Answer
$$\int^{\pi/2}_0\frac{3\sin x\cos x}{\sqrt{1+3\sin^2x}}dx=1$$
Work Step by Step
$$A=\int^{\pi/2}_0\frac{3\sin x\cos x}{\sqrt{1+3\sin^2x}}dx$$
We set $u=\sqrt{1+3\sin^2x}$, which means $$du=\frac{(1+3\sin^2x)'}{2\sqrt{1+3\sin^2x}}dx=\frac{6\sin x\cos x}{2\sqrt{1+3\sin^2x}}dx$$ $$du=\frac{3\sin x\cos x}{\sqrt{1+3\sin^2x}}dx$$
For $x=\pi/2$: $$u=\sqrt{1+3\sin^2(\pi/2)}=\sqrt{1+3\times1^2}=2$$
For $x=0$: $$u=\sqrt{1+3\sin^20}=\sqrt{1+3\times0^2}=1$$
Therefore, $$A=\int^2_1du$$ $$A=u\Big]^2_1$$ $$A=2-1=1$$
