Trig integral-1

It is given that $\theta = \arcsin \left(\dfrac x{\sqrt{1+x^2}}\right)$ , $\implies \sin \theta = \dfrac x{\sqrt{1+x^2}}$ . This means that $\tan \theta = x$ as $\dfrac x{\sqrt{1+x^2}} = \dfrac {\tan \theta}{\sqrt{1+\tan^2 \theta}} = \dfrac {\tan \theta}{\sec \theta} = \sin \theta$ . Then:

$\begin{aligned} g\left(\frac \pi 6 \right) - g(0) & = \int_0^\frac \pi 6 \sqrt{1+x^2}\ d\theta & \small \color{#3D99F6} \text{Since }x = \tan \theta \\ & = \int_0^\frac \pi 6 \sqrt{1+\tan^2 \theta}\ d\theta & \small \color{#3D99F6} \text{and }1+\tan^2 \theta = \sec^2 \theta \\ & = \int_0^\frac \pi 6 \sec \theta\ d\theta & \small \color{#3D99F6} \text{Multiply integrand by }\frac {\sec \theta + \tan \theta}{\sec \theta + \tan \theta} \\ & = \int_0^\frac \pi 6 \frac {\sec^2 \theta + \tan \theta \sec \theta}{\sec \theta + \tan \theta}\ d\theta \\ & = \ln (\tan \theta + \sec \theta)\ \bigg|_0^\frac \pi 6 \\ & = \ln \left(\frac 1{\sqrt 3} + \frac 2{\sqrt 3}\right) - \ln (0 + 1) \\ & = \ln (\sqrt 3) \end{aligned}$

Therefore $A = \boxed 3$ .

$\theta =arcsin\bigg(\dfrac{x}{\sqrt{1+x^{2}}}\bigg)$

$sin\theta =\dfrac{x}{\sqrt{1+x^{2}}}$

$cos\theta =\dfrac{1}{\sqrt{1+x^{2}}}$

$g(\theta)=\displaystyle \int \dfrac{1} {cos\theta} d\theta$

$g(\theta) =\displaystyle \int sec\theta d\theta$

$g(\theta)=\dfrac{1}{2} ln\dfrac{1+sin\theta}{1-sin\theta}$

$g(\dfrac{\pi}{6})-g(0)=\dfrac{1}{2} ln \dfrac {1+sin(\dfrac{\pi}{6})}{1-sin(\dfrac{\pi}{6}}) -\dfrac{1}{2} ln\dfrac{1+sin0}{1-sin0}$

$\dfrac{1}{2}( ln 3 - ln 1)$

$g(\dfrac{\pi}{6})-g(0)=ln(\sqrt{3})$