Solve integration with substitution without using hyperbolic functions

Question

I am stuck with this question: $L = \frac{1}{2}\int_{0}^{2}\sqrt{1+t^2}dt$ where you solve the integral by using substitution $t = \frac{1}{2}({e}^{u}-{e}^{-u})$. How do I solve this without using hyperbolic functions? The expected answer is $L = \frac{1}{2}\cdot\sqrt{5} + \frac{1}{4}\cdot\ln(2+\sqrt{5})$

Answer 1

Sketch of a way: with the change of variables $\sqrt{x^{2}+1}=\sqrt{\tan^{2}u+1}=\sqrt{\sec^{2}u}$, the integral can be written as $\frac{1}{2}\int_{0}^{\arctan(2)}\sec^{3}u\, {\rm d}u$. The reduction formula for $\sec^{m}u$ allows the integral to be rewritten as $\frac{\sqrt{5}}{2}+\frac{1}{4}\int_{0}^{\arctan(2)}\sec u\, {\rm d}u$ integration for secant and then substitution back give $\frac{\sqrt{5}}{2}+\frac{1}{4}\ln(2+\sqrt{5})$. You can complete the details.

Answer 2

Without trigonometric and hyperbolic functions:

$\begin{align} I=\int_0^2\sqrt{t^2+1}dt&=\left.t\sqrt{t^2+1}\right\rvert_{t=0}^{t=2}-\int_0^2t\frac{t}{\sqrt{t^2+1}}dt\hspace{1cm}\text{(Integration by parts)}\\ &=(2\sqrt{5}-0)-\int_0^2\frac{t^2}{\sqrt{t^2+1}}dt\\ &=2\sqrt{5}-\int_0^2 (\sqrt{t^2+1}-\frac{1}{\sqrt{t^2+1}})dt\\ &=2\sqrt{5}-\int_0^2 \sqrt{t^2+1} dt+\int_0^2\frac{1}{\sqrt{t^2+1}}dt\\ &=2\sqrt{5}-I+\int_0^2\frac{1}{\sqrt{t^2+1}}dt.\\ \end{align}$

So, we have

$\begin{align} 2I=2\sqrt{5}+\int_0^2\frac{1}{\sqrt{t^2+1}}dt &=2\sqrt{5}+\int_0^2\frac{1+\frac{t}{\sqrt{t^2+1}}}{t+\sqrt{t^2+1}}dt\\ &=2\sqrt{5}+\left(\left.\ln|t+\sqrt{t^2+1}|\right\rvert_0^2\right)\\ &=2\sqrt{5}+\ln(2+\sqrt{5}).\\ \end{align}$

Hence, $2I=2\sqrt{5}+\ln(2+\sqrt{5})$. Since $I=2L$, we have $4L=2\sqrt{5}+\ln(2+\sqrt{5})$ and $$L=\frac{1}{2}\sqrt{5}+\frac{1}{4}\ln(2+\sqrt{5}).$$

Answer 3

Letting $t=\tan \theta$ yields $$ L=\frac{1}{2} \int_0^{\tan ^{-1} 2} \sec ^3 \theta d \theta $$ $$ \begin{aligned} \int \sec ^3 \theta d \theta &=\int \sec \theta d(\tan \theta) \\ &=\sec \theta \tan \theta-\int \sec \theta \tan ^2 \theta d \theta \\ &=\sec \theta \tan \theta-\int \sec \theta\left(\sec ^2 \theta-1\right) d \theta \\ &=\frac{1}{2}(\sec \theta \tan \theta+\ln |\sec \theta+\tan \theta|)+C \end{aligned} $$ Hence $$ \begin{aligned} L &=\frac{1}{4}[\sec \theta \tan \theta+\ln |\sec \theta+\tan \theta|]_0^{\tan ^{-1} 2} \\ &=\frac{1}{2} \sqrt{5}+\frac{1}{4} \ln |2+\sqrt{5}| \end{aligned} $$

Answer 4

$$ \begin{aligned} \int_0^2 \sqrt{1 +t^2} d t &=\left[t \sqrt{1+t^2}\right]_0^2-\int_0^2 \frac{t^2}{\sqrt{1+t^2}} d t \\ &=2 \sqrt{5}-\int_0^2\left(\sqrt{1+t^2}-\frac{1}{\sqrt{1+t^2}}\right) d t \\ &=\frac{1}{2}\left(2 \sqrt{5}+\int_0^2 \frac{1}{\sqrt{1+t^2}} d t\right) \\ &=\sqrt{5}+\frac{1}{2} \int_0^{\tan ^{-1} 2} \sec \theta d \theta, \text { where } t=\tan \theta \\ &=\sqrt{5}+\frac{1}{2} \ln |\sec \theta+\tan \theta|_0^{\tan^{-1}2} \\ &=\sqrt{5}+\frac{1}{2} \ln (\sqrt{5}+2)\\ \therefore L&=\frac{1}{2} \sqrt{5}+\frac{1}{4} \ln (\sqrt{5}+2) \end{aligned} $$