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

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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


\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}{-18}=\frac{\mu}{-1}=\frac{1}{-25}, \lambda=\frac{18}{25} \quad ; \quad \mu=\frac{1}{25}


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


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