Question

Use the substitution $u=e^x$ to evaluate
${\int }_0^1\sec hxdx$

Answer 1

Formula: 1. $\sec hx=\frac{1}{\cos hx}=\frac{2}{e^x+e^{-x}}$

2. ${\int }_a^b\frac{1}{x^2+a^2}=|\frac{1}{a}ta{n}^{-1}x{|}_a^b$

${\int }_0^1\sec hxdx$

$\Rightarrow {\int }_0^1\frac{2}{e^x+e^{-x}}dx$

$\Rightarrow {\int }_0^1\frac{2e^x}{e^{2x}+1}dx...\left(i\right)$

Let, $u=e^x$

On differentiating $u$ w.r.t $X$, we get

$\frac{du}{dx}=e^x$

$\Rightarrow du=e^{x}dx$

Lower limit: at $x=0$, $u=e^0=1$

Upper limit: at $x=1$, $u=e^1=e$

Substituting the above values in equation (i), we obtain

${\int }_1^e\frac{2}{u^2+1}du$

$\Rightarrow 2\times |ta{n}^{-1}u{|}_1^e$

$\Rightarrow 2\times (ta{n}^{-1}e-ta{n}^{-1}1)$

$\Rightarrow 2(ta{n}^{-1}e-\frac{\pi }{4})$

Therefore, ${\int }_0^1\sec hxdx=2(ta{n}^{-1}e-\frac{\pi }{4})$