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### 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}

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