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Question:

Find the integral $$I=\int\dfrac{1}{\sin^5{x}+\cos^5{x}}dx$$

my solution: since \begin{align*}\sin^5{x}+\cos^5{x}&=(\sin^2{x}+\cos^2{x})(\sin^3{x}+\cos^3{x})-\sin^2{x}\cos^2{x}(\sin{x}+\cos{x})\\ &=(\sin{x}+\cos{x})(1-\sin{x}\cos{x})-(\sin{x}+\cos{x})(\sin^2{x}\cos^2{x})\\ &=(\sin{x}+\cos{x})[1-\sin{x}\cos{x}-\sin^2{x}\cos^2{x}) \end{align*}

then let $$\sin{x}+\cos{x}=t\Longrightarrow \sin{x}\cos{x}=\dfrac{1}{2}(t^2-2)$$

then $$\sin^5{x}+\cos^5{x}=t[1-\dfrac{1}{2}(t^2-2)-\dfrac{1}{4}(t^2-2)^2]=-\dfrac{1}{4}t(t^4-2t^2+4)$$ $$x+\dfrac{\pi}{4}=\arcsin{\dfrac{t}{\sqrt{2}}}\Longrightarrow dx=\dfrac{1}{\sqrt{2-t^2}}dt$$ so $$I=-4\int\dfrac{dt}{t\sqrt{2-t^2}(t^4-2t^2+4)}=-2\int\dfrac{1}{u\sqrt{2-u^2}(u^2-2u+4)}du$$ where $u=t^2$

Then I can't, because this wolf can't http://www.wolframalpha.com/input/?i=1%2F%28xsqrt%282-x%5E2%29%28x%5E2-2x%2B4%29%29dx

Thank you for you help

  • http://math.stackexchange.com/questions/595038/calculation-of-int-frac1-sin3-x-cos3-xdx-and-int-frac1-sin5-x-c – lab bhattacharjee Dec 10 '13 at 03:50
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    I have not checked your calculation, so am not solving the problem. If it is right, multiply top and bottom by $2t$. and let $u=t^2-4$. Then $du$ is sitting on top, and at the bottom you have something fairly but not entirely horrible, which you can use partial fractions on. – André Nicolas Dec 10 '13 at 04:04
  • @labbhattacharjee，But other post can't solve it.But Thank you –  Dec 10 '13 at 05:29
  • You don't expect anyone to derive these; do you? – Ali Dec 10 '13 at 06:38

1 Answers1

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since \begin{align*}\sin^5{x}+\cos^5{x}&=(\sin^2{x}+\cos^2{x})(\sin^3{x}+\cos^3{x})-\sin^2{x}\cos^2{x}(\sin{x}+\cos{x})\\ &=(\sin{x}+\cos{x})(1-\sin{x}\cos{x})-(\sin{x}+\cos{x})(\sin^2{x}\cos^2{x})\\ &=(\sin{x}+\cos{x})[1-\sin{x}\cos{x}-\sin^2{x}\cos^2{x}] \end{align*} so \begin{align*}\int\dfrac{1}{\sin^5{x}+\cos^5{x}}dx&=\dfrac{1}{(\sin{x}+\cos{x})(1-\cos{x}\sin{x}-\cos^2{x}\sin^2{x})}dx\\ &=\int\dfrac{1}{\sqrt{1-2y}(1+2y)(1-y-y^2)}dy (y=\cos{x}\sin{x})\\ &=\dfrac{4}{5}\int\dfrac{1}{\sqrt{1-2y}(1+2y)}dy+\dfrac{1}{5}\int\dfrac{1+2y}{\sqrt{1-2y}(1-y-y^2)}dy \end{align*} note $$\int\dfrac{1}{\sqrt{1-2y}(1+2y)}=\int\dfrac{1}{z^2-2}dz=\dfrac{1}{2\sqrt{2}}\ln{\left|\dfrac{z-\sqrt{2}}{z+\sqrt{2}}\right|}+C_{1} (z=\sqrt{1-2y})$$ and we $$\int\dfrac{1+2y}{\sqrt{1-2y}(1-y-y^2)}dy=-4\int\dfrac{z^2-2}{z^4-4z^2+1}dz$$ and \begin{align*} \int\dfrac{z^2-2}{z^4-4z^2+1}dz&=\int\dfrac{z^2-1}{z^4-4z^2+1}dz-\int\dfrac{1}{z^4-4z^2+1}dz\\ &=\int\dfrac{d(z+\dfrac{1}{z})}{(z+\dfrac{1}{z})^2-6}-\int\dfrac{1}{z^4-4z^2+1}dz \end{align*} we can easy have $$\int\dfrac{1}{z^4-4z^2+1}dz=\dfrac{1}{2}\left(\dfrac{1+z^2}{z^4-4z^2+1}+\int\dfrac{1-z^2}{z^4-4z^2+4}\right)$$ so $$\dfrac{1}{5}\int\dfrac{1+2y}{\sqrt{1-2y}(1-y-y^2)}dy=\dfrac{1}{5}\left(\dfrac{1}{\sqrt{2}}\ln{\left|\dfrac{z^2-\sqrt{2}z+1}{z^2+\sqrt{2}z+1}\right|}-\dfrac{1}{\sqrt{6}}\ln{\left|\dfrac{z^2-\sqrt{6}z+1}{z^2+\sqrt{6}z+1}\right|}\right)+C_{2}$$ so $$I=\dfrac{\sqrt{2}}{5}\ln{\left|\dfrac{z-\sqrt{2}}{z+\sqrt{2}}\right|}+\dfrac{1}{5}\left(\dfrac{1}{\sqrt{2}}\ln{\left|\dfrac{z^2-\sqrt{2}z+1}{z^2+\sqrt{2}z+1}\right|}-\dfrac{1}{\sqrt{6}}\ln{\left|\dfrac{z^2-\sqrt{6}z+1}{z^2+\sqrt{6}z+1}\right|}\right)+C$$ where $z=\cos{x}-\sin{x},C=\dfrac{4}{5}C_{1}+C_{2}$

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