A calculus problem by Jose Sacramento

$\begin{aligned} I & = \int_0^\frac \pi 2 \cos x \sqrt{1+\sin^2 x} \ dx & \small \color{#3D99F6} \text{Let }u = \sin x, \ du = \cos x \ dx \\ & = \int_0^1 \sqrt{1+u^2} du & \small \color{#3D99F6} \text{Let }u = \tan \theta, \ du = \sec^2 \theta \ d\theta \\ & = \int_0^\frac \pi 4 \sec^3 \theta \ d \theta & \small \color{#3D99F6} \text{By reduction formula (see below).} \\ & = \frac 12 \int_0^\frac \pi 4 \sec \theta \ d \theta + \frac {\sec \theta \tan \theta}2 \bigg|_0^\frac \pi 4 \\ & = \frac 12 \int_0^\frac \pi 4 \frac {\sec \theta \tan \theta + \sec^2 \theta}{\tan \theta + \sec \theta} \ d \theta + \frac 1{\sqrt 2} & \small \color{#3D99F6} \text{Let }v = \tan \theta + \sec \theta; \ dv = (\sec^2 \theta + \tan \theta \sec \theta) \ d\theta \\ & = \frac 12 \int_1^{1+\sqrt 2} \frac 1v \ dv + \frac 1{\sqrt 2} \\ & = \frac {\ln (1+\sqrt 2)}2 + \frac 1{\sqrt 2} \\ & \approx \boxed{1.148} \end{aligned}$

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Reduction formula:

$\begin{aligned} \int \sec^n x \ dx & = \frac {n-2}{n-1} \int \sec^{n-2} x \ dx + \frac {\sec^{n-2}x \tan x}{n-1} \end{aligned}$

First, let u equals to sinx, thenjoy we have sqrt (1+u^2) from 0 to 1. Then, let u equals to tan (v) and this gives us sec^3 (v) ....