Definite integral of $\sqrt{\frac{1}{\cos^2(x)}}$

Question

I've got problems with this integral:

$$\int_0^{\frac{\pi}{4}} \sqrt{\frac{1}{\cos^2(x)}} \, \mathrm{d}x$$

First I substitute $x=2\arctan(x)$ but this leads nowhere. Any hints for solving?

Answer 1

Hint: $\mathrm{d}x=\frac{\mathrm{d}\sin(x)}{\cos(x)}$ and $\cos^2(x)=1-\sin^2(x)$. Then partial fractions will be useful.

In other words, try $u=\sin(x)$, then $\cos^2(x)=1-u^2$ and $\frac1{\cos(x)}\frac{\mathrm{d}u}{\mathrm{d}x}=1$. Thus, $$ \begin{align} \int\frac1{\cos(x)}\,\mathrm{d}x &=\int\frac1{\cos^2(x)}\frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x\\ &=\int\frac1{1-u^2}\,\mathrm{d}u \end{align} $$

Answer 2

By just substituting $x=\arctan t$, from $\cos^2(\arctan t)=\frac{1}{1+t^2}$ it follows that: $$ \int_{0}^{\pi/4}\sqrt{\frac{1}{\cos^2(x)}}\,dx = \int_{0}^{1}\frac{dt}{\sqrt{1+t^2}}=\text{arcsinh}(1)=\color{red}{\log(1+\sqrt{2}).}$$

Answer 3

$$\sqrt{\dfrac1{\cos^2x}}=|\sec x|$$

In $\in[0,\pi/4],\sec x>0\implies|\sec x|=+\sec x$

Now use this

Answer 4

There is an easy trick for this integral:

$$\int\frac{dx}{\cos x}=\int\frac{\cos x}{\cos^2 x}dx=\int\frac{d\sin x}{1-\sin^2x}$$ and $$\int\frac{du}{1-u^2}=\text{artanh(u)}=\text{artanh}(\sin x).$$

Answer 5

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Note that $\ds{\totald{\bracks{\sec\pars{x} + \tan\pars{x}}}{x} = \sec\pars{x}\tan\pars{x} + \sec^{2}\pars{x}}$ such that:

\begin{align} \color{#f00}{\int_{0}^{\pi/4}\root{1 \over \cos^{2}\pars{x}}\,\dd x} & = \int_{0}^{\pi/4}\sec\pars{x}\,\dd x = \int_{0}^{\pi/4} {\sec^{2}\pars{x} + \sec\pars{x}\tan\pars{x} \over \sec\pars{x} + \tan\pars{x}} \,\dd x \\[3mm] & =\left.\vphantom{\LARGE A}% \ln\pars{\vphantom{\large A}\sec\pars{x} + \tan\pars{x}} \,\right\vert_{\ 0}^{\ \pi/4} = \ln\pars{\sec\pars{\pi/4} + \tan\pars{\pi/4} \over \sec\pars{0} + \tan\pars{0}} \\[3mm] & = \color{#f00}{\ln\pars{\root{2} + 1}} \approx 0.8814 \end{align}