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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.24 question 1 sub question (v) maths textbook solution

Answers (1)


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


\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}

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