Practise your integration

$\begin{aligned} \int_0^\frac \pi 4 \frac {\cos^3 x + \sin x}{\cos^2 x} dx & = \int_0^\frac \pi 4 \left(\cos x + \tan x \sec x \right) dx \\ & = \sin x + \sec x \ \bigg|_0^\frac \pi 4 \\ & = \frac 1{\sqrt 2} + \sqrt 2 - 0 - 1 \\ & \approx \boxed{1.12} \end{aligned}$

$\begin{aligned} \int_{0}^{\frac{\pi}{4}} \frac{\cos^3(x)+\sin(x)}{\cos^2(x)}dx &= \int_{0}^{\frac{\pi}{4}} \frac{\cos^3(x)}{\cos^2(x)}+\frac{\sin(x)}{\cos^2(x)}dx \\ &= \int_{0}^{\frac{\pi}{4}} \frac{\cos^3(x)}{\cos^2(x)}dx + \int_{0}^{\frac{\pi}{4}}\frac{\sin(x)}{\cos^2(x)}dx \\ &= \int_{0}^{\frac{\pi}{4}} \cos(x)dx + \int_{0}^{\frac{\pi}{4}}\frac{\sin(x)}{\cos^2(x)}dx \\ &= \sin(x)\bigg|_{0}^{\frac{\pi}{4}} + \int_{0}^{\frac{\pi}{4}}\frac{-(-\sin(x))}{\cos^2(x)}dx \end{aligned}$ We can let $\blue{u = \cos(x)}$ and $\orange{du = -\sin(x)dx}$ . Remember that as $x \to 0, u \to 1$ and as $x \to \frac{\pi}{4}, u \to \frac{1}{\sqrt{2}}$ : $\begin{aligned} \sin(x)\bigg|_{0}^{\frac{\pi}{4}} + \int_{0}^{\frac{\pi}{4}}\frac{-(\orange{-\sin(x)})}{\blue{\cos^2(x)}}\orange{dx} \implies \sin\left(\frac{\pi}{4}\right) - \sin(0) + \int_{u=1}^{u=\frac{1}{\sqrt{2}}}\frac{-\orange{1}}{\blue{u^2}}\orange{du} &= \frac{1}{\sqrt{2}} + \frac{1}{u} \bigg|_{1}^{\frac{1}{\sqrt{2}}} \\ &= \frac{1}{\sqrt{2}} + \sqrt{2} - 1 \\ &= \green{\boxed{\frac{3-\sqrt{2}}{\sqrt{2}}}} = \green{\boxed{1.121320343559643\dots}} \end{aligned}$