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

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

Answers (1)

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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

 

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