$$ \because $$ $$\cos x = 1 - {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}} - {{{x^6}} \over {6!}} + ...$$

and $$\sin x = x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}} - {{{x^7}} \over {7!}} + ....$$

$$ \therefore $$ $$\int_0^1 {x\cos xdx} \ge \int_0^1 {\left( {x - {{{x^3}} \over {2!}}} \right)} \,dx$$

$$ = \left[ {{{{x^2}} \over 2} - {{{x^4}} \over 8}} \right]_0^1 = {1 \over 2} - {1 \over 8} = {3 \over 8}$$

$$ \Rightarrow \int_0^1 {x\cos xdx\, \ge \,{3 \over 8}} $$

and, $$\int_0^1 {x\sin dx \ge \int_0^1 {\left( {{x^2} - {{{x^4}} \over 6}} \right)} \,dx} $$

$$\left[ {{{{x^3}} \over 3} - {{{x^5}} \over {30}}} \right]_0^1 = {1 \over 3} - {1 \over {30}} = {9 \over {30}} = {3 \over {10}}$$

$$ \Rightarrow \int_0^1 {x\sin xdx\, \ge \,{3 \over {10}}} $$

and, $$\int_0^1 {{x^2}\cos xdx\, \ge \,\int_0^1 {\left( {{x^3} - {{{x^5}} \over 2}} \right)\,dx} } $$

$$ = \left[ {{{{x^4}} \over 4} - {{{x^6}} \over {12}}} \right]_0^1 = {1 \over 4} - {1 \over {12}} = {2 \over {12}} = {1 \over 6}$$

$$ \therefore $$ $$\int_0^1 {{x^2}\cos xdx\, \ge \,{1 \over 6}} $$

and, $$\int_0^1 {{x^2}\sin xdx\, \ge \int_0^1 {\left( {{x^3} - {{{x^5}} \over 6}} \right)\,dx} } $$

$$ = \left[ {{{{x^4}} \over 4} - {{{x^6}} \over {36}}} \right]_0^1 = {1 \over 4} - {1 \over {36}} = {8 \over {36}} = {2 \over 9}$$

$$ \therefore $$ $$\int_0^1 {{x^2}\sin xdx\, \ge \,{2 \over 9}} $$