$$\eqalign{ & I = \int\limits_0^1 {\frac{{8\log \left( {1 + x} \right)}}{{1 + {x^2}}}} dx \cr & {\text{Put}}\,x = \tan \,\theta , \cr & \therefore \frac{{dx}}{{d\theta }} = {\sec ^2}\theta \Rightarrow dx = {\sec ^2}\theta \,d\theta \cr & \therefore I = 8\int\limits_0^{\frac{\pi }{4}} {\frac{{\log \left( {1 + \tan \,\theta } \right)}}{{1 + {{\tan }^2}\theta }}} .{\sec ^2}\theta \,d\theta \cr & I = 8\int\limits_0^{\frac{\pi }{4}} {\log \left( {1 + \tan \,\theta } \right)} \,d\theta .....{\text{(i)}} \cr & = 8\int\limits_0^{\frac{\pi }{4}} {\log \left[ {1 + \tan \,\left( {\frac{\pi }{4} - \theta } \right)} \right]} d\theta \cr & = 8\int\limits_0^{\frac{\pi }{4}} {\log \left[ {1 + \frac{{1 - \tan \,\theta }}{{1 + \tan \,\theta }}} \right]} d\theta \cr & = 8\int\limits_0^{\frac{\pi }{4}} {\log \left[ {\frac{2}{{1 + \tan \,\theta }}} \right]d\theta } \cr & = 8\int\limits_0^{\frac{\pi }{4}} {\left[ {\log 2 - \log \left( {1 + \tan \,\theta } \right)} \right]d\theta } \cr & I = 8.\left( {\log \,2} \right)\left[ x \right]_0^{\frac{\pi }{4}} - 8\int\limits_0^{\frac{\pi }{4}} {\log \left( {1 + \tan \,\theta } \right)d\theta } \cr & I = 8.\frac{\pi }{4}.\log \,2 - I\,\,\,\,\left[ {{\text{from equation (i)}}} \right] \cr & \Rightarrow 2I = 2\pi \,\log \,2, \cr & \therefore I = \pi \,\log \,2 \cr} $$

$$\eqalign{ & I = \int_0^\pi {{e^{{{\cos }^2}x\,}}\,} {\cos ^3}\left( {2n + 1} \right)x\,dx,\,n \in \,Z.....(1) \cr & = \int_0^\pi {{e^{{{\cos }^2}\left( {\pi - x} \right)}}} \,{\cos ^3}\left[ {\left( {2n + 1} \right)\left( {\pi - x} \right)} \right]dx \cr & {\text{Using }}\int_0^a {f\left( x \right)dx = } \int_0^a {f\left( {a - x} \right)dx} \cr & \therefore I = \int_0^\pi {{e^{{{\cos }^2}x}}} \,{\cos ^3}\left[ {\left( {2n + 1} \right)\pi - \left( {2n + 1} \right)x} \right]dx \cr & I = \int_0^\pi {\left( { - {e^{{{\cos }^2}x}}\,{{\cos }^3}} \right)\left( {2n + 1} \right)x\,dx.....(2)} \cr & {\text{Adding (1) and (2) we get}} \cr & 2I = 0\,\,\,\, \Rightarrow I = 0 \cr} $$

$$\eqalign{ & \frac{{dy}}{{dx}} = f'\left( x \right) \cr & \Rightarrow x\left( {{e^x} - 1} \right)\left( {x - 1} \right){\left( {x - 2} \right)^3}{\left( {x - 3} \right)^5} = 0 \cr} $$
Critical points are $$0,\,1,\,2,\,3$$
Consider change of sign of $$\frac{{dy}}{{dx}}$$  at $$x = 3.$$
$$x < 3,\,\frac{{dy}}{{dx}} = $$   negative and $$x > 3,\,\frac{{dy}}{{dx}} = $$   positive
Change is from negative to positive, hence minimum at $$x = 3.$$  Again minimum and maximum occur alternately.
$$\therefore $$  2nd minimum is at $$x = 1.$$

$$\mathop {Lim}\limits_{n \to \infty } \sum\limits_{r = 1}^n {\frac{1}{n}{e^{\frac{r}{n}}}} $$
[Using definite integrals as limit of sum]
$$ = \int\limits_0^1 {{e^x}dx = e - 1} $$

$$\eqalign{ & I = \int_0^\pi {\log \left( {1 + \cos \,x} \right)} dx \cr & = \int_0^\pi {\log \left( {2\,{{\cos }^2}\frac{x}{2}} \right)} dx \cr & = \int_0^\pi {\left( {\log \,2 + 2\,\log \,\cos \frac{x}{2}} \right)} dx \cr & = \int_0^\pi {\log \,2\,dx + 2\int_0^\pi {\log \,\cos \frac{x}{2}dx} } \cr & = \pi \,\log \,2 + 2\int_0^{\frac{\pi }{2}} {\left( {2\,\log \,\cos \,t} \right)} dt\,\,\,\,\,\left( {{\text{where }}\frac{x}{2} = t} \right) \cr & = \pi \,\log \,2 + 4\left( { - \frac{\pi }{2}\log \,2} \right) \cr & = \pi \,\log \,2 - 2\pi \,\log \,2 \cr & = - \pi \,\log \,2 \cr & \left[ {\int_0^{\frac{\pi }{2}} {\log \,\sin \,\theta \,d\theta = } \int_0^{\frac{\pi }{2}} {\log \,\cos \,\theta \,d\theta } = - \frac{\pi }{2}\log \,2} \right] \cr} $$

