#2015_ KVPY 2015 preparations-4

By using the property of definite integrals

$\displaystyle \int_a^c f(x) \,dx = \int_a^b f(x) \,dx + \int_b^c f(x) \,dx$

for $a<b<c$ , we can write down our integral as

$I = \displaystyle \int_1^3 \left\lfloor x \right\rfloor \cos { \left( \frac { \pi }{ 2 } \left( x-\left\lfloor x \right\rfloor \right) \right) dx } = \int_1^2 \left\lfloor x \right\rfloor \cos { \left( \frac { \pi }{ 2 } \left( x-\left\lfloor x \right\rfloor \right) \right) dx } + \int_2^3 \left\lfloor x \right\rfloor \cos { \left( \frac { \pi }{ 2 } \left( x-\left\lfloor x \right\rfloor \right) \right) dx }$

Now

$1 \le x < 2 \implies \left\lfloor x \right\rfloor = 1$

$2 \le x < 3 \implies \left\lfloor x \right\rfloor = 2$

Thus

$I = \displaystyle \int_1^2 \cos { \left( \frac { \pi }{ 2 } \left( x-1 \right) \right) dx } + \int_2^3 \cos { \left( \frac { \pi }{ 2 } \left( x-2 \right) \right) dx }$

Appropriate substitutions of

$\dfrac{\pi}{2} (x-1) = u \implies \dfrac{\pi}{2} \,dx = du \\ \dfrac{\pi}{2} (x-2) = v \implies \dfrac{\pi}{2} \,dx = dv$

yields

$\begin{aligned} I &= \displaystyle \dfrac{2}{\pi} \int_0^{{\pi}/{2}} \cos u \,du + \dfrac{4}{\pi} \int_0^{{\pi}/{2}} \cos v \,dv \\ &= \dfrac{2}{\pi} \sin u {\huge |}_0^{{\pi}/{2}} + \dfrac{4}{\pi} \sin v {\huge |}_0^{{\pi}/{2}} \\ &= \dfrac{6}{\pi} \\ & \approx \boxed{1.90986} \end{aligned}$