Integration via Substitution (1 Viewer)

ADrew · Aug 8, 2011

Wondering if some kind person could help a little with this difficult integral question. Thanks!

$Use{\hspace{1 mm}}the{\hspace{1 mm}}substitution\hspace{1 mm} u=x-\frac{1}{x}\hspace{1 mm}to \hspace{1 mm}show \int_{1}^{\frac{\sqrt{6}+\sqrt{2}}{2}}\frac{1+x^{2}}{1+x^{4}}dx=\frac{\pi }{4\sqrt{2}}$

Hermes1 · Aug 8, 2011

Posting solution in a second.....

Hermes1 · Aug 8, 2011

$\int \frac{1+x^{2}}{1+x^{4}}dx = \int \frac{x^{2}+1}{x^{2}(x^{2}+\frac{1}{x^{2}})}dx$
$\frac{du}{dx}=1+\frac{1}{x^{2}}$
$\frac{du}{dx}=\frac{x^{2}+1}{x^{2}}$
therefore
$\int \frac{x^{2}+1}{x^{2}(x^{2}+\frac{1}{x^{2}})}dx = \int \frac{1}{u^{2}+2}du$
noting that:
$x^{2}+\frac{1}{x^{2}}=(x-\frac{1}{x})^{2}+2$
i didnt change the limits i just solved it as a indefinite integral, but u can do the rest...

Hermes1 · Aug 8, 2011

$\int_{0}^{\sqrt{2}}\frac{1}{2+u^{2}}du$
$\int_{0}^{\sqrt{2}}\frac{1}{2+u^{2}}du = {\frac{1}{\sqrt{2}}\tan^{-1}\frac{u}{\sqrt{2}}}$ with limits 0 to root (2) and solve....

ADrew · Aug 8, 2011

Thanks so much Hermes1!

appreciate it!

Hermes1 · Aug 8, 2011

ADrew said:
Thanks so much Hermes1! appreciate it!

ur welcome

cssftw · Aug 9, 2011

fuuuarrkk... hermes you are a beast... what are you aiming for in the hsc 3u exam? 98/99?

Integration via Substitution (1 Viewer)

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