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

Answers (1)

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: $\int \frac{2 \sin x+3 \cos x}{3 \sin x+4 \cos x} \mathrm{~d} x$

Explanation:

$\text { Let } I=\int \frac{2 \sin x+3 \cos x}{3 \sin x+4 \cos x} \mathrm{~d} \mathrm{x}$

$\text { Let } 2 \sin x+3 \cos x=\lambda(3 \sin x+4 \cos x)+\mu(3 \cos x-4 \sin x)$

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;

$\frac{\lambda}{-6-12}=\frac{\mu}{-9+8}=\frac{1}{-16-9}$

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

Therefore,

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

$\therefore I=\int \frac{\frac{18}{25}(3 \sin x+4 \cos x)+\frac{1}{25}(3 \cos x-4 \sin x)}{3 \sin x+4 \cos x} d x$

$=\frac{18}{25} \int 1 d x+\frac{1}{25} \int \frac{3 \cos x-4 \sin x}{3 \sin x+4 \cos x} d x$

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 $t=3 \sin x+4 \cos x$

$I=\frac{18}{25} x+\frac{1}{25} \log |3 \sin x+4 \operatorname{Cos} x|+C$

Posted by

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