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  • need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.24 question 1 sub question (vii)

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

Answers (1)

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Answer: \frac{3}{25} x-\frac{4}{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{1}{3+4 \cot x} \mathrm{~d} \mathrm{x}

Explanation:

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

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

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

Solving this, we get

\frac{\lambda}{0+3}=\frac{\mu}{-4-0}=\frac{1}{9+16}, \lambda=\frac{3}{25}, \mu=\frac{-4}{25}

Therefore,

\therefore \sin x=3 / 25(3 \sin x+4 \cos x)-4 / 25(3 \cos x-4 \sin x)

        I=\int \frac{\frac{3}{25}(3 \sin x+4 \operatorname{cos} x)-\frac{4}{25}(3 \operatorname{cos} x-4 \operatorname{sin} x)}{3 \sin x+4 \cos x} d x

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

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

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

we get  (3 \cos x-4 \sin x) d x=d t

        \mathrm{I}=\frac{3}{25} \int 1 . d x-\frac{4}{25} \int \frac{1}{t} d x

        \mathrm{I}=\frac{3}{25} x-\frac{4}{25} \log |t|+c

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

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

 

 

Posted by

infoexpert26

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