Three trigonometry friends together

$\dfrac{\sin(x)\cos(x)}{\tan(x)} = \dfrac{\sin(x)\cos(x)}{\frac{\sin(x)}{\cos(x)}} = {\cos}^2(x)$ .
Now, $\cos(2x) = {\cos}^2(x) - {\sin}^2(x)\\ \cos(2x) = {2\cos}^2(x) - {\cos}^2(x) - {\sin}^2(x)\\ \cos(2x) = {2\cos}^(x) - \left({\cos}^2(x) + {\sin}^2(x)\right)\\ \cos(2x) = {2\cos}^2(x) - 1\\ {\cos}^2(x) = \dfrac{1 + \cos(2x)}{2}$

So, the integral reduces to $\int_{0}^{1} \dfrac{1 + \cos(2x)}{2} \ dx$ $= \dfrac{1}{2} \int_{0}^{1} 1 + \cos(2x) \ dx\\ \\ = \dfrac{1}{2}\left(\int_{0}^{1} 1 \ dx + \int_{0}^{1} \cos(2x) \ dx\right)\\ \\ = \dfrac{1}{2}\left(1 + \dfrac{\sin(2)}{2}\right)\\ \\ = \dfrac{2 + \sin(2)}{4}\\ \\ \approx \color{#3D99F6}{\boxed{0.73}}$

$\begin{aligned} I & = \int_0^1 \frac {\sin x \cos x} {\tan x} dx \\ & = \int_0^1 \frac {\sin x \cos x} {\dfrac {\sin x} {\cos x }} dx \\ & = \int_0^1 \color{#3D99F6}{\cos^2 x} \ dx \quad \quad \small \color{#3D99F6}{\cos (2x)=2\cos ^2 x-1} \\ & = \int_0^1 \color{#3D99F6} {\frac {1 +\cos (2x)} {2}} dx \\ & = \frac 12 \left(x+\frac {\sin(2x) } {2} \right) \bigg|_0^1 \\ & = \frac 12\left(1+\frac {\sin 2} {2} \right) \\& \approx \boxed {0.73}\end{aligned}$