Recently I've been working on problems including integration and I'm having some difficulties solving this integral :
$$\int_0^1 (1-x)\sqrt{4x-x^2} {dx}$$
I tried to rewrite it in this form:
$$\int_0^1 (1-x)x\sqrt{\frac{4-x}{x}} dx$$ and then subsitute $\frac{4-x}{x}=t^2$ and by this substitution I get a function way easier integrated of this form:
$$-32\int\frac{(t^2-3)t^2}{(1+t^2)^4}dt$$
I can integrate this function easily by simplifying it but the problem is that I cannot define its borders because $t$ is not defined for $x=0$ so this means that my substitution doesn't hold. But I don't have any other idea how to solve it so any help will be appreciated.Thank you!
Best Answer
The us try to get rid of the square root step-by-step: $$\begin{eqnarray*} \int_{0}^{1}(1-x)\sqrt{x}\sqrt{4-x}\,dx&\stackrel{x\mapsto z^2}{=}& 2\int_{0}^{1}z^2(1-z^2)\sqrt{4-z^2}\,dz\\&\stackrel{z\mapsto 2u}{=}&32\int_{0}^{1/2}u^2(1-4u^2)\sqrt{1-u^2}\,du\\&\stackrel{u\mapsto\sin\theta}{=}&32\int_{0}^{\pi/6}\sin^2\theta\cos^2\theta(1-4\sin^2\theta)\,d\theta\\&=&4\int_{0}^{\pi/6}\left[-1+\cos(2\theta)+\cos(4\theta)-\cos(6\theta)\right]\,d\theta.\end{eqnarray*}$$ I guess you may take it from here.