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Let $\displaystyle F(x)=\int_{0}^{\frac{\pi}{2}} \sqrt{1-x^2\sin^2 t} \; \;dt$.

Find the Maclaurin Series for $F(x)$. All integrals have to be completely evaluated in the final answer.

$\mathbf{Attempt}$ $$\begin{align} F(x)&=\int_{0}^{\frac{\pi}{2}}\left(1+(-x^2\sin^2t) \right)^{0.5}\; dt \\ &= \int_0^{\frac{\pi}{2}}\sum_{n=0}^{\infty} \begin{pmatrix}0.5 \\n \end{pmatrix} \left( -x^2\sin^2t\right)^n \; dt\\ &=\int_0^{\frac{\pi}{2}} \sum_{n=0}^{\infty}\begin{pmatrix}0.5 \\n\end{pmatrix}x^{2n}\left(0.5\right)^n\left(-2\sin^2t+1-1\right)^n \; dt\\ &=\int_0^{\frac{\pi}{2}} \sum_{n=0}^{\infty}\begin{pmatrix}0.5 \\n\end{pmatrix}\left(\frac{x^2}{2}\right)^n \left( \cos2t-1\right)^n \; dt\end{align} $$

And then I don't know how to proceed further.

NetUser5y62
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1 Answers1

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$$F(x)=\int_0^{\pi/2}(1-x^2\sin^2t)^{1/2}\,\mathrm dt $$ $$=\sum_{k=0}^\infty\binom{1/2}{k}(-1)^kx^{2k}\int_0^{\pi/2}\sin^{2k}t\,\mathrm dt$$ We know, through integration by part, that $$I_{2k}:=\int_0^{\pi/2}\sin^{2k}t\,\mathrm dt$$ $$=-\cos t\sin^{2k-1}t\Big|^{\pi/2}_0+\int_0^{\pi/2}(\cos t)(2k-1)\sin^{2k-2}t(\cos t)\,\mathrm dt$$ $$=\int_0^{\pi/2} (2k-1)(1-\sin^2t)\sin^{2k-2}t\,\mathrm dt$$ $$=(2k-1)I_{2k-2}-(2k-1)I_{2k}$$ Which means $$I_{2k}=\frac{2k-1}{2k}I_{2k-2}$$ Can you take it from here?

BAI
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