$$\int\left(\sqrt{4-9x^2}\right)^3\mathrm dx$$
I tried using $t=\frac23\sin{x}$ but I get to $\int(\cos{x})^4\mathrm dx$ which I don't know how to evaluate easily.
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Integreek
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GorillaApe
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$\cos^4x=\frac18\cos4x-\frac12\cos2x+\frac38$ – Integreek Dec 09 '24 at 13:21
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this is probably a duplicate – Sine of the Time Dec 09 '24 at 13:37
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See this for instance – Sine of the Time Dec 09 '24 at 13:40
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This question is similar to: Integrating $\int \sin^n{x} \ dx$. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Integreek Jan 30 '25 at 17:12
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$\textbf{Hint:}$
$$\cos(2x)=\cos^2(x)-\sin^2(x)=2\cos^2(x)-1$$
So:
$$\cos^2(x)=\frac{1+\cos(2x)}{2}$$
$$\cos^4(x)=\frac{(1+\cos(2x))^2}{4}$$
The same with $\cos^2(2x)$
$$\cos(4x)=\cos^2(2x)-\sin^2(2x)=2\cos^2(2x)-1$$
agha
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I had this in mind but seemed complex. Maybe there other substitutions too.. (initial) – GorillaApe Feb 04 '15 at 20:12