Further Calculus Q (1 Viewer)

MC Squidge · May 13, 2009

1) by substituting u=x-1/x, integrate (1+x^2)/(1+x^4)
2)find the volume generated by y=x/((x^2+16)^(1/2)) HINT let y=sin@

GUSSSSSSSSSSSSS · May 13, 2009

for question 2.....wat volume???

MC Squidge · May 13, 2009

between 4 and 0

GUSSSSSSSSSSSSS · May 13, 2009

i mean wat r you rotating around ...??? =S

MC Squidge · May 13, 2009

y axis

kurt.physics · May 13, 2009

MC Squidge said:
1) by substituting u=x-1/x, integrate (1+x^2)/(1+x^4)
2)

$\frac{du}{dx} = 1 + \frac{1}{x^2}$

$\therefore dx = \frac{du}{1+\frac{1}{x^2}}$

$\int \frac{1+x^2}{1+x^4}\,\, dx$

$= \int \frac{ \frac{1+x^2}{1+x^4}}{ 1+\frac{1}{x^2} }\,\, du$

$= \int \frac{1+x^2}{1+x^4} \times \frac{x^2}{1+x^2}\,\, du$

$= \int \frac{x^2}{1+x^4}\,\, du$

$u^2 = x^2 + \frac{1}{x^2} -2$

$\therefore u^2 + 2 = x^2 + \frac{1}{x^2}$

$= u^2 + 2 = \frac{1+x^4}{x^2}$

$\therefore \frac{x^2}{1+x^4} = \frac{1}{u^2+2}$

$= \int \frac{x^2}{1+x^4}\,\, du$

$= \int \frac{1}{u^2 + 2}\,\, du$

$= \frac{tan^{-1} \left( \frac{u}{\sqrt{2}}\right) }{\sqrt{2} }$

$= \frac{tan^{-1} \left( \frac{x-\frac{1}{x}}{\sqrt{2}}\right) }{\sqrt{2} }$

shaon0 · May 13, 2009

kurt.physics said:
$\frac{du}{dx} = 1 + \frac{1}{x^2}$

$\therefore dx = \frac{du}{1+\frac{1}{x^2}}$

$\int \frac{1+x^2}{1+x^4}\,\, dx$

$= \int \frac{ \frac{1+x^2}{1+x^4}}{ 1+\frac{1}{x^2} }\,\, du$

$= \int \frac{1+x^2}{1+x^4} \times \frac{x^2}{1+x^2}\,\, du$

$= \int \frac{x^2}{1+x^4}\,\, du$

$u^2 = x^2 + \frac{1}{x^2} -2$

$\therefore u^2 + 2 = x^2 + \frac{1}{x^2}$

$= u^2 + 2 = \frac{1+x^4}{x^2}$

$\therefore \frac{x^2}{1+x^4} = \frac{1}{u^2+2}$

$= \int \frac{x^2}{1+x^4}\,\, du$

$= \int \frac{1}{u^2 + 2}\,\, du$

$= \frac{tan^{-1} \left( \frac{u}{\sqrt{2}}\right) }{\sqrt{2} }$

$= \frac{tan^{-1} \left( \frac{x-\frac{1}{x}}{\sqrt{2}}\right) }{\sqrt{2} }$

you could've just divided every term by x^2.

Further Calculus Q (1 Viewer)

MC Squidge

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