Integration (1 Viewer)

lychnobity · Jul 25, 2009

kaz1 said:
$\int_{0}^{1}\frac{sin^{-1}x}{\sqrt{1+x}} dx$

By parts,

let u = sin^-1x and dv = $\frac{1}{\sqrt{1+x}} dx$

Timothy.Siu · Jul 25, 2009

kaz1 said:
$\int_{0}^{1}\frac{sin^{-1}x}{\sqrt{1+x}} dx$

just do it by parts
u=arsin x
u'=1/root(1+x^2)
v'=1/root(1+x)
v=2root(1+x)

I=2root(1+x)sin^-1x-2 S root[(1+x)/(1+x^2)]dx

thats what i got up to lol..

Drongoski · Jul 25, 2009

Timothy.Siu · Jul 25, 2009

oh i see, i differentiated arcsine wrong

kaz1 · Jul 25, 2009

Drongoski said:
$\100dpi \int \frac{sin^{-1}x}{\sqrt{1+x}}\ dx\ \ =\ \ 2\int sin^{-1}x\ d(1+x)^{\frac{1}{2}}\\ \\ \\ =\ \ 2\sqrt{1+x}\ sin ^{-1}x\ -\ 2\int (1+x)^{\frac{1}{2}}\cdot \frac{1}{\sqrt{1-x^{2}}}\ dx\\ \\ =\ \ 2\sqrt{1+x}\ sin^{-1}x\ -\ 2\int \sqrt{\frac{1+x}{(1+x)(1-x)}}\ dx\\ \\ =\ \ 2\sqrt{1+x}\ sin^{-1}x\ +\ 2\int (1-x)^{\frac{-1}{2}}\ d(1-x)\\ \\ =\ \ 2\sqrt{1+x}\ sin^{-1}x\ +\ \frac{2\sqrt{1-x}}{\frac{1}{2}}\ +\ C\\ \\ \\ \therefore \ \ \int_{0}^{1}\frac{sin^{-1}x}{\sqrt{1+x}}\ dx\ \ =\ \ \left [ 2\sqrt{1+x}\ sin^{-1}x\ +\ 4\sqrt{1-x} \right ]^{1}_{0}\\ \\ =\ \ (2\sqrt{2}sin^{-1}1\ +\ 0)\ -\ (0\ +\ 4\sqrt{1-0})\\ \\ =\ \ \sqrt{2}\ \pi \ -\ 4\ \ ???$

I don't get the first line. How did you turn 1/root(1+x) into 2root(1+x)

Timothy.Siu · Jul 25, 2009

kaz1 said:
I don't get the first line. How did you turn 1/root(1+x) into 2root(1+x)

its the same as by parts skipping a few steps. thats all.
root(1+x)^-1 dx=dx. d(2root(1+x))/dx=d(2root(1+x)

Drongoski · Jul 25, 2009

kaz1

I'm very used to taking differentials; i.e. df(x) which is simply f '(x)dx or the reverse. So

$\100dpi d\sqrt{1+x}\ =\ \frac{1}{2}\times \frac{1}{\sqrt{1+x}}\ dx\ \ \ or\ \ equivalently\ \ \frac{1}{\sqrt{1+x}}\ dx\ =\ 2\ d\sqrt{1+x}$

It's not quite correct to say I skip a few steps; I see things that way more or less immediately.

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Integration (1 Viewer)

kaz1

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lychnobity

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Timothy.Siu

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Drongoski

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Timothy.Siu

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kaz1

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Drongoski

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