Solving Integrals 2

Consider:

$I = \int_{0}^{1}\frac{\sin{x}}{\sqrt{x}}dx$ Implies: $I = 2\int_{0}^{1}\frac{\sin{x}}{2\sqrt{x}}dx$ Taking $x = z^2$ leads to: $dx = 2zdz$ . Changing the variables of integration gives: $I = 2\int_{0}^{1}\sin(z^2)dz$

By performing a series expansion of $\sin(z^2)$ gives: $I = 2\int_{0}^{1}\left(z^2 - \frac{z^6}{3!} +\frac{z^{10}}{5!} - \frac{z^{14}}{7!} ...\right)dz$

$I = 2\left(\frac{1}{3} - \frac{1}{7.3!} + \frac{1}{11.5!} - \frac{1}{15.7!}...\right) \approx 0.62(<0.66)$

$\boxed{I < \frac{2}{3}}$

The same procedure can be repeated for the integral defined as:

$J = \int_{0}^{1}\frac{\cos{x}}{\sqrt{x}}dx$ Implies: $J = 2\int_{0}^{1}\frac{\cos{x}}{2\sqrt{x}}dx$ Taking $x = z^2$ leads to: $dx = 2zdz$ . Changing the variables of integration gives: $J = 2\int_{0}^{1}\cos(z^2)dz$

By performing a series expansion of $\cos(z^2)$ gives: $J = 2\int_{0}^{1}\left(1 - \frac{z^4}{2!} +\frac{z^{8}}{4!} - \frac{z^{12}}{6!} ...\right)dz$

$J = 2\left(1 - \frac{1}{5.2!} + \frac{1}{9.4!} - \frac{1}{13.6!}...\right) \approx 1.809(<2)$

$\boxed{J < 2}$

$\begin{aligned} I & = \int_0^1 \frac {\sin x}{\sqrt x} dx & \small \color{#3D99F6} \text{Let }x = u^2 \implies dx = 2u\ du \\ & = \int_0^1 2 \sin (u^2)\ du & \small \color{#3D99F6} \text{Since }\sin (u^2) = u^2 - \frac {u^6}{3!} + \frac {u^{10}}{5!} - \cdots \\ & < \int_0^1 2u^2 \ du = \frac 23 \end{aligned}$

$\begin{aligned} J & = \int_0^1 \frac {\cos x}{\sqrt x} dx & \small \color{#3D99F6} \text{Let }x = u^2 \implies dx = 2u\ du \\ & = \int_0^1 2 \cos (u^2)\ du & \small \color{#3D99F6} \text{Since }\cos (u^2) = 1 - \frac {u^4}{2!} + \frac {u^8}{4!} - \cdots \\ & < \int_0^1 2 \ du = 2 \end{aligned}$

Therefore the answer is $\boxed{I < \frac 23 \text{ and }J < 2}$ .