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

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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}


\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

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