$\underset{0}{\overset{\mathit{\pi }}{\int }}\frac{\mathit{x}}{1+\mathit{s}\mathit{i}\mathit{n}\mathit{x}}$ dx
Using the property $\underset{0}{\overset{\mathit{a}}{\int }}$ f (x) dx = $\underset{0}{\overset{\mathit{a}}{\int }}$ f (a - x) dx
⇒ I = $\underset{0}{\overset{\mathit{\pi }}{\int }}\frac{\mathit{\pi }-\mathit{x}}{1+\mathit{s}\mathit{i}\mathit{n}\left(\mathit{\pi }-\mathit{x}\right)}$ dx
= $\underset{0}{\overset{\mathit{\pi }}{\int }}$ $\frac{\mathit{\pi }-\mathit{x}}{1+\mathit{s}\mathit{i}\mathit{n}\mathit{x}}$ dx ... (2)
Now adding (1) and (2), we get
2I = $\underset{0}{\overset{\mathit{\pi }}{\int }}$ $\frac{\mathit{x}}{1+\mathit{s}\mathit{i}\mathit{n}\mathit{x}}$ dx + $\underset{0}{\overset{\mathit{\pi }}{\int }}$ $\frac{\mathit{\pi }-\mathit{x}}{1+\mathit{s}\mathit{i}\mathit{n}\mathit{x}}$ dx
= $\underset{0}{\overset{\mathit{\pi }}{\int }}$ $\frac{\mathit{\pi }}{1+\mathit{s}\mathit{i}\mathit{n}\mathit{x}}$ dx
= π $\underset{0}{\overset{\mathit{\pi }}{\int }}$ $\frac{1}{1+\mathit{s}\mathit{i}\mathit{n}\mathit{x}}$ dx
= π $\underset{0}{\overset{\mathit{\pi }}{\int }}$ $\frac{\left(1-\mathit{s}\mathit{i}\mathit{n}\mathit{x}\right)}{\left(1-\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{x}\right)}$ dx
== π $\underset{0}{\overset{\mathit{\pi }}{\int }}$ $\frac{\left(1-\mathit{s}\mathit{i}\mathit{n}\mathit{x}\right)}{\left(\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{x}\right)}$ dx
= π $\left[\underset{0}{\overset{\mathit{\pi }}{\int }}\left(\frac{1}{\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{x}}-\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{x}}{\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{x}}\right)\mathit{d}\mathit{x}\right]$
= π $\left[\underset{0}{\overset{\mathit{\pi }}{\int }}\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{x}-\mathit{s}\mathit{e}\mathit{c}\mathit{x}\mathit{t}\mathit{a}\mathit{n}\mathit{x}\mathit{d}\mathit{x}\right]$
= π $\left[\underset{0}{\overset{\mathit{\pi }}{\int }}\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{x}\mathit{d}\mathit{x}-\underset{0}{\overset{\mathit{\pi }}{\int }}\mathit{s}\mathit{e}\mathit{c}\mathit{x}\mathit{t}\mathit{a}\mathit{n}\mathit{x}\mathit{d}\mathit{x}\right]$
= π $\left({\left[\mathit{t}\mathit{a}\mathit{n}\mathit{x}\right]}_0^{\mathit{\pi }}-{\left[\mathit{s}\mathit{e}\mathit{c}\mathit{x}\right]}_0^{\mathit{\pi }}\right)$
⇒ 2I = π (2)
⇒ I = π
So, $\underset{0}{\overset{\mathit{\pi }}{\int }}\frac{\mathit{x}}{1+\mathit{s}\mathit{i}\mathit{n}\mathit{x}}$ dx = π