@一苇之所如 大佬已经在评论区给出提示了，我这里把具体过程写写吧首先区间再现：令 t = \frac{\pi}{2} - x ，有 \mathrm dx = \mathrm d(-t) .那么\begin{split} I= \int_0^{\frac{\pi}{2}} \frac{\sin x\sqrt{\sin x \cos x}}{1 + \sin x \cos x}\mathrm {~d}x &=\int_{\frac{\pi}{2}}^0 \frac{\cos t\sqrt{\cos t\sin t}}{1 + \cos t\sin t }\mathrm {~d}(-t)\cr &=\int_0^{\frac{\pi}{2}} \frac{\cos t\sqrt{\sin t \cos t}}{1 + \sin t \cos t}\mathrm {~d}t \end{split} 即得\begin{split} 2I = \int_0^{\frac{\pi}{2}} \frac{(\sin x + \cos x)\sqrt{\sin x \cos x}}{1 + \sin x \cos x}\mathrm {~d}x \end{split} .又 (\sin x + \cos x)\mathrm {~d}x = \mathrm d(\sin x - \cos x) ，且 \sin x \cos x = \frac{1 - (\sin x - \cos x)^2}{2} .故令 \sin x - \cos x = t 可得\begin{split} 2I &= \int_0^{\frac{\pi}{2}} \frac{(\sin x + \cos x)\sqrt{\sin x \cos x}}{1 + \sin x \cos x}\mathrm {~d}x\cr &=\int_{-1}^{1} \frac{\sqrt{\frac{1 - t^2}{2}}}{1 + \frac{1 - t^2}{2}}\mathrm {~d}t\cr &=2\sqrt{2}\int_{0}^{1} \frac{\sqrt{1 - t^2}}{3 - t^2}\mathrm {~d}t\cr \end{split} .即\begin{split} I&=\sqrt{2} \underbrace{\int_0^1 \frac{\sqrt{1-t^2}}{3 - t^2}
\mathrm {~d}t}_{t\to \sin x}\cr &=\sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\cos^2 x}{3 - \sin^2x}
\mathrm {~d}x\cr &=\sqrt{2} \int_0^{\frac{\pi}{2}} \frac{3-\sin^2 x - 2}{3 - \sin^2x}
\mathrm {~d}x\cr &=\sqrt{2}\left(\frac{\pi}{2} - 2\int_0^{\frac{\pi}{2}} \frac{1}{3 - \sin^2x}
\mathrm {~d}x\right)\cr \end{split} 又\begin{split} \int_0^{\frac{\pi}{2}} \frac{1}{3 - \sin^2x}
\mathrm {~d}x &= 2\int_0^{\frac{\pi}{2}} \frac{1}{5+\cos 2x}
\mathrm {~d}x\cr &=\int_0^{\frac{\pi}{2}} \frac{1}{5+\cos 2x}
\mathrm {~d}(2x)\cr &=\int_0^{\pi} \frac{1}{5+\cos x}
\mathrm {~d}x\cr &=\left.\frac{1}{\sqrt{25 - 1}}\arccos \frac{5+\cos x}{1+5\cos x}\right|_0^{\pi}\cr &=\frac{\sqrt{6}}{12}\pi \end{split} 所以\begin{split} I &= \sqrt{2}\left(\frac{\pi}{2} - \frac{\sqrt{6}}{6}\pi\right)\cr &=\left(\frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{3}\right) \pi \end{split} \begin{align*} \int_0^{\frac{\pi}{2}}\dfrac{\sqrt{\sin^3x\cos x}}{1+\sin x\cos x}\mathrm{d}x&=\int_0^{\frac{\pi}{2}}\dfrac{{\tan}^{\frac{3}{2}}x}{\tan^2x+\tan x+1}\mathrm{d}x\\&=\int_0^\infty\dfrac{t^3}{t^4+t^2+1}\mathrm{d}\arctan t^2\\ &=\int_0^\infty\dfrac{2t^4}{(t^4+1)(t^4+t^2+1)}\mathrm{d}t \\&=\int_0^\infty\left(\dfrac{2t^2}{t^4+1}-\dfrac{2t^2}{t^4+t^2+1}\right)\mathrm{d}t\\ &=\int_0^\infty\left(\dfrac{t^2+1}{t^4+1}+\dfrac{t^2-1}{t^4+1}-\dfrac{t^2+1}{t^4+t^2+1}-\dfrac{t^2-1}{t^4+t^2+1}\right)\mathrm{d}t \end{align*} 根据以下积分：\begin{align*} \int\dfrac{x^2+1}{x^4+1}&=\int\dfrac{1+\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}}\mathrm{d}x=\int\dfrac{\mathrm{d}\left(x-\dfrac{1}{x}\right)}{\left(x-\dfrac{1}{x}\right)^2+2}\\ &=\dfrac{\sqrt{2}}{2}\int\dfrac{1}{\left(\dfrac{x-\dfrac{1}{x}}{\sqrt{2}}\right)^2+1}\mathrm{d}\dfrac{x-\dfrac{1}{x}}{\sqrt{2}} =\dfrac{\sqrt{2}}{2}\arctan\dfrac{x^2-1}{\sqrt{2}x}+C \end{align*} \begin{align*} \int\dfrac{x^2-1}{x^4+1}\mathrm{d}x&=\int\dfrac{1-\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}}\mathrm{d}x=\int\dfrac{\mathrm{d}\left(x+\dfrac{1}{x}\right)}{\left(x+\dfrac{1}{x}\right)^2-2}\\& =\dfrac{1}{2\sqrt{2}}\int\left[\dfrac{1}{\left(x+\dfrac{1}{x}\right)-\sqrt{2}}-\dfrac{1}{\left(x+\dfrac{1}{x}\right)+\sqrt{2}}\right]\mathrm{d}\left(x+\dfrac{1}{x}\right)\\\ \\ &=\dfrac{1}{2\sqrt{2}}\ln\dfrac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}+C \end{align*} \begin{align*} \int\dfrac{x^2+1}{x^4+x^2+1}&=\int\dfrac{1+\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}+1}\mathrm{d}x=\int\dfrac{\mathrm{d}\left(x-\dfrac{1}{x}\right)}{\left(x-\dfrac{1}{x}\right)^2+3}\\ &=\dfrac{\sqrt{3}}{3}\int\dfrac{1}{\left(\dfrac{x-\dfrac{1}{x}}{\sqrt{3}}\right)^2+1}\mathrm{d}\dfrac{x-\dfrac{1}{x}}{\sqrt{3}} =\dfrac{\sqrt{3}}{3}\arctan\dfrac{x^2-1}{\sqrt{3}x}+C \end{align*} \begin{align*} \int\dfrac{x^2-1}{x^4+x^2+1}\mathrm{d}x&=\int\dfrac{1-\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}+1}\mathrm{d}x=\int\dfrac{\mathrm{d}\left(x+\dfrac{1}{x}\right)}{\left(x+\dfrac{1}{x}\right)^2-1}\\& =\dfrac{1}{2}\int\left[\dfrac{1}{\left(x+\dfrac{1}{x}\right)-1}-\dfrac{1}{\left(x+\dfrac{1}{x}\right)+1}\right]\mathrm{d}\left(x+\dfrac{1}{x}\right)\\\ \\ &=\dfrac{1}{2}\ln\dfrac{x^2-x+1}{x^2+x+1}+C \end{align*} 因此： \begin{align*} &\int_0^{\frac{\pi}{2}}\dfrac{\sqrt{\sin^3x\cos x}}{1+\sin x\cos x}\mathrm{d}x\\&=\left[\dfrac{\sqrt{2}}{2}\arctan\dfrac{x^2-1}{\sqrt{2}x}+\dfrac{1}{2\sqrt{2}}\ln\dfrac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}-\dfrac{\sqrt{3}}{3}\arctan\dfrac{x^2-1}{\sqrt{3}x}-\dfrac{1}{2}\ln\dfrac{x^2-x+1}{x^2+x+1}\right]_0^{\infty}\\ &=\dfrac{\sqrt 2\pi}{4}+0-\dfrac{\sqrt3\pi}{6}-0-(-\dfrac{\sqrt 2\pi}{4})-0+(-\dfrac{\sqrt3\pi}{6})+0 \\&=\left(\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{3}}{3}\right)\pi \end{align*} }

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