a. Đặt \(x=sinu\Rightarrow dx=cosu.du\) ; \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\Rightarrow u=\dfrac{\pi}{3}\\x=1\Rightarrow u=0\end{matrix}\right.\)
\(A=\int\limits^0_{\dfrac{\pi}{3}}\sqrt{1-cos^2u}.sinudu=\int\limits^{\dfrac{\pi}{3}}_0-sin^2udu=\int\limits^{\dfrac{\pi}{3}}_0\left(\dfrac{cos2u}{2}-\dfrac{1}{2}\right)du\)
\(=\left(\dfrac{1}{4}sin2u-\dfrac{1}{2}u\right)|^{\dfrac{\pi}{3}}_0=\dfrac{\sqrt{3}}{8}-\dfrac{\pi}{6}\)
b.
Đặt \(x=\sqrt{3}tanu\Rightarrow dx=\dfrac{\sqrt{3}}{cos^2u}du\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow u=0\\x=-1\Rightarrow u=-\dfrac{\pi}{6}\end{matrix}\right.\)
\(B=\int\limits^0_{-\dfrac{\pi}{6}}\dfrac{1}{3+3tan^2u}.\dfrac{\sqrt{3}}{cos^2u}du=\int\limits^0_{-\dfrac{\pi}{6}}\dfrac{1}{\sqrt{3}}du=\dfrac{1}{\sqrt{3}}.\dfrac{\pi}{6}\)
