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explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.24 question 1 sub question (viii) maths

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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


\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}


2 \sin x+3 \cos x=\frac{18}{25}(3 \sin x+4 \cos x)+\frac{1}{25}(3 \cos x-4 \sin x)


    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


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