Integration (1 Viewer)

juantheron · Nov 14, 2012

Can We solve it in terms of elementry function.

$\displaystyle \mathbf{\int\frac{(1+x)\sin x}{(x^2+2x)\cos^2 x+(1+x)\sin 2x}dx}$

rolpsy · Nov 19, 2012

Mathematica refuses to integrate it

so I highly doubt it can be found in terms of elementary functions

math man · Dec 5, 2012

i can integrate if there is a minus sign on denom, not plus

juantheron · Dec 9, 2012

Sorry friends .

actual Question is

$\displaystyle \mathbf{\int\frac{(1+x)\sin x}{(x^2+2x)\cos^2 x-(1+x)\sin 2x}dx}$

math man · Dec 13, 2012

Then the answer is
$-\frac{1}{2}log[\frac{(x+1)cosx-sinx-1}{(x+1)cosx-sinx+1}] +c$

math man · Dec 13, 2012

working out is, complete the square on the bottom as:

$(x^{2}+2x+1)cos^{2}x-2(1+x)sinxcosx +sin^{2}x -cos^{2}x-sin^{2}x= \\ \\ ((x+1)cosx-sinx)^{2}-1$

and we note that
$\frac{d}{dx} (x+1)cosx-sinx= -(1+x)sinx$

so the integral is in the form

$\int \frac{f'(x)}{f^{2}(x)-1}dx =\frac{1}{2}log[\frac{f(x)-1}{f(x)+1}]+C$

juantheron · Dec 25, 2012

Thanks mathman got it

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Integration (1 Viewer)

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