问题及解答

求定积分

\[\int_{0}^{\frac{\pi}{2}}\sin^5\theta\cdot\cos^4\theta\mathrm{d}\theta.\]

\[
\begin{split}
\int_{0}^{\frac{\pi}{2}}\sin^5\theta\cdot\cos^4\theta\mathrm{d}\theta&=-\int_{0}^{\frac{\pi}{2}}\sin^4\theta\cdot\cos^4\theta\mathrm{d}\cos\theta\\
&=-\frac{1}{2^4}\int_{0}^{\frac{\pi}{2}}(\sin 2\theta)^4\mathrm{d}\cos\theta\\
&=-\frac{1}{16}\int_{0}^{\frac{\pi}{2}}\bigl[(\sin 2\theta)^2\bigr]^2\mathrm{d}\cos\theta\\
&=-\frac{1}{16}\int_{0}^{\frac{\pi}{2}}\bigl[1-\cos^2(2\theta)\bigr]^2\mathrm{d}\cos\theta\\
&=-\frac{1}{16}\int_{0}^{\frac{\pi}{2}}\Bigl[1-(2\cos^2\theta-1)^2\Bigr]^2\mathrm{d}\cos\theta\\
&\stackrel{t=\cos\theta}{=}-\frac{1}{16}\int_{1}^{0}\Bigl[1-(2t^2-1)^2\Bigr]^2\mathrm{d}t\\
&=\frac{1}{16}\int_0^1 \Bigl[(2t^2-1)^2-1\Bigr]^2\mathrm{d}t\\
&=\frac{1}{16}\int_0^1 \Bigl[4t^4-4t^2\Bigr]^2\mathrm{d}t\\
&=\int_0^1 (t^4-t^2)^2\mathrm{d}t\\
&=\int_0^1 (t^8-2t^6+t^4)\mathrm{d}t\\
&=[\frac{1}{9}t^9-\frac{2}{7}t^7+\frac{1}{5}t^5]\Bigr|_{0}^{1}\\
&=\frac{1}{9}-\frac{2}{7}+\frac{1}{5}\\
&=\frac{8}{315}
\end{split}
\]

\[
\begin{split}
\int_{0}^{\frac{\pi}{2}}\sin^5\theta\cdot\cos^4\theta\mathrm{d}\theta&=-\int_{0}^{\frac{\pi}{2}}\sin^4\theta\cdot\cos^4\theta\mathrm{d}\cos\theta\\
&=-\int_{0}^{\frac{\pi}{2}}(\sin^2\theta)^2\cdot\cos^4\theta\mathrm{d}\cos\theta\\
&=-\int_{0}^{\frac{\pi}{2}}(1-\cos^2\theta)^2\cdot\cos^4\theta\mathrm{d}\cos\theta\\
&\stackrel{t=\cos\theta}{=}-\int_1^0 (1-t^2)^2\cdot t^4\mathrm{d}t\\
&=\int_0^1 (t^2-1)^2\cdot t^4\mathrm{d}t\\
&=\int_0^1 (t^4-2t^2+1)\cdot t^4\mathrm{d}t\\
&=\int_0^1 (t^8-2t^6+t^4)\mathrm{d}t\\
&=\Bigl(\frac{1}{9}t^9-\frac{2}{7}t^7+\frac{1}{5}t^5\Bigr)\biggr|_{0}^{1}\\
&=\frac{1}{9}-\frac{2}{7}+\frac{1}{5}\\
&=\frac{8}{315}.
\end{split}
\]