Integrate by substitution method

Question

Need help on solving integrals using substitution. As I have only solved ones with Newton-Leibniz, I don't know how to solve this types:

$$\int_0^1 \frac{\sqrt{e^x}dx}{\sqrt{e^x+e^{-x}}}$$

Answer 1

Hint:

Substitute $t=e^x$ and $dt=e^x dx$

$$=\int \frac{dt}{\sqrt{t^2+1}}$$

Substitute $t=\tan p$ and $dt=\sec^2 p dp$. Then $\sqrt{t^2+1}=\sqrt{\tan^2 p+1}=\sec p$ and $p=\arctan t$

$$=\int\sec p dp=\ln|\tan p+\sec p|+\mathcal C=\dots$$

Now you should substitute back $p$ and $t$, and to evaluate from zero to one

Answer 2

$$ \begin{align} \int_0^1\frac{\sqrt{e^x}\,\mathrm{d}x}{\sqrt{e^x+e^{-x}}} &=\int_0^1\frac{e^x\,\mathrm{d}x}{\sqrt{e^{2x}+1}}\tag{1}\\ &=\int_1^e\frac{\mathrm{d}u}{\sqrt{u^2+1}}\tag{2}\\ &=\int_{\pi/4}^{\arctan(e)}\sec(\theta)\,\mathrm{d}\theta\tag{3}\\ &=\int_{\pi/4}^{\arctan(e)}\frac{\mathrm{d}\sin(\theta)}{1-\sin^2(\theta)}\tag{4}\\ &=\frac12\int_{\pi/4}^{\arctan(e)}\left(\frac1{1-\sin(\theta)}+\frac1{1+\sin(\theta)}\right)\mathrm{d}\sin(\theta)\tag{5}\\[3pt] &=\frac12\left.\log\left(\frac{1+\sin(\theta)}{1-\sin(\theta)}\right)\right]_{\pi/4}^{\arctan(e)}\tag{6}\\[6pt] &=\left.\log\left(\tan(\theta)+\sec(\theta)\right)\right]_{\pi/4}^{\arctan(e)}\tag{7}\\[6pt] &=\log\left(\frac{e+\sqrt{e^2+1}}{1+\sqrt2}\right)\tag{8} \end{align} $$ Explanation:
$(1)$: multiply by $\frac{\sqrt{e^x}}{\sqrt{e^x}}$
$(2)$: $u=e^x$
$(3)$: $u=\tan(\theta)$
$(4)$: rewrite integrand
$(5)$: partial fractions
$(6)$: integrate
$(7)$: simplify
$(8)$: evaluate

Answer 3

Substitute ${e^x} = \tan z$ .