1. Find the volume of the solid obtained by rotating the region bounded by the curve
y^2 - y^3 - x = 0, x = 0
about the line x = 0
2. Integral of (cos x + xsinx)/x(x+cosx) dx
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1. Find the volume of the solid obtained by rotating the region bounded by the curve
y^2 - y^3 - x = 0, x = 0
about the line x = 0
2. Integral of (cos x + xsinx)/x(x+cosx) dx
The intersection points are $\displaystyle (0,0)$ and $\displaystyle (0,1)$ . Then,Quote:
Originally Posted by devil10078
$\displaystyle V=\pi\displaystyle\int_0^1(y^2-y^3)^2\;dy=\ldots$
Fernando Revilla
$\displaystyle \displaystyle \frac{\cos{x} + x\sin{x}}{x(x+\cos{x})}=-\left(\frac{-\cos{x} - x\sin{x}}{x^2 + x\cos{x}}\right)$
$\displaystyle \displaystyle = -\left(\frac{2x + \cos{x} - x\sin{x} - 2x- 2\cos{x} }{x^2 + x\cos{x}}\right)$
$\displaystyle \displaystyle = -\left(\frac{2x + \cos{x} - x\sin{x}}{x^2 + x\cos{x}}\right) - \left(\frac{-2x - 2\cos{x}}{x^2 + x\cos{x}}\right)$
$\displaystyle \displaystyle = -\left(\frac{2x + \cos{x} - x\sin{x}}{x^2 + x\cos{x}}\right) + \frac{2(x + \cos{x})}{x(x + \cos{x})}$
$\displaystyle \displaystyle = \frac{2}{x} - \frac{2x + \cos{x} - x\sin{x}}{x^2 + x\cos{x}}$.
Therefore $\displaystyle \displaystyle \int{\frac{\cos{x} + x\sin{x}}{x(x + \cos{x})}\,dx} = \int{\frac{2}{x} - \frac{2x + \cos{x} - x\sin{x}}{x^2 + x\cos{x}}\,dx}$.
To integrate the second term, make the substitution $\displaystyle \displaystyle u = x^2 + x\cos{x}$.
good call, but it's actually easier than that, since
$\displaystyle \begin{aligned}
\int{\frac{\cos x+x\sin x}{x(x+\cos x)}\,dx}&=\int{\frac{x+\cos x+x\sin x-x}{x(x+\cos x)}\,dx} \\
& =\ln \left| x \right|-\int{\frac{1-\sin x}{x+\cos x}\,dx} \\
& =\ln \left| x \right|-\ln \left| x+\cos x \right|+k \\
& =\ln \left| \frac{x}{x+\cos x} \right|+k.
\end{aligned}$
