$$\int_{ - \pi }^\pi {{{2x\left( {1 + \sin \,x} \right)} \over {1 + {{\cos }^2}x}}} dx$$

$$ = \int_{ - \pi }^\pi {{{2x\,dx} \over {1 + {{\cos }^2}x}} + 2\int_{ - \pi }^\pi {{{x\,\sin x} \over {1 + {{\cos }^2}x}}} } dx$$

$$ = 0 + 4\int_0^\pi {{{x\sin x\,dx} \over {1 + {{\cos }^2}x}}} ;$$

$$\left[ \, \right.$$ as $$\int\limits_{ - a}^a {f\left( x \right)} dx = 0$$ $$\left. \, \right]$$

if $$f(x)$$ is odd

$$ = 2\int\limits_0^a {f\left( x \right)} dx$$ if $$f(x)$$ is even.

$$I = 4\int_0^\pi {{{\left( {\pi - x} \right)\sin \left( {\pi - x} \right)} \over {1 + {{\cos }^2}\left( {\pi - x} \right)}}} dx$$

$$I = 4\int_0^\pi {{{\left( {\pi - x} \right)\sin \,x} \over {1 + {{\cos }^2}x}}} $$

$$ \Rightarrow I = 4\pi \int_0^\pi {{{\sin x\,dx} \over {1 + {{\cos }^2}x}}} - 4\int {{{x\sin x\,dx} \over {1 + {{\cos }^2}x}}} $$

$$ \Rightarrow 2I = 4\pi \int_0^\pi {{{\sin x} \over {1 + {{\cos }^2}x}}} dx$$

put $$\cos x = t \Rightarrow - \sin xdx = dt$$

$$\therefore$$ $$I = - 2\pi \int\limits_1^{ - 1} {{1 \over {1 + {t^2}}}} dt$$

$$ = 2\pi \int\limits_{ - 1}^1 {{1 \over {1 + {t^2}}}} dt$$

$$ = 2\pi \left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1$$

$$ = 2\pi \left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}\left( { - 1} \right)} \right]$$

$$ = 2\pi \left[ {{\pi \over 4} - \left( {{{ - \pi } \over 4}} \right)} \right] = 2\pi {\pi \over 2} = {\pi ^2}$$

$$\int\limits_0^2 {\left[ {{x^2}} \right]} dx = \int\limits_0^1 {\left[ {{x^2}} \right]dx} + \int\limits_1^{\sqrt 2 } {\left[ {{x^2}} \right]} dx + $$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$$$\int\limits_{\sqrt 2 }^{\sqrt 3 } {\left[ {{x^2}} \right]} + \int\limits_{\sqrt 3 }^2 {\left[ {{x^2}} \right]} dx$$

$$ = \int\limits_0^1 {0dx} + \int\limits_1^{\sqrt 2 } {1dx} + \int\limits_{\sqrt 2 }^{\sqrt 3 } {2dx} + \int\limits_{\sqrt 3 }^2 {3dx} $$

$$ = \left[ x \right]_1^{\sqrt n } + \left[ {2x} \right]_{\sqrt 2 }^{\sqrt 3 } + \left[ {3x} \right]_{\sqrt 3 }^2$$

$$ = \sqrt 2 - 1 + 2\sqrt 3 - 2\sqrt 2 + 6 - 3\sqrt 3 $$

$$ = 5 - \sqrt 3 - \sqrt 2 $$

$$I = \int\limits_0^{10\pi } {\left| {\sin x} \right|} dx$$

$$ = 10\int\limits_0^\pi {\left| {\sin x\,} \right|} \,dx$$

$$ = 10\int\limits_0^\pi {\sin \,x\,dx} $$

$$\left[ \, \right.$$ as $$\left| {\sin x} \right|$$ is periodic with period $$\pi $$

and $$sin$$ $$x > 0$$ if $$0 < x$$ $$ < \pi $$ $$\left. \, \right]$$

$$I = 20\int\limits_0^{\pi /2} {\sin x} \,dx$$

$$ = 20\left[ { - \cos \,x_0^{\pi /2}} \right] = 20$$

$${I_n} + {I_{n + 2}}$$

$$ = \int\limits_0^{\pi /4} {{{\tan }^n}} \,x\left( {1 + {{\tan }^2}x} \right)dx$$

$$ = \int\limits_0^{\pi /4} {{{\tan }^n}} x\,{\sec ^2}x\,dx$$

$$ = \left[ {{{{{\tan }^{n + 1}}x} \over {n + 1}}} \right]_0^{\pi /4}$$

$$ = {{1 - 0} \over {n + 1}} = {1 \over {n + 1}}$$

$$\therefore$$ $${I_n} + {I_{n + 2}} = {1 \over {n + 1}}$$

$$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } n\left[ {{I_n} + {I_{n + 2}}} \right]$$

$$ = \mathop {\lim }\limits_{n \to \infty } \,n.{1 \over {n + 1}} = \mathop {\lim }\limits_{n \to \infty } {n \over {n + 1}}$$

$$ = \mathop {\lim }\limits_{n \to \infty } {n \over {n\left( {1 + {1 \over n}} \right)}} = 1$$