therefore \sin x=3 / 25(3 \sin x+4 \cos x)-4 / 25(3 \cos x-4 \sin x)$
$I=\int \frac{\frac{3}{25}(3 \sin x+4 \operatorname{cos} x)-\frac{4}{25}(3 \operatorname{cos} x-4 \operatorname{sin} x)}{3 \sin x+4 \cos x} d x$
$\mathrm{I}=\int \frac{3}{25} \frac{(3 \sin \mathrm{x}+4 \cos \mathrm{x})}{(3 \sin x+4 \cos x)} \mathrm{d} \mathrm{x}-\frac{4}{25} \int \frac{3 \cos \mathrm{x}-4 \sin \mathrm{x}}{3 \sin x+4 \cos x} d x$
$1=\frac{3}{25} \int 1 . d x-\frac{4}{25} \int \frac{3 \operatorname{cos} x-4 \sin x}{3 \sin x+4 \cos x} d x$
put $3sin \: x + 4 cos \; x = t$ in second integral and differentiate both sides,
we get $(3 \cos x-4 \sin x) d x=d t$
$\mathrm{I}=\frac{3}{25} \int 1 . d x-\frac{4}{25} \int \frac{1}{t} d x$
$\mathrm{I}=\frac{3}{25} x-\frac{4}{25} \log |t|+c$
put $t=3 \sin x+4 \cos x$
$I=\frac{3}{25} x-\frac{4}{25} \log |3 \sin x+4 \operatorname{cos} x|+c$

Indefinite Integrals exercise 18.24 question 1 sub question (viii)

Answer: $\frac{18}{25} x+\frac{1}{25} \log |3 \operatorname{sin} x+4 \cos x|+C$
Hint: use the formula in which ,
Put Numerator = λ denominator+ μ (derivative of denominator)
Given: $\int \frac{2 \tan x+3}{3 \tan x+4} d x$
Explanation:
$\text { Let } I=\int \frac{2 \tan x+3}{3 \tan x+4} d x$
$=\int \frac{2\left(\frac{\sin x}{\cos x}\right)+3}{3\left(\frac{\sin x}{\cos x}\right)+4} \mathrm{~d} \mathrm{x}$
$=\int \frac{\frac{2 \operatorname{sin} x+3 \operatorname{cos} x}{\operatorname{cos} x}}{\frac{3 \operatorname{sin} x+4 \operatorname{cos} x}{\operatorname{cos} x}} d x$
$=\int \frac{2 \sin x+3 \cos x}{3 \sin x+4 \cos x} d x$
$\text { Let, } 2 \sin x+3 \cos x=\lambda(3 \sin x+4 \cos x)+\mu(3 \cos x-4 \sin x)$
Equating co- efficient of Cos x and Sin x, We get,
$\begin{array}{ll} 4 \lambda+3 \mu=3 & \text { i.e. } 4 \lambda+3 \mu-3=0 \\ 3 \lambda-4 \mu=2 & \text { i.e. } 3 \lambda-4 \mu-2=0 \end{array}$
Solving these, we get;
$\begin{aligned} &\frac{\lambda}{-6-12}=\frac{\mu}{-9+8}=\frac{1}{-16-9} \\ &\frac{\lambda}{-18}=\frac{\mu}{-1}=\frac{1}{-25} \\ &\lambda=\frac{18}{25} \quad, \quad \mu=\frac{1}{25} \end{aligned}$
Therefore,
$2 \sin x+3 \cos x=\frac{18}{25}(3 \sin x+4 \cos x)+\frac{1}{25}(3 \cos x-4 \sin x)$
Therefore,
$I=\frac{\frac{18}{25}(3 \sin x+4 \cos x)+\frac{1}{25}(3 \operatorname{Cos} x-4 \sin x)}{3 \sin x+4 \cos x} \mathrm{dx}$
$=\frac{18}{25} \int 1 d x+\frac{1}{25} \int \frac{3 \cos x-4 \sin x}{3 \sin x+4 \cos x} \mathrm{~d} \mathrm{x}$
$=\frac{18}{25} x+\frac{1}{25} \log |3 \operatorname{Sin} x+4 \operatorname{Cos} x|+C$

