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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 61

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 61

Answers (1)

Answer:

               \frac{1}{6} \log (\cos x-1)+\frac{1}{2} \log (\cos x+1)-\frac{2}{3} \log \left(\cos x+\frac{1}{2}\right)+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{1}{\sin x+\sin 2 x} d x \\

Explanation:

Let

\begin{aligned} &I=\int \frac{1}{\sin x+\sin 2 x} d x \\ &I=\int \frac{1}{\sin x+2 \sin x \cos x} d x \\ &I=\int \frac{1}{\sin x(1+2 \cos x)} d x \end{aligned}

Put

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

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

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

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

At t=- 1

\begin{aligned} &1=-2 B\left(\frac{-1}{2}\right) \\ &B=1 \end{aligned}

At t= 1

\begin{aligned} &1=2 A\left(\frac{3}{2}\right) \\ &A=\frac{1}{3} \end{aligned}

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

Put in (1) using (2)
\begin{aligned} &I=\frac{1}{2} \cdot \frac{1}{3} \int \frac{1}{t-1} d t+\frac{1}{2} \int \frac{1}{t+1} d t-\frac{1}{2} \cdot \frac{4}{3} \int \frac{1}{t+\frac{1}{2}} d t \\ &I=\frac{1}{6} \log (t-1)+\frac{1}{2} \log (t+1)-\frac{2}{3} \log \left(t+\frac{1}{2}\right)+C \\ &I=\frac{1}{6} \log (\cos x-1)+\frac{1}{2} \log (\cos x+1)-\frac{2}{3} \log \left(\cos x+\frac{1}{2}\right)+C \end{aligned}

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