I stumbled upon this integration $$\int_0^a \left(x-\sqrt{a^2-x^2}\right)^3\,dx.$$ And I have no idea on how to proceed with it. It would be very helpful if someone would provide me with some hints. Thank you.
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Did you try with $x=a\sin t$.? – Nosrati Mar 07 '17 at 16:46
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After the substitution $x=a\sin t$ we are left with
$$ a^4\int_{0}^{\pi/2}(\sin t-\cos t)^3 \cos(t)\,dt =a^4\sqrt{2}^3\int_{0}^{\pi/2}\sin^3\left(t-\frac{\pi}{4}\right)\cos(t)\,dt$$
that equals
$$ 2a^4\int_{-\pi/4}^{\pi/4}\sin^3(t)\left[\cos(t)-\sin(t)\right]\,dt = -4a^4\int_{0}^{\pi/4}\sin^4(t)\,dt$$
since the integral of an odd, integrable function over a symmetric interval is zero.
The last integral is elementary and leads to
$$ \int_{0}^{a}\left(x-\sqrt{a^2-x^2}\right)^3\,dx = \color{red}{\frac{8-3\pi}{8}\,a^4}.$$
Jack D'Aurizio
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Sorry, $\int_{0}^{a}\left(x-\sqrt{a^2-x^2}\right)^3,dx = \color{blue}{\dfrac{8-3\pi}{8},a^4}$. – Nosrati Mar 08 '17 at 03:37
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@JackD'Aurizio I have one last question. Can you explain this step of yours.$a^4\int_{0}^{\pi/2}(\sin t-\cos t)^3 \cos(t),dt =a^4\sqrt{2}^3\int_{0}^{\pi/2}\sin^3\left(t-\frac{\pi}{4}\right)\cos(t),dt$ – Ayan Shah Mar 09 '17 at 04:14
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@AyanShah: I am exploiting the formulas for $\sin(\theta\pm\pi/4)$ and $\cos(\theta\pm\pi/4)$ and, at last, the substitution $t\mapsto t+\pi/4$. – Jack D'Aurizio Mar 09 '17 at 12:47
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@Jack D'Aurizio In the last step when you integrate $-4a^4\int_{0}^{\pi/4}\sin^4(t),dt$ you got $\frac{8-3\pi}{8},a^4$, but when i used the reduction formula for $sin^n x$ i did not get the constant 8. Can you explain why this happened? – Ayan Shah Mar 10 '17 at 14:10
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@AyanShah: which reduction formula are you referring to? I just wrote $\sin^4(t)$ as $\frac{3}{8}-\frac{1}{2} \cos(2t)+\frac{1}{8}\cos(4t)$ and performed termwise integration. – Jack D'Aurizio Mar 10 '17 at 14:13
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@JackD'Aurizio link . I used the 3rd result from this image. Is it the case that it only works for $pi/2$ – Ayan Shah Mar 10 '17 at 14:15
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@AyanShah: there is no particular reason for expecting it works for any integration range, I guess. – Jack D'Aurizio Mar 10 '17 at 14:18
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@Jack D'Aurizio Actually this is not the full problem that i meant to solve. It was "Show that $\int_c( (x-y)^3,dx+(x-y)^3dy) = 3pi a^4$ taken along the circle $x^2 + y^2=a^2$ in counter clockwise direction. page1 and page2 is my effort which was possible through your help. But somehow the answer that i am getting does not match with what is required to be shown. – Ayan Shah Mar 10 '17 at 14:51
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@AyanShah: what about a change of variables? Maybe like $x+y = \sqrt{2} u,; x-y=\sqrt{2} v$, leading to a straightforward integral? You over-complicated things by directly switching to polar coordinates, and incorrectly performed the change of coordinates. – Jack D'Aurizio Mar 10 '17 at 14:53
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@JackD'Aurizio I took the first quadrant of the circle and multiplied it by 4. – Ayan Shah Mar 10 '17 at 14:54
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@AyanShah: that is a different, simpler problem and I think I helped enough here. I am willing to politely disengage. – Jack D'Aurizio Mar 10 '17 at 14:55
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HINTS:
$$(x-y)^3=x^3-3x^2y+3xy^2-y^3$$
Use the substitution $x=a\sin(y)$ for the terms involving $\sqrt{a^2-x^2}$ and $(\sqrt{a^2-x^2})^3$.
Mark Viola
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lease let me know how I can improve this answer too. As always, I really want to give you the best answer I can. If the answer was not useful, I am happy to delete it. -Mark – Mark Viola Apr 12 '17 at 17:12
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Hint: Let $x=a\sin t$ then $$\int_0^a (x-(\sqrt{a^2-x^2}))^3dx=a^4\int_0^{\frac{\pi}{2}}(\sin t-\cos t)^3\cos t\,dt$$
Nosrati
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Can you give me hints as to what i am to do after this. Integration by parts after expanding doesnot seem to be working – Ayan Shah Mar 07 '17 at 17:20
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Uh. yes. Simplify integrand to $\sin t\cos t-\sin t\cos^3t-\cos2t-\frac14\cos4t-\frac34$. – Nosrati Mar 07 '17 at 18:29