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: $I=\log |3 \operatorname{Cos} x+2 \operatorname{Sin} x|+2 x+C$
Hint: use the formula in which ,
Put Numerator = A denominator+ B (derivative of denominator)
Given: $\int\frac{8 \cot x+1}{3 \cot x+2} \mathrm{~d} \mathrm{x}$
Explanation:
$\text { Let } \mathrm{I}=\int \frac{8 \cot x+1}{3 \cot x+2} \mathrm{~d} \mathrm{x}$
$I=\int \frac{8 \frac{\cos x}{\sin x}+1}{3 \frac{\cos x}{\sin x}+2} \mathrm{dx}$
$=\int \frac{\frac{8 \cos x+\sin x}{\sin x}}{\frac{3 \cos x+2 \sin x}{\sin x}} \mathrm{~d} \mathrm{x}$
$=\int \frac{8 \cos x+\sin x}{3 \cos x+2 \sin x} \mathrm{~d} \mathrm{x}$
So, we will take the steps as directed:
$\begin{aligned} &\sin x+8 \cos x=A \frac{d}{d x}(3 \cos x+2 \sin x)+B(3 \cos x+2 \sin x) \\ &\sin x+8 \cos x=A(-3 \sin x+2 \cos x)+B(3 \cos x+2 \sin x) \end{aligned}$
$\begin{aligned} &\left(\frac{d}{d x} \cos x=-\sin x\right) \\ &\sin x+8 \cos x=\sin x(2 B-3 A)+\operatorname{cos} x(2 A+3 B) \end{aligned}$
Comparing both the sides,
$\begin{array}{ll} 2 \mathrm{~B}-3 \mathrm{~A}=1 & \text { ...(i) } \\ 3 \mathrm{~B}+2 \mathrm{~A}=8 & \text { ...(ii) } \end{array}$
Multiplying (i) by 2 and (ii) by 3 then add;
$4B-6A+9B+6A=2+24$
On solving for A, B , We have A =1 , B = 2
Thus, I can be expressed as:
$I=\int \frac{(-3 \sin x+2 \operatorname{cos} x)+2(3 \cos x+2 \sin x)}{3 \operatorname{cos} x+2 \sin x} d x$
$\begin{aligned} &\mathrm{I}=\int \frac{(-3 \sin x+2 \cos x)+2(3 \cos x+2 \sin x)}{3 \cos x+2 \sin x} \mathrm{dx} \\ &\mathrm{I}=\int \frac{(-3 \sin x+2 \cos x)}{3 \cos x+2 \sin x} \mathrm{dx}+\int \frac{2(3 \cos x+2 \sin x)}{3 \cos x+2 \sin x} \mathrm{dx} \end{aligned}$
$\text { Let, } I_{1}=\int \frac{(-3 \sin x+2 \operatorname{Cos} x)}{3 \cos x+2 \sin x} \mathrm{dx} \quad \text { and } I_{2}=\int \frac{2(3 \cos x+2 \sin x)}{3 \cos x+2 \sin x} \mathrm{dx}$
$\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2} \quad \text { ... Eq. } (iii)$
$I_{1}=\int \frac{(-3 \sin x+2 \cos x)}{3 \cos x+2 \sin x} \mathrm{~d} \mathrm{x}$
$\begin{aligned} &\text { Let, } 3 \cos x+2 \sin x=\mu \\ &(-3 \sin x+2 \cos x)=d \mu \end{aligned}$
So, $I_{1}$ reduce to
$\mathrm{I}_{1}=\int \frac{d \mu}{\mu}=\log |\mu|+\mathrm{c}_{1}$
Therefore ,
$\begin{aligned} &I_{1}=\log |3 \operatorname{Cos} x+2 \sin x|+c_{1} \quad \ldots \text { Eq. (iv) }\\ &\text { As } I_{2}=\int 2 \frac{(3 \cos x+2 \sin x)}{3 \cos x+2 \sin x} \mathrm{dx}\\ &I_{2}=2 \int d x=2 x+c_{2} \quad \ldots\text { Eq. (v) } \end{aligned}$
From Eq. (iii) , (iv) and (v) we have
$I=\log | 3 \operatorname{cos} x+2 \sin x \mid+c_{1}+2 x+c_{2}$
Therefore,
$I=\log |3 \operatorname{Cos} x+2 \sin x|+2 x+c \quad\left[c_{1}+c_{2}=c\right]$

