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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 11

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 11

Answers (1)

Answer:

            \log \left|\frac{(\sin x+2)^{4}}{(\sin x+1)^{2}}\right|+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

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

Explanation:

Let

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

Let

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

\begin{aligned} &\frac{2 y}{(1+y)(2+y)}=\frac{A}{1+y}+\frac{B}{2+y} \\ &\frac{2 y}{(1+y)(2+y)}=\frac{A(2+y)+B(1+y)}{(1+y)(2+y)} \\ &2 y=(2 A+B)+y(A+B) \end{aligned}

Comparing coefficient

2=A+B                    (1)

2A+B=0                  (2)

Subtract equation (1) from equation (2)

A=-2

Equation (1)

2=-2+B\\B=4

Now

\begin{aligned} &\frac{2 y}{(y+1)(2+y)}=\frac{-2}{y+1}+\frac{4}{y+2} \\ &I=-2 \int \frac{1}{y+1} d y+4 \int \frac{1}{y+2} d y \\ &I=-2 \log |y+1|+4 \log |y+2|+C \end{aligned}

\begin{aligned} &I=\log \left|\frac{(y+2)^{4}}{(y+1)^{2}}\right|+C \\ &I=\log \left|\frac{(\sin x+2)^{4}}{(\sin x+1)^{2}}\right|+C\quad\quad\quad\quad \quad[\because y=\sin x] \end{aligned}

 

Posted by

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