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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 60

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 60

Answers (1)

Answer:

                \frac{1}{10} \log (\cos x-1)-\frac{1}{2} \log (\cos +1)+\frac{2}{5} \log \left(\frac{3}{2}+\cos x\right)+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{1}{\sin x(3+2 \cos x)} d x

Explanation:

Let

I=\int \frac{1}{\sin x(3+2 \cos x)} d x

\begin{aligned} &\text { Put }\\ &\operatorname{cox}=t\\ &-\sin x d x=d t\\ &d x=\frac{-1}{\sin x} d t\\ &I=\int \frac{-1}{\sin ^{2} x(3+2 t)} d t=\int \frac{-1}{\left(1-\cos ^{2} x\right)(3+2 t)} d t\\ &I=\int \frac{1}{\left(t^{2}-1\right)(3+2 t)} d t \end{aligned}

I=\frac{1}{2} \int \frac{1}{(t-1)(t+1)\left(\frac{3}{2}+t\right)} d t                (1)

\frac{1}{(t-1)(t+1)\left(\frac{3}{2}+t\right)}=\frac{A}{t-1}+\frac{B}{t+1}+\frac{C}{\frac{3}{2}+t}

\begin{aligned} &\frac{1}{(t-1)(t+1)\left(\frac{3}{2}+t\right)}=\frac{A(t+1)\left(\frac{3}{2}+t\right)+B(t-1)\left(\frac{3}{2}+t\right)+C\left(t^{2}-1\right)}{(t-1)(t+1)\left(\frac{3}{2}+t\right)} \\ &1=A(t+1)\left(\frac{3}{2}+t\right)+B(t-1)\left(\frac{3}{2}+t\right)+C\left(t^{2}-1\right) \end{aligned}

\begin{aligned} &\text { At } t=-1 \\ &1=-2 B\left(\frac{1}{2}\right) \\ &B=-1 \\ &\text { At } t=1 \\ &1=2 A\left(\frac{5}{2}\right) \end{aligned}

\begin{aligned} &A=\frac{1}{5} \\ &\text { At } t=\frac{-3}{2} \\ &1=C\left(\frac{9}{4}-1\right) \\ &C=\frac{4}{5} \end{aligned}

Put in (1)

\begin{aligned} &I=\frac{1}{2} \times \frac{1}{5} \int \frac{1}{t-1} d t+\frac{(-1)}{2} \int \frac{1}{t+1} d t+\frac{1}{2} \times \frac{4}{5} \int \frac{1}{\frac{3}{2}+t} d t \\ &I=\frac{1}{10} \log (t-1)-\frac{1}{2} \log (t+1)+\frac{2}{5} \log \left(\frac{3}{2}+t\right)+C \\ &I=\log (\cos x-1)-\frac{1}{2} \log (\cos x+1)+\frac{2}{5} \log \left(\frac{3}{2}+\cos x\right)+C \end{aligned}

 

Posted by

infoexpert27

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