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provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.24 question 1 sub question (x)

Answers (1)


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}


\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;


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}


        I=\log |3 \operatorname{Cos} x+2 \sin x|+2 x+c \quad\left[c_{1}+c_{2}=c\right]    

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