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I was reading Advanced Integration Techniques, and found that$$int_{0}^{1}sqrt{x}sqrt{1-x},mathrm dx =frac{pi}{8}$$

The book provides one method using residue theorem and Laurent expansion. However, I wonder if there are other techniques that I can use to evaluate this integral.

The most direct method would be solving the integral and plug in the limits.
$$int_{0}^{1}sqrt{x}sqrt{1-x},mathrm dx = left[frac{arcsin(2x-1)+sqrt{(1-x)x}(4x-2)}{8}right]_{0}^{1}=frac{pi}{8}$$

Write $x=sin^2 t$ so $dx=2sin tcos t dt$. Your integral becomes $$int_0^{pi/2}2sin^2 tcos^2 t dt=int_0^{pi/2}frac12 sin^2 2t dt=int_0^{pi/2}frac{1-cos 4t}{4}dt=left[frac{t}{4}-frac{1}{16}sin 4tright]_0^{pi/2}=frac{pi}{8}.$$

Let $sqrt{x-x^2}=y$.

Thus, $ygeq0$ and $$x^2-x+y^2=0$$ or
$$left(x-frac{1}{2}right)^2+y^2=left(frac{1}{2}right)^2,$$ which is a semicircle with radius $frac{1}{2}.$

Thus, our integral it's $$frac{1}{2}pileft(frac{1}{2}right)^2=frac{pi}{8}.$$

Another technique just for fun (and in the meanwhile, happy new year!). We have
$$ frac{1}{sqrt{1-x}}stackrel{L^2(0,1)}{=}2sum_{ngeq 1}P_n(2x-1) $$
hence by Bonnet's recursion formulas and symmetry
$$ sqrt{1-x}=2sum_{ngeq 0}frac{1}{(1-2n)(2n+3)}P_n(2x-1) $$
$$ sqrt{x}=2sum_{ngeq 0}frac{(-1)^n}{(1-2n)(2n+3)}P_n(2x-1) $$
and these FL expansions lead to
$$ int_{0}^{1}sqrt{x(1-x)},dx = color{blue}{4sum_{ngeq 0}frac{(-1)^n}{(1-2n)^2 (2n+1)(2n+3)^2}}. $$
Partial fraction decomposition and telescopic series convert the RHS into
$$ frac{1}{2}sum_{ngeq 0}frac{(-1)^n}{2n+1} = frac{1}{2}int_{0}^{1}sum_{ngeq 0}(-1)^n x^{2n},dx = frac{1}{2}int_{0}^{1}frac{dx}{1+x^2}=frac{pi}{8}.$$
As a by-product, we got a nice, rapidly convergent representation for $pi$ (eight times the blue one).

Don't like the Legendre base of $L^2(0,1)$? Fine, let us go with the Chebyshev one (with respect to a different inner product). Our integral is
$$ int_{0}^{1}x(1-x)frac{dx}{sqrt{x(1-x)}} $$
and the orthogonality relation for the Chebyshev base is
$$ int_{0}^{1}T_m(2x-1)T_n(2x-1)frac{dx}{sqrt{x(1-x)}}=frac{pi}{2}delta(m,n)(1+delta(m)). $$
Since $x(1-x)$ decomposes as $color{blue}{frac{1}{8}}T_0(2x-1)-frac{1}{8}T_2(2x-1)$, the value of our integral is $frac{pi}{8}$.

J.G. has the elementary method down. If you see a $1-x^2$ term inside your integrand, it might be wise to give a trig substitution a try. In this case, letting $x=sin^2theta$ works out beautifully.

There’s another less elementary way by utilizing the beta function and the Gamma function. Let me know if you need a proof (I proudly have an elementary proof of it).

Beta Function:$$operatorname{B}left(m,nright)=intlimits_0^1mathrm dx,x^{m-1}(1-x)^{n-1}=frac {Gamma(m)Gamma(n)}{Gamma(m+n)}$$

The integral under question is then simply$$begin{align*}mathfrak{I} & =operatorname{B}left(frac 32,frac 32right)\ & =frac {1}{2}left(frac {sqrt{pi}}2right)^2end{align*}$$
Thus$$intlimits_0^1mathrm dx, sqrt{x(1-x)}color{blue}{=frac {pi}8}$$