Integration

**great_math** · October 20th 2008, 06:46 AM

Please help me with this one:

$\int \log \frac{a}{x}\sin^{-1} (x)\ dx<br />$

**dax918** · October 20th 2008, 01:36 PM

Interesting. I spent over an hour on this one but I haven't been able to finish it. The only method I can think of is integrating by parts

int (u dv) = u.v - int (v du)

Knowing that

int Arcsin x = sqrt(1-x^2) + x.Arcsin(x) and

int Log(a/x) = x + xLog(a/x)

so, if this is correct you just have to pick the right values and continue solving the next integral. I haven't given up on this one so I'll keep trying to solve it.

**Mathstud28** · October 21st 2008, 03:23 PM

Originally Posted by great_math

Please help me with this one:

$\int \log \frac{a}{x}\sin^{-1} (x)\ dx<br />$

This cannot be computed nicely

$\int\ln\left(\frac{a}{x}\right)\arcsin(x)dx$
$=\int\bigg[\arcsin(x)\ln(a)-\ln(x)\arcsin(x)\bigg]dx$

Now we are only interested in the second integral. First let

$\arcsin(x)=\varphi\Rightarrow{dx=\cos(\varphi)d\va rphi}$

Giving us

$\int\ln(x)\arcsin(x)dx\overbrace{\mapsto}^{\arcsin (x)=\varphi}\int\ln(\sin(\varphi))\varphi\cos(\var phi)d\varphi$

Integration by parts gives that this is equivalent to

$\varphi\bigg[\sin(\varphi)\ln(\sin(\varphi))-\sin(\varphi))-\int\bigg[\sin(\varphi)\ln(\sin(\varphi))-\sin(\varphi)\bigg]d\varphi$

Now for the second integral we only care about the first term...so in that we let
$x=2\arctan(\psi)$

After simplification this gives us

$\int\sin(\varphi)\ln(\sin(\varphi))d\varphi\overbr ace{\mapsto}^{\varphi=2\arctan(\psi)}\int\frac{2\p si}{1+\psi^2}\ln\left(\frac{2\psi}{1+\psi^2}\right )\frac{2}{1+\psi^2}d\psi$

This looks kind of messy but isnt too bad...do all the undone work...there is your integral

**Krizalid** · October 21st 2008, 03:51 PM

I think it can be computed nicely,

$\int{\ln (x)\arcsin x\,dx}=\int{(x\ln x-x)'\arcsin x\,dx},$ hence the integral equals $x(\ln x-1)\arcsin x-\int{\frac{x\ln x-x}{\sqrt{1-x^{2}}}\,dx}.$ Now one only cares about $\int\frac{x\ln x}{\sqrt{1-x^2}}\,dx,$ hence, do the partial integration and we get,

$\int{\frac{x\ln x}{\sqrt{1-x^{2}}}\,dx}=-\int{\left( \sqrt{1-x^{2}} \right)'\ln x\,dx}=-\left( \sqrt{1-x^{2}}\ln x-\int{\frac{\sqrt{1-x^{2}}}{x}\,dx} \right),$

where the last integral is elementary.

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