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

Answer: $\frac{4}{25} x+\frac{3}{25} \log |4 \cos x+3 \operatorname{Sin} x|+C$

Hint: use the formula in which ,

Put Numerator = λ denominator+ μ (derivative of denominator)

Given: $\int \frac{1}{4+3 \tan x} d x$

Explanation:

$\text { Let } \mathrm{I}=\int \frac{1}{4+3 \tan x} \mathrm{dx}$

\begin{aligned} &=\int \frac{1}{4+3\left(\frac{\sin x}{\cos x}\right)} \mathrm{dx} \\ &=\int \frac{\cos x}{4 \cos x+3 \sin x} \mathrm{dx} \end{aligned}

$\cos x=\lambda(4 \cos x+3 \sin x)+\mu(-4 \sin x+3 \cos x)$

Equating co- efficient of Cos x and Sin x, We get,

\begin{aligned} &4 \lambda+3 \mu-1=0 \ \\ &3 \lambda-4 \mu-0=0 \end{aligned}

Solving these, we get;

\begin{aligned} &\frac{\lambda}{0-4}=\frac{\mu}{-3+0}=\frac{1}{-16-9} \quad \lambda=\frac{4}{25} \quad, \quad \mu=\frac{3}{25} \\ &\therefore \cos x=\frac{4}{25}(4 \cos x+3 \sin x)+\frac{3}{25}(-4 \sin x+3 \cos x) \end{aligned}

$\therefore \mathrm{I}=\int \frac{\frac{4}{25}(4 \cos x+3 \sin x)+\frac{3}{25}(-4 \sin x+3 \cos x)}{4 \cos x+3 \sin x} \mathrm{dx}$

$=\int \frac{4}{25} d x+\frac{3}{25} \cdot \int \frac{-4 \sin x+3 \cos x}{4 \cos x+3 \sin x} \mathrm{~d} \mathrm{x}$

put $4cos x + 3 sin x = t$  in second integral and differentiate both sides,

we get, $(-4 \sin x+3 \cos x) d x=d t$

$I=\int \frac{4}{25} d x+\frac{3}{25} \cdot \int \frac{1}{t} \mathrm{dt}$

$=\frac{4}{25} x+\frac{3}{25} \log |t|+C$

$\text { put } t=4 \cos x+3 \sin x$

$I=\frac{4}{25} x+\frac{3}{25} \log |4 \cos x+3 \operatorname{sin} x|+C$