If $\underset{20}{\overset{40}{\int }}\frac{\sin x}{\sin x+\mathrm{s}\mathrm{i}\mathrm{n}\left(60-x\right)}dx=k,$ then the value of $\frac{k}{4}$ is equal to

As, $k=\underset{20}{\overset{40}{\int }}\frac{\sin x}{\sin x+\mathrm{s}\mathrm{i}\mathrm{n}\left(60-x\right)}dx$  .... (i)

Applying $\left(a+b-x\right)$ property, we get,

$k=\underset{20}{\overset{40}{\int }}\frac{\sin \left(60-x\right)}{\mathrm{s}\mathrm{i}\mathrm{n}\left(60-x\right)+\sin x}dx$  ... (ii)

Adding (i) and (ii), we get, 

 $2k=\underset{20}{\overset{40}{\int }}1\mathrm{ }dx={\left[x\right]}_{20}^{40}$

$\Rightarrow 2k=40-20$

$\Rightarrow k=10$

$\therefore \frac{k}{4}=2.5$