Integral 1

The substitution $t = \tan^2x$ gives $I \; = \; \int_0^{\frac12\pi} \sin x \cos x \sqrt{\tan x} \ln(\tan x)\,dx \; = \; \tfrac14\int_0^\infty \frac{t^{\frac14}}{(t+1)^2}\ln t\, dt$ and hence $\begin{aligned} I & = \; \tfrac14\left(\frac{\partial B}{\partial a}(a,b) - \frac{\partial B}{\partial b}(a,b)\right)\Big|_{a=\frac54,b=\frac34} \\ & = \tfrac14B(a,b)(\psi(a) - \psi(b)) \Big|_{a=\frac54,b=\frac34} \; = \; \tfrac14B\big(\tfrac54,\tfrac34\big)\big[\psi\big(\tfrac54\big) - \psi\big(\tfrac34\big)\big] \\ & = \frac{\Gamma(\frac54)\Gamma(\frac34)}{4\Gamma(2)} \big[4 + \psi(\tfrac14) - \psi(\tfrac34)\big] \; = \; \frac{\Gamma(\frac14)\Gamma(\frac34)}{16}\big[4 + \psi(\tfrac14) - \psi(\tfrac14) - \pi\big] \\ & = \frac{\pi\sqrt{2}}{16}(\pi-4) \; =\; \frac{\pi(4-\pi)\sqrt{2}}{16} \end{aligned}$ making the answer $100\times16 + 10\times4 + 2 = \boxed{1642}$ .

Use the $u$ -sub, $u=\sqrt{\tan{x}}\to\,du=\frac{\sec^2{x}}{2\sqrt{\tan{x}}}\,dx$ :

$I=\int_{0}^{\pi/2} {\sin{x}\cos{x}\sqrt{\tan{x}}\ln{\left(\tan{x}\right)}\,dx}=4\int_{0}^{\pi/2} {\frac{\sec^2{x}\tan^2{x}}{2\sec^4{x}\sqrt{\tan{x}}}\ln{\left(\sqrt{\tan{x}}\right)}\,dx}=4\int_{0}^{\infty} {\frac{u^4 \ln{u}}{{\left(u^4 + 1\right)}^2}\,du}$

This integral can be solved using complex integration, with a keyhole contour; but a simpler, strictly-real solution introduces the integral:

$F(a)=\int_{0}^{\infty} {\frac{\,dx}{x^a+1}}=\int_{0}^{\infty} {\int_{0}^{\infty} {e^{-t}e^{-tx^a}\,dt\,dx}}=\frac{1}{a}\Gamma\left(\frac{1}{a}\right)\Gamma\left(1-\frac{1}{a}\right)=\frac{\pi}{a}\csc{\left(\frac{\pi}{a}\right)}$

Then, by Feynman's trick of 'differentiation under the integral sign':

$F'(a)=\frac{\,d}{\,da}\int_{0}^{\infty} {\frac{\,dx}{x^a+1}}=\int_{0}^{\infty} {\frac{x^a \ln{x}\,dx}{\left(x^a+1\right)^2}}=\frac{1}{a}\left(\frac{\pi}{a}\csc{\left(\frac{\pi}{a}\right)}\right)\left(1-\frac{\pi}{a}\cot{\left(\frac{\pi}{a}\right)}\right)$

Thus, the integral is:

$I=4F'(4)=4\left(\frac{1}{4}\left(\frac{\pi}{4}\csc{\left(\frac{\pi}{4}\right)}\right)\left(1-\frac{\pi}{4}\cot{\left(\frac{\pi}{4}\right)}\right)\right)=\boxed{\frac{1}{16}\pi\left(4-\pi\right)\sqrt{2}}$

and the answer is $100\cdot16+10\cdot4+2=1642$ .