Indefinite Integrals exercise 18.24 question 1 sub question (xi)

Answer: $I=\frac{9}{41} \log |4 \operatorname{Cos} x+5 \sin x|+\frac{40 x}{41}+C$
Hint: use the formula in which
Put Numerator = λ denominator+ μ (derivative of denominator)
Given: $I=\int \frac{4 \sin x+5 \cos x}{5 \sin x+4 \cos x} \mathrm{~d} \mathrm{x}$
Explanation:
$\text { let, } I=\int \frac{4 \sin x+5 \cos x}{5 \sin x+4 \cos x} \mathrm{~d} \mathrm{x}$
So, we will take the steps as described
$4 \sin x+5 \cos x=A(5 \cos x-4 \sin x)+B(4 \cos x+5 \sin x)$
$\left(\frac{d}{d x} \operatorname{cos} x=-\sin x\right)$
$4 \sin x+5 \cos x=(5 B-4 A) \sin x+(5 A+4 B) \cos x$
Comparing both sides, we have
$\begin{aligned} &5 \mathrm{~B}-4 \mathrm{~A}=4 \quad \ldots (i)\\ &4 B+5 A=5 \quad \ldots(ii) \end{aligned}$
Multiplying (i) by 5 and (ii) by 4 and then add;
$25 B-20 A+16 B+20 A=20+20$
On solving for A and B , we have
$A=\frac{9}{41}, B=\frac{40}{41}$
Thus, I can be expressed as
$\mathrm{I}=\int \frac{\frac{9}{41}(5 \cos x-4 \sin x)+\frac{40}{41}(4 \cos x+5 \sin x)}{4 \cos x+5 \sin x} d x$
$I=\int \frac{9}{41}\left(\frac{5 \cos x-4 \sin x}{4 \cos x+5 \sin x}\right) \mathrm{d} \mathrm{x}+\int \frac{40}{41}\left(\frac{4 \cos x+5 \sin x}{4 \cos x+5 \sin x}\right) \mathrm{dx}$
$\begin{aligned} &\text { Let, } I_{1}=\frac{9}{41} \int\left(\frac{5 \cos x-4 \sin x}{4 \cos x+5 \sin x}\right) \mathrm{d} \mathrm{x} \\ &\mathrm{I}_{2}=\frac{40}{41} \int\left(\frac{4 \cos x+5 \sin x}{4 \cos x+5 \sin x}\right) \mathrm{d} \mathrm{x} \end{aligned}$
$\begin{array}{ll} I=I_{1}+I_{2} & \ldots \text { equ. }(iii) \end{array}$
$I_{1}=\frac{9}{41} \int\left(\frac{5 \cos x-4 \sin x}{4 \cos x+5 \sin x}\right) d x$
$\begin{aligned} &\text { Let } 4 \cos x+5 \sin x=u \\ &\Rightarrow(-4 \sin x+5 \cos x) d x=d u \end{aligned}$
So, $I_{1}$ reduces to
$I_{1}=\frac{9}{41} \int \frac{d u}{u}=\frac{9}{41} \log |u|+c_{1}$
Therefore,
$I_{1}=\frac{9}{41} \log |4 \operatorname{Cos} x+5 \sin x|+c_{1} \quad \ldots \text { Eq. (iv) }$
$\text { As, }I_{2}=\frac{40}{41} \int \frac{4 \cos x+5 \sin x}{4 \cos x+5 \sin x} \mathrm{dx}$
$I_{2}=\frac{40}{41} \int d x=\frac{40}{41} \times+c_{2} \quad \text { ... Eq. }(v)$
From Equ (iii) , (iv) , (v), we have
$I=\frac{9}{41} \log |4 \cos x+5 \sin x|+c_{1}+\frac{40}{41} x+c_{2}$
Therefore,
$I=\frac{9}{41} \log |4 \cos x+5 \sin x|+\frac{40}{41} x+c \quad\left[c_{1}+c_{2}=c\right]$

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

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