Question

Please solve this with proper steps ​

Answer 1

$\displaystyle\longrightarrow\sf{\int\dfrac{\cos(2x)+\sin x}{1+2\sin x}\ dx=\int\dfrac{1-2\sin^2x+\sin x}{1+2\sin x}\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac{\cos(2x)+\sin x}{1+2\sin x}\ dx=\int\dfrac{1-2\sin^2x+2\sin x-\sin x}{1+2\sin x}\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac{\cos(2x)+\sin x}{1+2\sin x}\ dx=\int\dfrac{1+2\sin x-\sin x-2\sin^2x}{1+2\sin x}\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac{\cos(2x)+\sin x}{1+2\sin x}\ dx=\int\dfrac{1+2\sin x-\sin x(1+2\sin x)}{1+2\sin x}\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac{\cos(2x)+\sin x}{1+2\sin x}\ dx=\int\dfrac{(1+2\sin x)(1-\sin x)}{1+2\sin x}\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac{\cos(2x)+\sin x}{1+2\sin x}\ dx=\int(1-\sin x)\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac{\cos(2x)+\sin x}{1+2\sin x}\ dx=\int1\ dx-\int\sin x\ dx}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\int\dfrac{\cos(2x)+\sin x}{1+2\sin x}\ dx=x+\cos x+c}}}$