4U Integration. (1 Viewer)

x jiim · Jan 29, 2010

Most of the sheet we got was fairly alright, but got stuck on the last few gross questions.

∫ √((3-x)/(2+x)) dx

∫ ((logx)^7)/x dx

Thanks :]

Drongoski · Jan 29, 2010

x jiim said:
Most of the sheet we got was fairly alright, but got stuck on the last few gross questions.

∫ √((3-x)/(2+x)) dx

∫ ((logx)^7)/x dx

Thanks :]

2nd one dead easy.

the integral = int of (log x)^7 wrt log x

= (1/(7+1)) (log x)^7+1 + C

(you may prefer to use the substitution: let u = log x )

= (log x)^8/8 + C

x jiim · Jan 29, 2010

Drongoski said:
2nd one dead easy.

the integral = int of (log x)^7 wrt log x

= (1/(7+1)) (log x)^7+1 + C

(you may prefer to use the substitution: let u = log x )

= (log x)^8/8 + C

oh, whoops

any ideas on the first one?
the answer's apparently

√(6+x-x^2) + 5/2.sin^-1 (2x-1)/5 + k o_____o

Drongoski · Jan 29, 2010

$\100dpi \int \sqrt{\frac{3-x}{2+x}}\ dx\ \ = \ \ \int \frac{\sqrt{3-x}\sqrt{3-x}}{\sqrt{3-x}\sqrt{2+x}}\ dx\ \ =\ \ \int \frac{3-x}{\sqrt{6+x-x^{2}}}\ dx\\ \\ =\ \ \int \frac{\frac{1}{2}(1-2x)+\frac{5}{2}}{\sqrt{same}}\ dx\ \ \\ \\ =\ \frac{1}{2}\int (6+x-x^{2})^{-\frac{1}{2}}d(6+x-x^{2})\ \ + \frac{5}{2}\int \frac{1}{\sqrt{(\frac{5}{2})^{2}-(x-\frac{1}{2})^{2}}}\ dx\\ \\ =\ \frac{1}{2}\frac{(6+x-x^{2})^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\ +\ \frac{5}{2}sin^{-1}\frac{(2x-1)}{5}\ +\ C\\ \\ \ =\ \ \sqrt{6+x-x^{2}}\ +\ \frac{5}{2}sin^{-1} \frac{(2x-1)}{5}\ +\ C$

I've skipped some of latter steps . . . tiring typing out the details in LaTeX

study-freak · Jan 29, 2010

Sgt Grumbles · Jan 29, 2010

Wolframalpha is also really helpful

duckcowhybrid · Jan 29, 2010

Jim you could just ask someone from our class on MSN for the solution he wrote on the board. Regardless both methods posted above work and I cbf writing out the official solution.

study-freak · Jan 29, 2010

study-freak said:
$\\\int \sqrt{\frac{3-x}{2+x}}dx$ Let $u=x-0.5,$ $du=dx \\=\int \sqrt{\frac{-u+2.5}{u+2.5}}du \\=\int \frac{2.5-u}{\sqrt{2.5^2-u^2}}du \\=2.5sin^{-1}(\frac{u}{2.5})+\sqrt{2.5^2-u^2}+C \\=\frac{5}{2}sin^{-1}(\frac{2u}{5})+\sqrt{\frac{25}{4}-u^2}+C \\=\frac{5}{2}sin^{-1}(\frac{2x-1}{5})+\sqrt{\frac{25}{4}-(x^2-x+\frac{1}{4})}+C \\=\frac{5}{2}sin^{-1}(\frac{2x-1}{5})+\sqrt{6+x-x^2}+C$

Well I let u=x-0.5 because when the numerator is rationalised, it allows us to make the denominator into root of difference of two squares. Then, standard integrals can be used to integrate the expression.

x jiim · Jan 30, 2010

Thanks everyone :]

4U Integration. (1 Viewer)

x jiim

zimbardooo.

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x jiim

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