$$\eqalign{ & I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\left( {\sin \,x + \cos \,x} \right)}^2}}}{{\sqrt {1 + \sin \,2x} }}dx} \cr & {\text{We know }}\left[ {{{\left( {\sin \,x + \cos \,x} \right)}^2} = 1 + \sin \,2x} \right],{\text{ so}} \cr & I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\left( {\sin \,x + \cos \,x} \right)}^2}}}{{\left( {\sin \,x + \cos \,x} \right)}}dx} \cr & {\text{or }}\,I = \int\limits_0^{\frac{\pi }{2}} {\left( {\sin \,x + \cos \,x} \right)dx} \cr & \left[ {\because \,\sin \,x + \cos \,x > 0\,\,if\,0 < x < \frac{x}{2}} \right] \cr & {\text{or }}\,I = \left[ { - \cos \,x + \sin \,x} \right]_0^{\frac{\pi }{2}} = 2 \cr} $$

$$\eqalign{ & I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} \cr & = \int\limits_0^1 {\left( {1 - x} \right){{\left( {1 - 1 + x} \right)}^n}dx} \cr & = \int\limits_0^1 {\left( {1 - x} \right){x^n}dx} \cr & = \left[ {\frac{{{x^{n + 1}}}}{{n + 1}} - \frac{{{x^{n + 2}}}}{{n + 2}}} \right]_0^1 \cr & = \frac{1}{{n + 1}} - \frac{1}{{n + 2}} \cr} $$

Let $$P = \mathop {{\text{Lim}}}\limits_{n \to \infty } {\left\{ {\frac{{n!}}{{{{\left( {kn} \right)}^n}}}} \right\}^{\frac{1}{n}}}$$
Taking $$\log $$  of both the sides at the base $$e$$
$$\eqalign{ & {\log _e}P = \mathop {{\text{Lim}}}\limits_{n \to \infty } \frac{1}{n}{\log _e}\left\{ {\frac{{n!}}{{{{\left( {kn} \right)}^n}}}} \right\} \cr & = \mathop {{\text{Lim}}}\limits_{n \to \infty } \frac{1}{n}{\log _e}\left\{ {\frac{1}{{kn}}.\frac{2}{{kn}}.\frac{3}{{kn}}......\frac{n}{{kn}}} \right\} \cr & = \mathop {{\text{Lim}}}\limits_{n \to \infty } \frac{1}{n}\left[ {\log \left( {\frac{1}{{kn}}} \right) + \log \left( {\frac{2}{{kn}}} \right) + ...... + \log \left( {\frac{n}{{kn}}} \right)} \right] \cr & = \mathop {{\text{Lim}}}\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^n {\log \left( {\frac{r}{{kn}}} \right)} \cr & = \int\limits_0^1 {\log } \left( {\frac{x}{k}} \right)dx \cr & = \int\limits_0^1 {\left( {\log \,x - \log \,k} \right)dx} \cr & = \int\limits_0^1 {\log } \,x\,dx - \int\limits_0^1 {\log \,k\,dx} \cr & = \left[ {x\,\log \,x - x} \right]_0^1 - \log \,k\left[ x \right]_0^1 \cr & = \left[ {0 - 1 - 0 + 0} \right] - \log \,k \cr & = - 1 - \log \,k \cr & = - \left( {\log \,e + \log \,k} \right) \cr & = - \log \left( {ek} \right) \cr & = \log \frac{1}{{ek}} \cr & \left[ {{\text{Value of}}\,x\,\log \,x\,{\text{at}}\,x = 0\,{\text{is}}\,\mathop {{\text{Lim}}}\limits_{x \to {0^ + }} x\,\log \,x = 0} \right] \cr & \therefore \,P = \frac{1}{{ek}} = {k^{ - 1}}{e^{ - 1}} \cr} $$

$${\text{Let }}I = \int_{{e^{ - 1}}}^{{e^2}} {\left| {\frac{{{{\log }_e}x}}{x}} \right|dx} $$
We know that for $$\frac{1}{e} < x < 1,\,{\log _e}x < 0$$     and hence $$\frac{{{{\log }_e}x}}{x} < 0$$
and for $$1 < x < {e^2},\log \,x > 0$$     and hence $$\frac{{{{\log }_e}x}}{x} > 0$$
$$\eqalign{ & \therefore I = \int_{\frac{1}{e}}^1 {\left[ { - \frac{{{{\log }_e}\,x}}{x}} \right]dx + } \int_1^{{e^2}} {\frac{{{{\log }_e}\,x}}{x}dx} \cr & = - \frac{1}{2}\left[ {{{\left( {{{\log }_e}\,x} \right)}^2}} \right]_{\frac{1}{e}}^1 + \frac{1}{2}\left[ {{{\left( {{{\log }_e}\,x} \right)}^2}} \right]_1^{{e^2}} \cr & = \frac{1}{2} + 2 \cr & = \frac{5}{2} \cr} $$

$$\eqalign{ & {\text{Put }}x = 2\cos \,2\theta \cr & {\text{Then }}I = \int_{\frac{\pi }{4}}^0 {\cot \,\theta .\left( { - 4\sin \,2\theta } \right)d\theta } \cr & \therefore \,I = 8\int_0^{\frac{\pi }{4}} {{{\cos }^2}\theta \,d\theta } \cr & = 4\int_0^{\frac{\pi }{4}} {\left( {1 + \cos \,2\theta } \right)} d\theta \cr & = 4\left[ {\theta + \frac{{\sin \,2\theta }}{2}} \right]_0^{\frac{\pi }{4}} \cr & = 4\left[ {\frac{\pi }{4} + \frac{1}{2}} \right] \cr & = \pi + 2 \cr} $$